NEET Physics

Mechanical Properties of Solids

Question 1. The stress-strain curves are drawn for two different materials X and Y. It is observed that the ultimate strength point and the fracture point are close to each other for material X but are far apart for material Y. We can say that materials X and Y are likely to be (respectively):

1. ductile and brittle
2. brittle and ductile
3. brittle and plastic
4. plastic and ductile

Answer: 2. brittle and ductile

As the given factors point and ultimate strength point are close for material X₁ X is brittle and both points are apart for material ϒ Hence Cl is ductile X is brittle and Y is ductile.

Question 2. Given below are two statements: One is labeled as
Assertion and the other is labeled as Reason (R).
Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring.
Reason (R): A coil spring of copper has more tensile strength than a steel spring of the same dimensions.
In the light of the above statements, choose the most appropriate answer from the options given below:

1. Both (A) and (R) are true and (R) is not the correct explanation of (A).
2. (A) is true but (R) is false
3. (A) is false but (R) is true
4. Both (A) and (R) are true and (R) is the correct explanation of (A).

Answer: 2. (A) is true but (R) is false

From the concept of Hook’s law, we have,

Normal Stress, (CT) = Ee

where E = Young’s Modulus of Elasticity and e is a strain which gives linear deformation of the object.

Also, Shear Stress, (x) = Gy

where G = Shear Modulus of Elasticity of the material and y is a shear strain which gives angular deformation of the object.

When we stretch a spring, the length of the wire does not change but experiences an angular twist. Hence shear modulus is used to determine the stretching of a spring.

The Assertion is True. Also, we know that for a given dimension, Young’s Modulus of Elasticity of steel is more than the Copper hence from equation (1). We can say that the tensile strength of Steel is more than that of Cu.

∴ The Reason is False.

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Question 3. A wire of length L, area of cross-section A is hanging from a field support. The length of the wire changes to L₁ when mass M is suspended from its free end. The expression for Young’s modulus is :

1. \(\frac{\mathrm{Mg}\left(\mathrm{L}_1-\mathrm{L}\right)}{A L}\)
2. \(\frac{\mathrm{Mg} \mathrm{L}}{A L_1}\)
3. \(\frac{\mathrm{MgL}}{A\left(L_1-L\right)}\)
4. \(\frac{\mathrm{Mg} \mathrm{L}}{A L}\)

Answer: 1. \(\frac{\mathrm{Mg}\left(\mathrm{L}_1-\mathrm{L}\right)}{A L}\)

Change in length, \(\Delta L=L_1-L\), Area, =A, force,

F = mg

Younger modules,\(\gamma =\frac{\text { Normal Stress }}{\text { Longitudinal Strain }}\)

⇒ \(\gamma =\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}\)

∴ \(\gamma=\frac{\left(\frac{\mathrm{Mg}}{A}\right)}{\left(\frac{\mathrm{L}_1-\mathrm{L}}{L}\right)}=\frac{\mathrm{Mg}\left(\mathrm{L}_1-\mathrm{L}\right)}{A \mathrm{~L}}\)

Question 4. When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is :

1. MgL
2. \(\frac{1}{2} M g l\)
3. \(\frac{1}{2} M g L\)
4. Mgl

Answer: 2. \(\frac{1}{2} M g l\)

Cross-Sectional Area = A

Original length = L

Elongation = l

⇒ \(\frac{\text { Energy stored }}{\text { Volume }} =\frac{1}{2} \text { stress } \times \text { strain }\)

⇒ \(\frac{E}{V} =\frac{1}{2} \times \frac{F}{A} \times \frac{\text { elongation }}{\text { length }}\)

⇒ \(\frac{E}{A L} =\frac{1}{2} \times \frac{m g}{A^{\prime}} \times \frac{l}{L}\)

∴ \(\mathrm{E} =\frac{1}{2} m g l\)

Question 5. Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by At on applying a force F, how much force is needed to stretch the second wire by the same amount?

1. 4F
2. 6F
3. 9F
4. F

Answer: 3. 9F

For wire (1), \(\Delta l=\left(\frac{F}{A \gamma}\right) 3 l\)  → Equation 1

For wire (2), \(\frac{F^{\prime}}{3 A}=\gamma \frac{\Delta l}{l}\)

⇒ \(\Delta l=\left(\frac{F^{\prime}}{3 A Y}\right) l\)  →   Equation 2

From eq. (1) and eq. (2),

we get, \(\Delta l =\frac{F}{A \gamma} 3 l-\left(\frac{F^{\prime}}{3 A Y}\right) l\)

∴ \(F^{\prime}\) =9 F

Question 6. The bulk modulus of a spherical object is B. If it is subjected to uniform pressure P, the fractional decrease in radius is:

1. \(\frac{P}{B}\)
2. \(\frac{B}{3 P}\)
3. \(\frac{3 P}{B}\)
4. \(\frac{P}{3 B}\)

Answer: 4. \(\frac{P}{3 B}\)

Bulk modulus (B) is the ratio of normal stress to volumetric strain,

B=\(\frac{\text { Normal stress }}{\text { Volumetric strain }}\)

=\(\frac{F / A}{\Delta V / V}=\frac{P}{\Delta V / V} \)

⇒ \(\frac{\Delta V}{V}=\frac{P}{B}\) [ P –  pressure on the object]  → Equation 1

Now from the question of fractional decrease in volumes

⇒ \(\frac{\Delta V}{V}=3 \frac{\Delta R}{R}\)    →  Equation 2

⇒ \({\left[V=\frac{4}{3} \pi R^3\right]}\)

From eq. (1) and (2), we get

∴ \(\frac{\Delta V}{V}=\frac{3 \Delta R}{R} = \frac{P}{B}\)

Question 7. The Young’s modulus of steel is twice that of brass. Two wires of the same length and the same area of cross-section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of:

1. 1: 2
2. 2: 1
3. 4: 1
4. 1: 1

Answer: 2. 2: 1

Young’s modulus, \(\gamma=\frac{W. I}{A \Delta l}\)

I, A and \(\Delta l\) are same for both the wires

⇒ \(\gamma \propto W\)

⇒ \(\frac{\gamma_s}{\gamma_b}=\frac{W_s}{W_b}=\frac{2}{1}\)

∴ \(\mathrm{W}_s: \mathrm{W}_b=\) 2: 1

Question 8. The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 x 10^-11 Pa^-1 and the density of water is 10³ kg/m³. What fractional compression of water will be obtained at the bottom of the ocean?

1. 1.2 x 10^-2
2. 1.4×10^-2
3. 0.8×10^-2
4. 1.0×10^-2

Answer: 1. 1.2 x 10^-2

Depth of ocean, d=2700 m

Density of water, \(\rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}\)

Compressibility of water, \(\mathrm{K}=45.4 \times 10^{-11} \mathrm{~Pa}^{-1}\)

⇒ \(\frac{\Delta V}{V}\)=?

Excess pressure at the bottom, \(\Delta \mathrm{P}=\rho g d\)

=\(10^3 \times 10 \times 2700=27 \times 10^6 \mathrm{~Pa}\)

We know, \(\mathrm{B} =\frac{\Delta P}{(\Delta V / V)} \)

⇒ \(\frac{\Delta V}{(V)} =\frac{\Delta P}{B}=\mathrm{K} . \Delta \mathrm{P}\)

⇒ \((K \left.=\frac{1}{B}\right)\)

=\(45.4 \times 10^{-11} \times 27 \times 10^6\)

=\(1.2 \times 10^{-2}\)

Question 9. The copper of fixed volume V is drawn into a wire of length l. When the wire is subjected to a constant force F, the extension produced in the wire is Δl. Which of the following graphs is a straight line?

1. \( \Delta l versus \frac{1}{l} \)
2. \( \Delta l versus l^2 \)
3. \( \Delta l versus \frac{1}{l^2} \)
4. \( \Delta l versus l \)

Answer: 2. \(\Delta l versus l^2\)

⇒ \({ Young’s modulus }=\frac{\text { Longitudinal Stress }}{\text { Longitudinal Strain }}\)

Where, F= force applied

A = Area of cross-section

⇒ \(\gamma=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}\)  →  Equation 1

⇒ \(\Delta l=\text { Change in length }\)

l= Original length And

V=A \(\times l=\text { Constant }\)

A = \(\frac{V}{l}\)  →  Equation 2

From (1) and (2),

we get, \(\gamma =\frac{F \times l \times l}{\Delta l \times V}\)

⇒ \(\Delta l =\frac{F}{V \times Y} \times l^2\)

⇒ \(\Delta l \propto l^2\)

∴ Where \(\mathrm{F}, \mathrm{V}, \gamma\) are constant

Question 10. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

1. Length = 50 cm, diameter = 0.5 mm
2. Length =100 cm, diameter = 1 mm
3. Length = 200 cm, diameter = 2 mm
4. Length = 300 cm, diameter = 3 mm

Answer: 1. Length = 50 cm, diameter = 0.5 mm

We know that, \(\Delta L=\frac{F L}{A Y}\)

⇒ \(\Delta L \propto \frac{L}{d^2}\) { Since A=\(\frac{\pi d^2}{4}\}\)

This confirms that \(\Delta\) L will be the maximum for the wire when tension \(\frac{L}{A}\) maximum.

Question 11. In the ratio of diameters, lengths, and Young’s modulus, of steel and copper wires shown in the figure, are p, q, and s respectively, then the corresponding ratio of increase in their length would be:

1. \(\frac{7 q}{\left(7 s p^2\right)}\)
2. \(\frac{7 q}{\left(5 s p^2\right)}\)
3. \(\frac{2 q}{(5 s p)}\)
4. \(\frac{7 q}{(5 s p)}\)

Answer: 2. \(\frac{7 q}{\left(5 s p^2\right)}\)

We know that,

Young’s modulus,\(\gamma =\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}\)

=\(\frac{\Delta \mathrm{FL}}{\pi \mathrm{D}^2 \Delta \mathrm{L}}\)  →  Equation 1

\(\Delta \mathrm{L} =\frac{\Delta \mathrm{FL}}{\pi \mathrm{D}^2 \mathrm{~L}}\)

⇒ According to the questions,

⇒ \(\frac{\Delta \mathrm{L}_{\mathrm{S}}}{\Delta \mathrm{L}_{\mathrm{C}}}=\frac{\mathrm{F}_{\mathrm{S}}}{\mathrm{F}_{\mathrm{C}}} \cdot \frac{\mathrm{L}_{\mathrm{S}}}{\mathrm{L}_{\mathrm{C}}} \cdot \frac{\mathrm{D}_{\mathrm{C}}{ }^2 \mathrm{~L}_{\mathrm{S}}}{\mathrm{D}_{\mathrm{S}}{ }^2 \mathrm{~L}_{\mathrm{C}}}\)  → Equation – 2

Here, \(F_S=(5 m+2 m)\) g=7 m g

⇒ \(\mathrm{F}_{\mathrm{C}}=5 \mathrm{mg},\)=?,

⇒ \(\frac{\mathrm{D}_{\mathrm{S}}}{\mathrm{D}_{\mathrm{C}}}\)=p,

⇒ \(\frac{\mathrm{L}_{\mathrm{S}}}{\mathrm{L}_{\mathrm{C}}}=\mathrm{s}\)

Putting in eq. (2), we get,

⇒ \(\frac{\Delta \mathrm{L}_{\mathrm{S}}}{\Delta \mathrm{L}_{\mathrm{C}}} =\frac{7 \mathrm{mg}}{5 \mathrm{mg}} \times q+\left(\frac{1}{\mathrm{P}}\right)^2\left(\frac{1}{\mathrm{~S}}\right)\)

=\(\frac{7 q}{5 p^2 \mathrm{~S}}\)

Gravitation

Question 1. A mass falls from a height ‘h’ and its time of fall ‘t’ is recorded in terms of time T of a simple pendulum. On the surface of Earth, it is found that t = 2 T. The entire set-up is taken on the surface of another planet whose mass is half of Earth and whose radius the same. The same experiment is repeated and corresponding times as t’ and T’:

1. \(t^{\prime}=\sqrt{2} \mathrm{~T}^{\prime}\)
2. \(t^{\prime}>2 \mathrm{~T}^{\prime}\)
3. \(t^{\prime}<2 \mathrm{~T}^{\prime}\)
4. \(t^{\prime}=2 \mathrm{~T}\)

Answer: 1. \(t^{\prime}=\sqrt{2} \mathrm{~T}^{\prime}\)

Force surface of Earth time taken is falling h distance t=\(\sqrt{\frac{2 h}{g}} \text { and } T=2 \pi \sqrt{\frac{l}{g}}\)

Given, t=\(2 T\)

⇒ \(\frac{t}{T}\)=2

For surface of other planet, \(g^{\prime}=\frac{g}{2}\)

Time taken in falling h distance, \(t^{\prime}=\frac{2 h}{g}=\sqrt{2} t\)

and \(T=2 \pi \sqrt{\frac{1}{g^{\prime}}}=\sqrt{2} T\)

Here \(\frac{t^{\prime}}{T^{\prime}}=\frac{\sqrt{2} t}{\sqrt{12} T}\)=2

∴ \(t^{\prime}=2 T^{\prime}\)

Question 2. The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are D_A, K_B and K_C respectively, AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

1. \(t^{\prime}=\sqrt{2} \mathrm{~T}^{\prime}\)
2. \(t^{\prime}>2 \mathrm{~T}^{\prime}\)
3. \(t^{\prime}<2 \mathrm{~T}^{\prime}\)
4. \(t^{\prime}=2 \mathrm{~T}\)

Answer: 2. \(t^{\prime}>2 \mathrm{~T}^{\prime}\)

Point A is perihelion and C is aphelion.

So,\( V_A>V_B>V_C\)

As kinetic energy \(\mathrm{K}=(1 / 2) m v^2\) or \(K \propto v^2\)

So, \(K_A>K_B>K_C\)

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Question 3. A planet moving along an elliptical orbit is closest to the sun at a distance of r₁ and farthest away at a distance of r₂. If V₁ and V₂ are the liner velocities at these points respectively, then the ratio \(\frac{v_1}{v_2}\) is:

1. \(\frac{r_2}{r_1}\)
2. \(\left(\frac{r_2}{r_1}\right)^2\)
3. \(\frac{r_1}{r_2}\)
4. \(\left(\frac{r_1}{r_2}\right)^2\)

Answer: 1. \(\frac{r_2}{r_1}\)

Using the law of conservation of angular moments \(m_1 v_1 =m_2 v_2 \)

⇒ \(m r_1 v_1 =m r_2 v_2 \)

⇒ \(r_1 v_1 =r_2 v_2 \)

∴ \(\frac{v_1}{v_2} =\frac{r_2}{r_1}\)

Question 4. The figure shows the elliptical orbit of a plant m about the sun S. The shaded area SCD is twice the shaded area SAB, if t₁ is the time for the planet to move from C to D and t₂ is the time to move from A to B, then:

1. \(t_1>t_2\)
2. \(t_1=4 t_2\)
3. \(t_1=2 t_2\)
4. \(t_1=t_2\)

Answer: 3. \(t_1=2 t_2\)

Applying Kepler’s law of planetary motion. \(\frac{d A}{d t}\) = constant

⇒ \(\frac{A_1}{t_1} =\frac{A_2}{t_2} \)

⇒ \( t_1 =\frac{A_1}{A_2} t_2 \)

Since, \(A_1 =2 A_2\)

Hence, \(t_1 =2 t_2\)

Question 5. The period of revolution of the planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun?

1. 5
2. 4
3. 3
4. 2

Answer: 2. 4

According to Kelper’s third law,

⇒ \(T^2 \propto r^3\)

Where, T= period of revolution

r= Semi-major axis

⇒ \(\frac{T_A^2}{T_B^2} =\frac{r_A^3}{r_B^3}\)

⇒ \( \frac{r_A}{r_B} =\left(\frac{T_A}{T_B}\right)^{2 / 3} \)

=\( (8)^{\frac{2}{3}}\)

=\(2^{3 \times \frac{2}{3}}\) =4

or \(r_A =4 r_B\)

Question 6. The distances of the two planets from the sun are 10¹³ and 10¹² m respectively. The ratio of periods of these two planets is:

1. \(\frac{1}{\sqrt{10}}\)
2. 100
3. 10 \(\sqrt{10}\)
4. 1 \(\sqrt{10}\)

Answer: 3. 10 \(\sqrt{10}\)

According to Kepler’s third law (or the law of periods) \(T^2 \propto r^3\)

Here, \(r_1=10^{13} \mathrm{~m}, r_2=10^{12} \mathrm{~m}\)

⇒ \(\frac{T_1^3}{T_2^2}=\frac{r_1^3}{r_2^3}=\frac{\left(10^{13}\right)^3}{\left(10^{12}\right)^3}\)

or \(\frac{T_1^2}{T_2^2}=\frac{10^{39}}{10^{36}}=10^3 \)

or \(\frac{T_1}{T_2}=10 \sqrt{10}\)

Question 7. A planet is moving in an elliptical orbit around the Sun. If. T, V, E and L stand respectively for their kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, which of the following is correct?

1. T is conserved.
2. V is always positive.
3. E is always negative.
4. T is conserved but the direction of vector L changes continuously.

Answer: 3. E is always negative.

In a circular or elliptical orbital motion of a planet, angular momentum is conserved. In an attractive field, potential energy and total energy are negative. Kinetic energy increases with an increase in velocity. If the motion is in a plane, the direction of L does not change.

Question 8. The largest and the shortest distance of the Earth from the sun are r₁ and r₂. Its distance from the sun when it is perpendicular to the major axis of the orbit drawn from the sun is

1. \(\frac{r_1+r_2}{4}\)
2. \(\frac{r_1+r_2}{r_1-r_2}\)
3. \(\frac{2 r_1 r_2}{r_1+r_2}\)
4. \(\frac{r_1+r_2}{3}\)

Answer: 3. \(\frac{2 r_1 r_2}{r_1+r_2}\)

Applying the properties of ellipse,

we have, \(\frac{2}{R} =\frac{1}{r_1}+\frac{1}{r_2}\)

=\(\frac{r_1+r_2}{r_1 r_2}\)

∴ \(\mathrm{R} =\frac{2 r_1 r_2}{r_1+r_2}\)

Question 9. Two astronauts are floating in gravitation-free space after having lost contact with their spaceship. The two will:

1. keep floating at the same distance between them
2. move towards each other
3. move away from each other
4. will become stationary

Answer: 2. move towards each other

Both the astronauts are in the condition of weightless. The gravitational force between them pulls towards each other

Question 10. Kepler’s third law states that the square of the period of revolution (7) of a planet around the sun, is proportional to the third power of the average distance r between the sun and planet is T² = Kr³ here K is constant. If the masses of the sun and planet are M and m respectively, then as per Newton’s law of gravitation force of attraction between them is:\(\mathrm{F}=\frac{\mathrm{GM} m}{r^2}\), here G gravitational constant. The relation between G and K is described as:

1. \(\mathrm{GK}=4 \pi^2\)
2. \(\mathrm{GMK}=4 \pi^2\)
3. \(\mathrm{K}=\mathrm{G}\)
4. K=\(\frac{1}{G}\)

Answer: 2. \(\mathrm{GMK}=4 \pi^2\)

Let the mass of the Sun = M

and mass of planet = m

Then the gravitational force between them, \(\mathrm{F}=\frac{\mathrm{GM} m}{r^2}\)

\(\mathrm{F}=\frac{\mathrm{GM} m}{r^2}\)

where, r= distance between sun and planet.

The above force provided the centripetal force,

⇒ \(\frac{\mathrm{GM} m}{r^2}=\frac{m v^2}{2}\)

⇒ \(\sqrt{\frac{\mathrm{GM}}{r}}\)=v

We know that the period of a planet is,

T =\(\frac{2 \pi r}{v}\)

T =\(\frac{2 \pi r}{\sqrt{\frac{\mathrm{GM}}{r}}}\)

⇒ \(T^2=\frac{4 \pi^2 r^2}{\frac{\mathrm{GM}}{r}}=\frac{4 \pi^2 r^3}{\mathrm{GM}}\) → Equation 1

According to Kepler’s third law, \(T^2 \propto r^3 \)  →  Equation 2

⇒ \(T^2=k r^3 \)

From eq. (1) and (2),

⇒ \(\frac{4 \pi^2 r^3}{\mathrm{GM}} =k r^3 \)

∴ \(\mathrm{GMK} =4 \pi^2\)

Question 11. The dependence of the intensity of gravitational field (E) of Earth distance (r) from the centre of Earth is correctly represented by:

Answer: 2.

Option (B) is the graph of dependence of the intensity of gravitational field (E) of Earth with distance (r) from the centre of Earth

Question 12. Which one of the following plots represents the variation of the gravitational field on a particle with distance r due to a thin spherical shell of radius R (r is measured from the centre of the spherical shell)?

Answer: 2.

Question 13. A body projected vertically from the Earth reaches a height equal to the Earth’s radius before returning to the Earth. The power exerted by the gravitational force is greatest:

1. at the instant just before the body hits the Earth
2. it remains constant all through
3. at the instant just after the body is projected
4. at the height position of the body

Answer: 1. at the instant just before the body hits the Earth

We know that, P=\(\vec{F} \cdot \vec{v}=F v \cos \theta\)

Power will be maximum when velocity and \(cos \theta\) will be maximum.

Question 14. Two spheres of masses m and M are situated in the air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be:

1. 3F
2. F
3. \(\frac{F}{3}\)
4. \(\frac{F}{9}\)

Answer: 2. F

The gravitational force does not depend upon the medium in which objects are placed.

Question 15. Gravitational force is required for:

1. stirring of liquid
2. convection
3. conduction
4. radiation

Answer: 2. convection

Fluid sections become lighter and flow upward as temperature rises, while heavier and denser fluid portions migrate lower. As a result, particles move up and down depending on their weight and gravity. As a result, the existence of a gravitational field is essential in natural convection heat transfer

Question 16. A body of weight 72 N moves from the surface of Earth at a height half of the radius of Earth, then the gravitational force exerted on it will be:

1. 36 N
2. 32 N
3. 144 N
4. 50 N

Answer: 2. 32 N

⇒ \(\mathrm{F}_{\text {surface }} =F \frac{M m}{R_e^2}\)

⇒ \(\mathrm{~F}_{R_e / 2} =G \frac{M m}{\left(R_e+R_e / 2\right)^2}\)

=\(\frac{4}{9} \times F_{\text {surface }}\)

=\(\frac{4}{9} \times 72=32 \mathrm{~N}\)

Question 17. Two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational attraction. The speed v of each particle is:

1. \(\frac{1}{2} \sqrt{\frac{G m}{R}}\)
2. \(\sqrt{\frac{4 G m}{R}}\)
3. \(\frac{1}{2 R} \sqrt{\frac{1}{G m}}\)
4. \(\sqrt{\frac{G m}{R}}\)

Answer: 1. \(\frac{1}{2} \sqrt{\frac{G m}{R}}\)

Force between the two masses, F = \(-G \frac{m m}{4 R^2}\)

This force will provide the necessary centripetal force for the masses to go around a circle, then,

⇒ \(\frac{G m m}{4 R^2} =\frac{m v^2}{R}\)

⇒ \(v^2 =\frac{G m}{4 R}\)

v =\(\frac{1}{2} \sqrt{\frac{G m}{R}}\)

Question 18. The Earth (mass = 6 x 1024 kg) revolves around the sun with an angular velocity of 2 x 10^-7 rad/s in a circular orbit of radius 1.5 x 10⁸ km. The force exerted by the sun on the Earth, in Newton is:

1. 36 x 10²¹
2. 27 x 10³⁹
3. zero
4. 18 x 10²³

Answer: 1. 36 x 10²¹

Give: mass(m) =\(6 \times 10^{24} \mathrm{~kg}\)

angular velocity \((\omega) =2 \times 10^{-7} \mathrm{rad} / \mathrm{s}\) and

radius(r) =\(1.5 \times 10^8 \mathrm{~km}\)

=\(1.5 \times 10^{11} \mathrm{~m}\)

Force exerted on the Earth =\(\mathrm{mRw}^2\)

=\(\left(6 \times 10^{24}\right) \times\left(1.5 \times 10^{11}\right) \times\left(2 \times 10^{-7}\right)^2\)

=\(36 \times 10^{21} \mathrm{~N}\)

Question 19. If the gravitational force between two objects were proportional to 1 /R (and not as I /R²), where R is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v, proportional to:

1. R
2. R° (independent of R)
3. 1/R²
4. 1/R

Answer: 2. R° (independent of R)

Centripetal force (F)=\(\frac{m v^2}{R}\) and the gravitational force \((\mathrm{F})=\frac{G M m}{R}=\frac{G M m}{R}, (where R^2 \rightarrow R )\).

Since \(\frac{m v^2}{R}=\frac{G M m}{R}\) therefore v=\(\sqrt{G M}\).

Thus velocity v is independent of R.

Question 20. If the mass of the Earth were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?

1. The period of a simple pendulum on the Earth would decrease
2. Walking on the ground would become more difficult
3. Raindrops will fall faster
4. ‘g’ on the Earth will not change

Answer: 1. The Period of a simple pendulum on the Earth would decrease

Let \(M_e\) =riginal mass of Earth

G =Gravitational constant

Then from question, Mass of Earth \(M_E^{\prime}=\frac{M_E}{10}\)

New gravitational constant \(G^{\prime}\)=10 G

We know that, Acceleration due to gravity,

g=\(\frac{G M_E}{R^2}\)

⇒ \(M_E\) = mass of Earth

R = radius of Earth

New acceleration due to gravity u,

⇒ \(g^{\prime} =\frac{G^{\prime} M_E}{R^2}\)

=\(\frac{10 M_E G}{R L}\)

⇒ \(g^{\prime} =10 \mathrm{~g}\)

We also know that time pursued by a simple pendulum is,

T=\(2 \pi \sqrt{\frac{1}{g}}\)

⇒ \(T \propto \sqrt{\frac{1}{g}}\)

∴ This confirms that the period of the pendulum also decreases with the increase in g.

Question 21. A spherical planet has a mass of M_P and a diameter of D_P. A particle of mass m falling freely near the surface of this planet will experience acceleration due to gravity is

1. \(\frac{\Delta G M_P}{D_P^2}\)
2. \(\frac{G M_p m}{D_P^2}\)
3. \(\frac{G M_P m}{D_P^2}\)
4. \(\frac{4 G M_P m}{D_P^2}\)

Answer: 1. \(\frac{\Delta G M_P}{D_P^2}\)

We know that, Gravitational Force, F =\(\frac{G M_e m}{R^2}\)

F =\(\frac{G M_P m}{\left(D_P / 2\right)}\)

=\(\frac{4 G M_P m}{D_P^2}\)

And, F=m a

a =\(\frac{F}{m}\)

=\(\frac{\Delta G M_P m}{P^2}\)

=\(\frac{\Delta G M_P}{\Delta p^2}\)

Question 22. Imagine a new planet having the same density as that of Earth but it is 3 times bigger than the Earth in size. If the acceleration due to gravity on the surface of Earth is g and that on the surface of the new planet is g’ then:

1. \(g^{\prime}=\frac{g}{9}\)
2. \(g^{\prime}\)=27 g
3. \(g^{\prime}\)=9 g
4. \(g^{\prime}\)=3 g

Answer: 4. \(g^{\prime}\)=3 g

g =\(\frac{G M}{R^2}=\frac{G \frac{4}{3} \pi R^2 \rho}{R^2}\)

=\(\frac{4}{3} G \pi R \rho\)

∴ \(g^{\prime} =\frac{4}{3} G \pi(3 R) \rho\) =3 g

Question 23. The density of a newly discovered plant is twice that of Earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the Earth. If the radius of the Earth is R, the radius of the planet would be:

1. 2R
2. 4R
3. \(\frac{1}{4} R\)
4. \(\frac{1}{2} R\)

Answer: 4. \(\frac{1}{2} R\)

We know that, \(\frac{G M_p}{R_p^2} =\frac{G M_e}{R_e^2}\)

⇒ \(\frac{G \times \frac{4}{3} \pi R_p^3 \rho_p}{R_p^2} =\frac{G \times \frac{4}{3} \pi R_e^3 \rho_e}{R_e^2}\)

we get, Given,

⇒ \(R_p \rho_p =R_e \rho_e \)

⇒ \(\rho_p =2 \rho_e\)

then \(R_p \times 2 \rho_e =R_e \rho_e\)

∴ \(R_p =\frac{R_e}{2}=\frac{R}{2}\)

Question 24. The acceleration due to gravity on planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2 m on the surface of A. What is the height of the jump by the same person on the plane B?

1. 2/9 m
2. 18 m
3. 6 m
4. 2/3 m

Answer: 2. 18 m

Height Of Jump on the planet B

= \(\frac{g_A}{g_B} \times\)Height Of Jump on the planet A

= 2 x 9 = 18m

Question 25. What will be the formula of mass of the Earth in terms of g, R And G?

1. \(G \frac{R}{g}\)
2. \(g \frac{R^2}{G}\)
3. \(g^2 \frac{R}{G}\)
4. \(G \frac{g}{R}\)

Answer: 2. \(g \frac{R^2}{G}\)

As We Know, F =\(\frac{G M m}{R^2}\)

mg =\(\frac{G M m}{R^2}\)

g =\(\frac{G M}{R^2}\)

M =\(\frac{g R^2}{G}\)

Where, M = mass of earth

g = Gravitational Acceleration

R = Radius of Earth

G = Gravitational constant

Question 26. The acceleration due to gravity and mean density of the Earth \(rho\) are related by which of the following relations? (where G is the gravitational constant and R is the radius of the Earth.)

1. \(\rho=\frac{3 g}{4 \pi G R}\)
2. \(\rho=\frac{3 g}{4 \pi G R^3}\)
3. \(\rho=\frac{4 \pi g R^2}{3 G}\)
4. \(\rho=\frac{4 \pi g R^3}{3 G}\)

Answer: 1. \(\rho=\frac{3 g}{4 \pi G R}\)

Acceleration Due to Gravity g =\(G \times \frac{M}{R^2}\)

=G \(\frac{(4 / 3) \pi R^3 \times \rho}{R^2}\)

=G \(\times \frac{4}{3} \pi R \times \rho\)

or \(\rho =\frac{3 g}{4 \pi G R}\)

Question 27. The radius of Earth is about 6400 km and that of Mars is 3200 km. The mass of the Earth is about 10 times the mass of Mars. An object weighs 200 N on the surface of the Earth. Its weight on the surface of Mars will be:

1. 20 N
2. 8 N
3. 80 N
4. 40 N

Answer: 3. 80 N

Given: radius of Earth \(\left(R_e\right)\)=6400 \(\mathrm{~km}\); the radius of Mars \(\left(R_m\right)=3200 \mathrm{~km}\); the mass of Earth \(\left(M_e\right)=10 M_m\) and weight of the object on Earth \(/left(W_e)=200 \mathrm{~N}\).

⇒ \(\frac{W_m}{W_e} =\frac{m g_m}{m g_e}\)

=\(\frac{M_m}{M_e} \times\left(\frac{R_e}{R_m}\right)^2\)

=\(\frac{1}{10} \times(2)^2=\frac{2}{5}\)

∴ \(W_m=W_e \times \frac{2}{5}=200 \times 0.4=80 \mathrm{~N}\)

Question 28. A particle of mass ‘w’ is projected with a velocity v = kV_e (k < 1) from the surface of the Earth. (v_e = Escape velocity) The maximum height above the surface reached by the particle is

1. \(\mathrm{R}\left(\frac{\mathrm{K}}{1-k}\right)^2\)
2. \(\mathrm{R}\left(\frac{\mathrm{K}}{1+k}\right)^2\)
3. \(\frac{\mathrm{R}^2 k}{1+k}\)
4. \(\frac{\mathrm{R} k^2}{1-k^2}\)

Answer: 4. \(\frac{\mathrm{R} k^2}{1-k^2}\)

Given, v=\(\mathrm{K} v_e\), H =?

At max height H \(\mathrm{KE} =\mathrm{PE}\)

⇒ \(\mathrm{KE} =\frac{1}{2} m v^2\)

=\(\frac{1}{2} m \mathrm{~K}^2 \mathrm{~V}_e{ }^2 \)

⇒ \(\mathrm{PE} =\frac{\mathrm{GM} m}{\mathrm{R}+\mathrm{H}}\)

⇒ \(\mathrm{v}_e{ }^2 =\frac{2 \mathrm{GM}}{\mathrm{R}}\)

⇒ \(\frac{\mathrm{K}^2}{\mathrm{R}} =\frac{1}{\mathrm{R}+\mathrm{H}}\)

⇒ \(\mathrm{R}+\mathrm{H} =\frac{\mathrm{R}}{\mathrm{K}^2}\)

H =\(R\left(\frac{1}{\mathrm{~K}^2}-1\right)\)

=\(R\left(\frac{1-\mathrm{K}^2}{\mathrm{~K}^2}\right)\)

Question 29. The dependence of acceleration due to gravity g on the distance f from the centre of Earth assumed to be a sphere of radius R of uniform density is as shown in the figure below:

Answer: 4.

g \(\propto r(\text { if } r<R)\)

And \(\mathrm{g} \propto \frac{1}{r^2}(\text { If } r>R)\)

The correct figure is D.

Question 30. What is the depth at which the value of acceleration due to gravity becomes 1/n times the value at the surface of Earth (Radius of Earth = R)?

1. \(\frac{\mathrm{R}}{n^2}\)
2. \(\frac{\mathrm{R}(n-1)}{n}\)
3. \(\frac{\mathrm{Rn}}{(n-1)}\)
4. R.n

Answer: 2. \(\frac{\mathrm{R}(n-1)}{n}\)

Radius of Earth = R

Let at depth d, gravitational acceleration become \(\frac{g}{n}\)

⇒ \(g_d =g_o\left(1-\frac{d}{R}\right)\)

⇒ \(\frac{g_o}{n} =g_o\left(1-\frac{d}{R}\right)\)

⇒ \(\frac{1}{n} =1-\frac{d}{R}\)

⇒ \(\frac{d}{R} =1-\frac{1}{n}\)

⇒ \(\frac{d}{R} =\left(\frac{n-1}{n}\right)\)

d =\(\left(\frac{n-1}{n}\right)\) R .

Question 31. A body weighs 72 N on the surface of the Earth. What is the gravitational force on it, at a height equal to half of the radius of the Earth?

1. 32N
2. 30N
3. 24N
4. 48N

Answer: 1. 32N

Given, W = mg = 72 N

On the surface of the Earth. At a height equal to half of the radius of Earth.h=\(\frac{\mathrm{R}}{2}\)

Acceleration due to gravity,

⇒ \(g^{\prime} =g\left(\frac{\mathrm{R}}{\mathrm{R}+h}\right)^2\)

= \(\mathrm{g}\left(\frac{\mathrm{R}}{\mathrm{R}+\frac{\mathrm{R}}{2}}\right)^2\)

= \(g\left(\frac{4 \mathrm{R}^2}{9 \mathrm{R}^2}\right)\)

⇒ \(g^{\prime} =g \times \frac{4}{9}\)

⇒ \(m g^{\prime} =m g \times \frac{4}{9}\)

= \(\frac{4}{9} \times 72 \)

∴ \(m g^{\prime} =32 \mathrm{~N}\)

Question 32. A body weighs 200 N on the surface of the Earth. How much will it weigh halfway down to the centre of the Earth?

1. 200N
2. 250N
3. 100N
4. 150N/kg

Answer: 3. 250N

Below the surface of Earth, acceleration due to gravity is: \(g^{\prime}=\left(1-\frac{d}{R}\right)\)   →   Equation 1

Where, d= depth, R: is the radius of Earth g = acceleration due to gravity Multiply m in both sides of the equations

⇒ \(Given, m g^{\prime}=m g\left(1-\frac{d}{R}\right)\)  →   Equation 2

Given, W=m g=200

d=\(\frac{R}{2}\)

Putting these values in eq. (2), we get

⇒ \(m g^{\prime} =200\left(1-\frac{R}{\frac{2}{R}}\right)=200\left(1-\frac{1}{2}\right)\)

=\(200 \times \frac{1}{2}\)

=100 N

∴ The body will weigh 100 N halfway down to the centre of Earth.

Question 33. The acceleration due to gravity at a height of 1 km above the Earth is the same as at a depth d below the surface of the Earth. Then:

1. d=\(\frac{1}{2} \mathrm{~km}\)
2. d=1 \(\mathrm{~km}\)
3. d=\(\frac{3}{2} \mathrm{~km}\)
4. d=2 \(\mathrm{~km}\)

Answer: 4. d=2 \(\mathrm{~km}\)

Given that acceleration due to gravity at depth d is,

⇒ \(g^{\prime}=g\left(1-\frac{d}{R}\right)\)  →  Equation 1

Where, R=radius of Earth

And Acceleration due to gravity at height h is,

⇒ \(g^{\prime \prime}=g\left(1-\frac{2 h}{R}\right)\)   →  Equation 2

According to question,\( g^{\prime} =g^{\prime \prime}\) and h=1 \(\mathrm{~km}\)

⇒ \(g\left(1-\frac{d}{R}\right) =g\left(1-\frac{2 h}{R}\right)\)

(putting h=1 )

⇒ \(g\left(1-\frac{d}{R}\right) =g\left(1-\frac{2}{R}\right)\)

⇒ \(\frac{d}{R} =\frac{2}{R}\)

d =2 km

Question 34. Starting from the centre of the Earth having radius R, the variation of due to gravity) is shown by:

Answer: 2.

We know that acceleration due to gravity below \(g_d=g\left(1-\frac{d}{R}\right)\)

and acceleration due to gravity at height h is \(g_h=g\left(\frac{R^2}{(R+h)^2}\right)\)

From the above equation, it is clear that the value of g increases from the centre of the maximum at the surface and then decreases

g=\(\left(\frac{G M_e}{R_e^3}\right) r, \text { for } 0<r \leq R_e\)

⇒ \(g \propto r \)

⇒ \(g=\frac{G M_e}{r^2}\), for \(r<R_e \)

∴ \(g \propto \frac{1}{r^2}\)

Question 35. The height at which the weight of a body becomes 1/16th its weight on the surface of Earth (radius R), is:

1. 5 R
2. 15 R
3. 3 R
4. 4 R

Answer: 3. 3 R

According to the question, \(\frac{G M \not m}{(R+h)^2}=\frac{1}{16} \frac{G M \not h}{R^2}\)

Where, R= Radius of Earth

⇒ \(\frac{1}{(\mathrm{R}+h)^2}=\frac{1}{16 R^2}\)

⇒ \(\frac{R}{R+h}=\frac{1}{4}\)

⇒ \(\frac{R+h}{R}\)=4

∴ h = 3R

Question 36. A body of mass 60 g experiences a gravitational force of 3.0 N when placed at a particular point. The magnitude of the gravitational field intensity at that point is:

1. 15 N/kg
2. 20N/kg
3. 150N/kg
4. 0.05 N/kg

Answer: 1. 15 N/kg

Given: Mass of the body, m= 60 g = 0.06 kg

Gravitational force, F =3.0 N

The gravitational field intensity is written as; \(I_g=\frac{F}{m}\) →  Equation 1

Now, On putting all the given values in equation (1) we have,

⇒ \(I_g=\frac{3.0}{0.06}\)

∴ \(I_g=50 \mathrm{~N} / \mathrm{kg}\)

Question 37. A particle of mass M is situated at the centre of a spherical shell of the same mass and radius a. ‘The magnitude of the gravitational potential at a point situated at \(\frac{a}{2}\) distance from the centre, will be:

1. \(\frac{2 G M}{a}\)
2. \(\frac{3 G M}{a}\)
3. \(\frac{4 G M}{a}\)
4. \(\frac{G M}{a}\)

Answer: 2. \(\frac{3 G M}{a}\)

⇒ \(v_P =v_{\text {sphere }}+v_{\text {particle }}\)

⇒ \(v_P =\frac{G Y}{a}+\frac{G Y}{a / 2} \)

=\(\frac{G Y}{a}+\frac{2 G y}{a}=\frac{3 G Y}{a}\)

Question 38. The work done to raise a mass m from the surface of the Earth to a height h, which is equal to the radius of the Earth, is:

1. \(\frac{3}{2} m g R\)
2. mgR
3. 2 \(\mathrm{mgR}\)
4. \(\frac{1}{2} m g R\)

Answer: 4. \(\frac{1}{2} m g R\)

Work done = Change in potential energy

=\(u_f-u_i=\frac{-G M m}{(R+h)}-\left(\frac{-G M m}{R}\right)\)

where, \(\mathrm{M}\) is the mass of Earth and \(\mathrm{R}\) is the radius of Earth.

⇒ \(\mathrm{W}=G M m\left[\frac{1}{R}-\frac{1}{(R+h)}\right]\)

Now, h=R

W=\(G M m\left[\frac{1}{R}-\frac{1}{2 R}\right]=\frac{G M m}{2 R}\)

W=\(\frac{m g R}{2} g=\frac{G M}{R^2}\)

Question 39. At what height from the surface of Earth are the gravitation potential and the value of g – 5.4 * 10⁷ J kg^-2 and 6.0 ms^-2 respectively? [Take, the radius of Earth as 6400 km.]

1. 1600 km
2. 1400 km
3. 2000 km
4. 2600 km

Answer: 4. 2600 km

We know that Gravitational potential at height h from the surface of Earth is v=\(-\frac{G M}{R+h}\)

Here, v=\(-\frac{G M}{R+h}=-5.4 \times 10^7\)  →  Equation 1

Given an acceleration due to gravity of height ft from the surface of Earth,

g=\(\frac{G M}{(R+h)^2}\)=6  →  Equation 2

Divide eq. (1) by eq. (2),

=\(\frac{5.4 \times 10^7}{(R+h)}\)=6

⇒ R+h=9000 km =9000-6400

h =2600 km

Question 40. An infinite number of bodies, each of mass 2 kg are situated on the x-axis at distances 1 m, 2 m, 4 m, and 8 m, respectively from the origin. The resulting gravitational potential due to this system at origin will be:

1. -G
2. \(-\frac{8}{3} G\)
3. \(-\frac{4}{3} G\)
4. -4 G

Answer: 4. -4 G

The resulting gravitational potential due to the system again is,

V =\(-2 G\left[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots\right]\)

=\(-2 G\left[1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots\right]\)

V =\(-2 G\left(1-\frac{1}{2}\right)^{-1} \)

V =\(\frac{-2 G}{\left(1-\frac{1}{2}\right)}\)

V =\(\frac{-2 G}{\frac{1}{2}}\)=-4 G

V =-4 G

Question 41. A body of mass ‘m’ is taken from the Earth’s surface to a height equal to twice the radius (R) of the Earth. The change in potential energy of the body will be:

1. 3 mgR
2. \(\frac{1}{3} m g R\)
3. mg 2R
4. \(\frac{2}{3} m g R\)

Answer: 4. \(\frac{2}{3} m g R\)

Gravitational potential energy at any point at a distance r from the centre of the Earth is, U=\(-\frac{G M m}{r}\)

where M and m are masses of the Earth and the body respectively.

At the surface of the Earth, r =R ;

⇒ \(U_i =-\frac{G M m}{R}\)

At a height of h from the surface,

r=R+h=R+2 R=3 R

h = 2R (Given)

⇒ \(U_f =-\frac{G M m}{3 R}\)

Change in potential energy,

⇒ \(\Delta U =U_f-U_i\)

=\(-\frac{G M m}{3 R}-\left(-\frac{G M m}{R}\right)\)

=\(\frac{G M m}{R}\left(1-\frac{1}{3}\right)\)

=\(\frac{2}{3} \frac{G M m}{R}=\frac{2}{3} m g R\)

∴ \((g=\frac{G M}{R^2}\)

Question 42. A body of mass m is placed on Earth’s surface which is taken from Earth’s surface to a height of h = 3R, then the change in the gravitational potential energy is:

1. \(\frac{m g R}{4}\)
2. \(\frac{2}{3} m g R\)
3. \(\frac{3}{4} m g R\)
4. \(\frac{m g R}{2}\)

Answer: 3. \(\frac{3}{4} m g R\)

change in Gravitational Potential energy

= final energy – initial energy

=\(-\frac{G M m}{4 R}+\frac{\mathrm{G} M m}{R}\)

=\(-\frac{G M m}{4 R}\left[1-\frac{1}{4}\right]\)

=\(\frac{3}{4} \frac{G M m}{4 R}\)

=\(\frac{3}{4} \frac{G M}{R^2} m R=\frac{3}{4} g m R\)

Question 43. A particle of mass m is thrown upwards from the surface of the Earth, with a velocity u. The mass and the radius of the Earth are, respectively, M and R. G is gravitational contact and g is acceleration due to gravity on the the surface of the Earth. The minimum value of u so that the particle does not return to Earth, is:

1. \(\sqrt{\frac{2 G M}{R}}\)
2. \(\sqrt{\frac{2 G M}{R^2}}\)
3. \(\sqrt{2 g R^2}\)
4. \(\sqrt{\frac{2 G M}{R^2}}\)

Answer: 1. \(\sqrt{\frac{2 G M}{R}}\)

We know that Escape velocity, \(v_e=\sqrt{2 g R}\)

since, GM =\(9 \mathrm{R}^2 \)

=\(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}^2} \times \mathrm{R}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\)

Question 44. The ratio of escape velocity on Earth (v_e) to the escape velocity at a planet (v_p) whose radius and mean density are twice that of Earth is:

1. 1: 2 \(\sqrt{2}\)
2. 1: 4
3. 1: \(\sqrt{2}\)
4. 1: 2

Answer: 1. 1: 2 \(\sqrt{2}\)

Since, escape velocity, \(v_e =\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{2 G}{R}\left(\frac{4}{3} \pi R^3 \rho\right)}\)

⇒ \(v_e =R \sqrt{\rho}\)

∴ Ratio =\(1: 2 \sqrt{2}\)

Question 45. A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would Earth (mass = 5.98 x 10²⁴ kg) have to be compressed to be a black hole?

1. 10^-9 m
2. 10^-6 m
3. 10^-2 m
4. 100 m

Answer: 3. 10^-2 m

In this question compare escape relatives with the velocity of light c.

Escape velocity \(v_e=\sqrt{\frac{2 G M}{R}}\)

R= New radius of the Earth

Since,\( v_e=c\)

c=\(\sqrt{\frac{2 G M}{R}}\)

⇒ \(c^2=2 \frac{G M}{R}\)

R =\(\frac{2 G M}{c^2}\)

=\(\frac{2 \times 6.07 \times 10^{-11} \times 6 \times 10^{24}}{\left(3 \times 10^8\right)^2}\)

=\(\frac{2 \times 6.07 \times 10^{-11} \times 6 \times 10^{24}}{9 \times 10^{16}}\)

=\(\frac{4 \times 6.07}{3} \times 10^{-3}\)

=0.889 \(\times 10^{-2}\)

⇒ \(10^{-2} \mathrm{~m}\)

Here, gravitational constant G=\(6.07 \times 10^{-11}\)

Mass of Earth M=\(6 \times 10^{14} \mathrm{~kg}\)

Question 46. The radius of the planet is twice the radius of Earth. Both have almost equal average mass densities v_p and v_E are escape velocities of the planet and the Earth respectively, then:

1. v_p = 1.5 v_E
2. v_p = 2 v_E
3. v_E= 3 v_p
4. v_E= 1.5 v_p

Answer: 2. v_p = 2 v_E

According to the question,

⇒ \(R_P=2 R_E, \rho_E=\rho_P\)

Escape velocity of Earth, \(v_E =\sqrt{\frac{2 G M_E}{R_E}}\)

=\(\sqrt{\frac{2 G}{R_E}\left(\frac{4}{3} \pi R_E{ }^3 \rho_E\right)}\)

=\(R_E \sqrt{\frac{8}{3} \pi G \rho_E}\)  → Equation 1

Similarly, the Escape velocity of the planet

⇒ \(v_P=\sqrt{\frac{2 G M_P}{R_P}}\)

⇒ \(v_P=\sqrt{\frac{2 G}{R_P}\left(\frac{4}{3} \pi R_p^3 \rho_p\right)}\)

⇒ \(v_P=R_P \sqrt{\frac{8}{3} \pi G \rho_p}\)  →  Equation 2

from eq. (1) and eq. (2), we get,

⇒ \(\frac{v_E}{v_P} =\frac{R_E}{R_P} \sqrt{\frac{\rho_E}{\rho_P}}\)

=\(\frac{R_E}{2 R_E} \sqrt{\frac{\rho_Q}{\rho_E}}=\frac{1}{2}\)

∴ \(v_p =2 v_E\)

Question 47. A particle of mass W is kept at rest at a height 3R from the surface of Earth, where ‘R’ is the radius of Earth and ‘M is the mass of Earth. The minimum speed with which it should be projected, so that it does not return, is (g is the acceleration due to gravity on the surface of Earth):

1. \(\left(\frac{G M}{2 R}\right)^{\frac{1}{2}}\)
2. \(\left(\frac{g R}{4}\right)^{\frac{1}{2}}\)
3. \(\left(\frac{2 g}{R}\right)^{\frac{1}{2}}\)
4. \(\left(\frac{G M}{R}\right)^{\frac{1}{2}}\)

Answer: 1. \(\left(\frac{G M}{2 R}\right)^{\frac{1}{2}}\)

Escape velocity \(v_e=\sqrt{\frac{2 \mathrm{GM}}{(\mathrm{R}+h)}}\)

From the question h=3 R

⇒ \(v_e^{\prime} =\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}+3 \mathrm{R}}}\)

=\(\sqrt{\frac{2 G M}{4 R}}\)

=\(\sqrt{\frac{G M}{2 R}}=\sqrt{\frac{g R}{2}}\)

∴ where, g =\(\frac{G M}{R^2}\)

Question 48. The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is:

1. \(\frac{1}{2}\)
2. \(\sqrt{2}\)
3. \(\frac{1}{\sqrt{2}}\)
4. \(\frac{1}{3}\)

Answer: 3. \(\frac{1}{\sqrt{2}}\)

We know that Escape velocity,

⇒ \(v_e=\sqrt{2 g R}\)

Escape velocity of the body from the platform,

⇒ \(\mathrm{P} \cdot \mathrm{E}+\mathrm{K} \cdot \mathrm{E}\) =0

⇒ \(-\frac{G M m}{2 R}+\frac{1}{2} m v^2\) =0

v =\(\sqrt{\frac{G M}{R^2}+R}=\sqrt{g R}\)

This confirms that:

fv =\(\frac{v}{\sqrt{2}}\)

f =\(\frac{1}{\sqrt{2}}\)

Question 49. Which velocity should a particle be projected so that its height becomes equal to the radius of Earth?

1. \(\left(\frac{G M}{R}\right)^{1 / 2}\)
2. \(\left(\frac{8 G M}{R}\right)^{1 / 2}\)
3. \(\left(\frac{2 G M}{R}\right)^{1 / 2}\)
4. \(\left(\frac{4 G M}{R}\right)^{1 / 2}\)

Answer: 1. \(\left(\frac{G M}{R}\right)^{1 / 2}\)

Use , \(v^2=\frac{2 g n}{1+\frac{h}{R}}\) given h= R

v=\(\sqrt{g R}=\sqrt{\frac{G M}{R}}\)

Question 50. For a planet having mass equal to the mass of the Earth the radius is one-fourth of the radius of the Earth. The escape velocity for this planet will be :

1. 11.2 km/s
2. 22.4 km/s
3. 5.6 km/s
4. 44.8 km/s

Answer: 2. 22.4 km/s

⇒ \(v_e =\sqrt{2 g R_e}=\sqrt{\frac{2 G M}{R_e}}\)

⇒ \(R_p =\frac{1}{4} R_e\)

∴ \(v_p =2 v_e=2 \times 11.2=22.4 \mathrm{~km} / \mathrm{s}\)

Question 51. The escape velocity of a sphere of mass m is given by (G = Universal gravitational constant; M_e = Mass of the earth and R_e = Radius of the earth)₁

1. \(\sqrt{\frac{2 G M_e m}{R_e}}\)
2. \(\sqrt{\frac{2 G M_e}{R_e}}\)
3. \(\sqrt{\frac{G M_e}{R_e}}\)
4. \(\sqrt{\frac{2 G M_e+R_e}{R_e}}\)

Answer: 2. \(\sqrt{\frac{2 G M_e}{R_e}}\)

To escape Earth’s gravitational pull and reach infinity, the starting kinetic energy must be somewhat more than the gravitational potential energy of the body on the surface.

The total energy of the body just after projecting is: \(\mathrm{E}=\frac{1}{2} m v^2-\frac{G M_e m}{R_e}\)

This total energy must now be slightly greater than zero for the minimum speed to be achieved when kinetic energy equals gravitational potential energy.

⇒ \(\frac{1}{2} m v^2 =\frac{G M_e m}{R_e}\)

⇒ \(\frac{1}{2} v^2 =\frac{G M_e}{R_e}\)

v =\(\sqrt{\frac{2 G M_e}{R_e}}\)

Question 52. The escape velocity of a body on the surface of the Earth is 11.2 km/s. If the Earth’s mass increases to twice its present value and the radius of the Earth becomes half, the escape velocity becomes:

1. 22.4 km/s
2. 44.8 km/s
3. 5.6 km/s
4. 11.2 km/s

Answer: 1. 22.4 km/s

Escape velocity of a body \(\left(v_e\right)=11.2 \mathrm{~km} / \mathrm{s}\);

New mass of the Earth \(M_e^{\prime}=2 M_e \) and the new radius of the Earth \(R_e^{\prime}=0.5 R_e\)

Escape velocity \(\left(v_e\right)=\sqrt{\frac{2 G M_e}{R_e}} \propto \sqrt{\frac{M_e}{R_e}}\)

⇒ \(\frac{v_e}{v_e^{\prime}} =\sqrt{\frac{M_e}{R_e} \times \frac{0.5 R_e}{2 M_e}}\)=\(\sqrt{\frac{1}{4}}=\frac{1}{2}\)

∴ \(v_e^{\prime} =2 v_e=22.4 \mathrm{~km} / \mathrm{s}\)

Question 53. The escape velocity from the Earth is 11.2 km/s. If a body is to be projected in a direction making an angle of 45° to the vertical, then the escape velocity is :

1. 11.2×2 km/s
2. 11.2 km/s
3. \(\frac{11.2}{\sqrt{2}} \mathrm{~km} / \mathrm{s}\)
4. \(11.2 \sqrt{2} \mathrm{~km} / \mathrm{s}\)

Answer: 2. 11.2 km/s

As, \(v_e=\sqrt{2 g R}=\sqrt{\frac{2 G M}{R}}\)

Hence, escape velocity does not depend on the angle of projection. Escape velocity will remain the same.

Hence, the escape velocity is 11.2 Km/s.

Question 54. For a satellite escape velocity is 11 km/s. If the satellite is launched at an angle of 60° with the vertical, then the escape velocity will be

1. \(11 \mathrm{~km} / \mathrm{s}\)
2. \(11 \sqrt{3} \mathrm{~km} / \mathrm{s}\)
3. \(\frac{11}{\sqrt{3}} \mathrm{~km} / \mathrm{s}\)
4. \(33 \mathrm{~km} / \mathrm{s}\)

Answer: 3. \(\frac{11}{\sqrt{3}} \mathrm{~km} / \mathrm{s}\)

Escape velocity on Earth is the minimum velocity with which the body has to be projected vertically upwards from the surface of the Earth. So, that it just crosses the gravitational field of Earth and never returns on its own. The escape velocity of the Earth is given by

⇒ \(v_e =\sqrt{\frac{2 G M}{R}}=\sqrt{2 g R}\)

=\(\sqrt{\frac{8 \pi \rho G R^2}{3}}\)

It is clear from the above equation that the value of a body’s escape velocity is independent of its mass (m) and angle of projection from the planet’s surface. As a result, the escape velocity remains unchanged. Hence, v = 11 km/s Projected velocity.

Question 55. The escape velocity from the Earth’s surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and the same mass density is:

1. v
2. 2v
3. 3v
4. 4v

Answer: 2. 2v

Given the escape velocity of Earth = v

Radius of other planet = \(4 \mathrm{R}_e\)

Mass density is same = d

Mass of other planet =\(\frac{4}{3} \pi\left(4 \mathrm{R}_{\mathrm{e}}\right)^3 \cdot d \)

Mass of Earth =\(\frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^3 \cdot d \)

⇒ \(\mathrm{M}=\frac{\mathrm{M}_{\mathrm{P}}}{\mathrm{M}_e}=\frac{64 \mathrm{R}_e^3}{\mathrm{R}_e^3}\)=64

⇒ \(\mathrm{M}_P=64 \mathrm{M}_{\mathrm{e}}\)

We know, \(v_e=\sqrt{\frac{2 \mathrm{GM}_e}{\mathrm{R}_e}}\)

⇒ \(v_P=\sqrt{\frac{2 \mathrm{G}\left(64 \mathrm{M}_e\right)}{4 \mathrm{R}_e}}\)

=\(4 \sqrt{\frac{2 \mathrm{GM}_e}{\mathrm{R}_e}}\)

⇒ \(\frac{v_P}{v_e}\)=4

∴ \(v_P=4 v_e \)

Question 56. The period of a geostationary satellite is 24 h, at a height of 6RE (R_E is the radius of Earth) from the surface of Earth. The period of another satellite whose height is 2.5 R_E from the surface will be:

1. \(6 \sqrt{2} h\)
2. \(12 \sqrt{2} h\)
3. \(\frac{24}{2.5} h\)
4. \(\frac{12}{2.5} h\)

Answer: 1. \(6 \sqrt{2} h\)

From Kepler’s third law we know that period:

⇒ \(\mathrm{T}^2 \propto r^3\)

⇒ \(\mathrm{~T} =\sqrt[2]{r^3}\)  →  Equation 1

Where, r= radius of satellite’s orbit

Now from question, \(r_1=6 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}, \mathrm{T}_1=24 \mathrm{~h}\)

⇒ \(r_2=2.5 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}, \mathrm{T}_2\)=?

From eq. (1) we can write,

⇒ \(\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)=\left(\frac{r_1}{r_2}\right)^{3 / 2}\)

Putting the given values,

⇒ \(\frac{24}{\mathrm{~T}_2} =\left(\frac{6 \mathrm{R}_{\mathrm{E}} 7 \mathrm{R}_{\mathrm{E}}}{2.5 \mathrm{R}_{\mathrm{E}} 7 \mathrm{R}_{\mathrm{E}}}\right)^{3 / 2}=\left(\frac{7 \mathrm{R}_{\mathrm{E}}}{3.5 \mathrm{R}_{\mathrm{E}}}\right)^{3 / 2}\)

=\(\left(\frac{7}{3.5}\right)^{3 / 2}=(2)^{3 / 2}\)

⇒ \(\mathrm{T}_2 =\frac{24}{(2)^{3 / 2}}\)

=\(\frac{24}{2 \sqrt{2}}=6 \sqrt{2} h\)

Question 57. A remote sensing satellite of Earth revolves in a circular orbit at a height of 0.25 x 10⁶ m above the surface of Earth. If Earth’s radius is 6.38 x 10⁶ m and g = 9.8 ms^-2, then the orbital speed of the satellite is:

1. 7.76 km^-1
2. 8.56 km^-1
3. 9.13 km^-1
4. 6.67 km^-1

Answer: 1. 7.76 km^-1

According to question, Height of a satellite, h=\(0.25 \times 10^6 \mathrm{~m}\)

Radius of Earth, \(R_E=6.38 \times 10^6 \mathrm{~m}\)

For the satellite revolving around the Earth

⇒ \(v_0=\sqrt{\frac{G M_E}{R_E}}=\sqrt{\frac{G M_E}{R_E\left[1+\frac{h}{R_E}\right]}}\)

=\(\sqrt{\frac{g R_E}{1+\frac{h}{R_E}}}\)

Putting value, \(v_0 =\sqrt{\frac{9.8 \times 6.38 \times 10^6}{1+\frac{0.25 \times 10^6}{6.38 \times 10^6}}}=\sqrt{60 \times 10^6}\)

=\(7.76 \times 10^3 \mathrm{~ms}^{-1}=7.76 \mathrm{~km} / \mathrm{s}\)

Question 58. A satellite S is moving in an elliptical orbit around the Earth. The mass of the satellite is very small compared to the mass of the Earth. Then:

1. the angular momentum of S about the centre of the Earth changes in direction, but its magnitude remains constant
2. the total mechanical energy of S varies periodically with time
3. the linear momentum of S remains constant in magnitude
4. the acceleration of S is always directed towards the centre of the Earth

Answer: 4. the acceleration of S is always directed towards the centre of the Earth

The mass of satellites is very small. The Centre of mass of the system coincides with the centre of the Earth

Question 59. A geostationary satellite is orbiting the Earth at a height of 5R above the surface of the Earth, R being the radius of the Earth. The period of another satellite in hours at a height of 2R from the surface of the Earth is:

1. 5
2. 10
3. \(6 \sqrt{2}\)
4. \(\frac{6}{\sqrt{2}}\)

Answer: 3. \(6 \sqrt{2}\)

Using Kepler’s third law as \(\mathrm{T}^2 \propto r^3\)

⇒ \(T_1^2 \propto r_1^3 \) and \( T_2^2 \propto r_2^3\)

⇒ \(\frac{T_2^2}{T_1^2}=\frac{r_2^3}{r_1^3}=\frac{(3 R)^3}{(6 R)^3}=\frac{1}{8}\)

⇒ \(T_2^2=\frac{1}{8} T_1^2\)

∴ \(T_S=\frac{24}{2 \sqrt{2}}=6 \sqrt{2} h\)

Question 60. If \(v_e\) is escaping velocity and \(v_0\) is the orbital velocity of a satellite for orbit close to Each surface, then these are related by:

1. \(v_0=\sqrt{2} v_e\)
2. \(v_0=v_e\)
3. \(v_e=\sqrt{2} v_0\)
4. \(v_e=\sqrt{2} v_0\)

Answer: 3. \(v_e=\sqrt{2} v_0\)

From the question, \(v_{\text {escape }}=\sqrt{\frac{2 G M}{R}}=v_e\)

⇒ \(v_{\text {orbital }}=\sqrt{\frac{G M}{R}}=v_0\)

From the above equations,

∴ \(v_e=\sqrt{2} v_0\)

Question 61. The radii of the circular orbits of two satellites A and B of the Earth are 4R and R, respectively. If the speed of satellite A is 3v, then the speed of satellite B will be:

1. 3v/4
2. 6 v
3. 12v
4. 3v/2

Answer: 2. 6 v

According to the question,

Radius of satellite A = 4R

and radius of satellite B = R

Speed of satellite A = 3v

Speed of satellite B =?

We know that orbital velocity of satellite is, v =\(\sqrt{\frac{G M}{r}}\)

⇒ \(\frac{v_A}{v_B} =\sqrt{\frac{r_{\mathrm{B}}}{r_{\mathrm{A}}}}\)

=\(\sqrt{\frac{R}{4 R}}=\frac{1}{2}=\frac{3 v}{v_B}\)

∴ \(v_B =6 v\)

Question 62. A ball is dropped from a spacecraft revolving around the Earth at a height of 120 km. What will happen to the ball?

1. it will fall to the Earth gradually
2. it will go very far in the space
3. it will continue to move with the same speed along the original orbit of the spacecraft
4. it will move with the same speed, tangentially to the spacecraft

Answer: 3. it will continue to move with the same speed along the original orbit of the spacecraft

Since no external torque is applied therefore according to the law of conservation of angular momentum, the ball will continue to move with the same angular velocity along the original orbit of the spacecraft.

Question 63. The escape velocity from the surface of the Earth is \(v_e\). The escape velocity from the surface of a planet whose mass and radius are three times those of the Earth will be:

1. \(v_e\)
2. \(3 v_e\)
3. \(9 v_e\)
4. \(\frac{1}{3 v_{\mathrm{e}}}\)

Answer: 1. \(v_e\)

Escape velocity on the surface of the Earth is given by

⇒ \(v_e =\sqrt{2 g R_{\mathrm{e}}}\)

=\(\sqrt{\frac{2 G M_{\mathrm{e}}}{R_{\mathrm{e}}}}  g=\frac{G M_{\mathrm{e}}}{R_e^2}\)

where, \(M_e\)= mass of Earth

⇒\(R_{\mathrm{e}}\)= radius of the Earth

G= gravitational constant

⇒ \(v_{\mathrm{e}} \propto \sqrt{\frac{M_{\mathrm{e}}}{R_{\mathrm{e}}}}\)

If \(v_p\) is escape velocity from the surface of the planet, then,

⇒ \(\frac{v_e}{v_p}=\sqrt{\frac{M_e}{R_e}} \times \sqrt{\frac{R_P}{M_P}}\)

where, \(M_p\) is mass of the planet and \(R_p\) is radius of the planet.

but \(R_p=3 R_e\) { (given) }

and \(M_p=3 M_e\)

⇒ \(\frac{v_e}{v_p} =\sqrt{\frac{M_e}{R_e}} \times \sqrt{\frac{3 R_e}{3 M_e}}\)

=\(\frac{1}{1}\)

or \(v_p=v_e\)

Question 64. A satellite I of mass m is at a distance r from the surface of the Earth. Another satellite B of mass 2 z is at a distance of 2 r from the Earth’s surface. Their periods are in the ratio of

1. 1:2
2. 1:16
3. 1:32
4. \(1: 2 \sqrt{2}\)

Answer: 4. \(1: 2 \sqrt{2}\)

According to Kepler’s third law, the square of the period of revolution of a planet around the sun is directly proportional to the cube of the major- axis of its elliptical orbit i.e. \( T^2 \alpha r^3\)

where T= time taken by the planet to go once around the sun.

r= semi-major axis of the elliptical orbit

⇒ \(\frac{T_1^2}{T_2^2}=\frac{(r)^3}{(2 r)^3}=\frac{1}{8}\)

∴ \(\frac{T_1}{T_2}=\frac{1}{2 \sqrt{2}}\)

Question 65. The additional kinetic energy to be provided to a satellite of mass ia revolving around a planet of mass M, to transfer it from a circular orbit of radius \(R_1\) to another of radius \(R_2\left(R_2>R_1\right)\) is:

1. \(G m M\left(\frac{1}{R_1^2}-\frac{1}{R_2^2}\right)\)
2. \(G m M\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
3. \(2 G m M\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
4. \(\frac{1}{2} G m M\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

Answer: 4. \(\frac{1}{2} G m M\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

The kinetic energy of the orbit of the satellite,

K E =\(\frac{\mathrm{GM} m}{2 \mathrm{R}_1}+\left(-\frac{\mathrm{GMm}}{2 \mathrm{R}_2}\right)\)

=\(\frac{\mathrm{GM} m}{2}\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]\)

Question 66. Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final-initial) of an object of mass ru when taken to a height ft from the surface of Earth (of radius R) is given by:

1. \(\frac{G M m}{R+h}\)
2. \(\frac{G M m h}{R(R+h)}\)
3. mg
4. \(\frac{G M m}{R+h}\)

Answer: 2. \(\frac{G M m h}{R(R+h)}\)

⇒ \({ P.E. })_{\mathrm{A}} =-\frac{G M m}{R}\)

⇒ \({ P.E. })_{\mathrm{B}} =-\frac{G M m}{R+h}\)

⇒ \(\Delta U =(\mathrm{P} . \mathrm{E})_{\mathrm{B}}-(\mathrm{P} . \mathrm{E})_{\mathrm{A}}\)

=\(\frac{G M m}{R+h}+\frac{G M m}{R}\)

=\(\frac{G M m h}{R(R+h)}\)

Question 67. A satellite of mass z is orbiting the Earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g₀, the value of acceleration due to gravity at the Earth’s surface is:

1. \(\frac{m g_0 \mathrm{R}^2}{2(\mathrm{R}+h)}\)
2. \(-\frac{m g_0 \mathrm{R}^2}{2(\mathrm{R}+h)}\)
3. \(\frac{2 m g_0 \mathrm{R}^2}{\mathrm{R}+h}\)
4. \(-\frac{2 m g_0 \mathrm{R}^2}{\mathrm{R}+h}\)

Answer: 2. \(-\frac{m g_0 \mathrm{R}^2}{2(\mathrm{R}+h)}\)

Total energy =K.E +P.K.

=\(\frac{G M m}{2(R+h)}-\frac{G M m}{(R+h)}\)

=\(\frac{-G M m}{2(R+h)}-\frac{G M m}{(R+h)}\)

=\(\frac{-m g^{\prime} R^2}{2(R+h)}\)

∴ \(\left(\text { where, } g^{\prime}\right.\left.=\frac{G M}{R^2}\right)\)

Question 68. Two satellites of Earth, S₁ and S₂ are moving in the same orbit. The mass of S₁ of four times the mass of S₂. Which one of the following statements is true?

1. The potential energies of Earth and satellite in the two cases are equal.
2. S₁ and S₂ are moving at the same speed
3. The kinetic energies of the two satellites are equal
4. The period of S₁ is four times that of S₂.

Answer: 2. S₁ and S₂ are moving at the same speed

⇒ \({K.E.}=\frac{G M m}{2 r}\)

Kinetic energies are unequal,

⇒ \(\mathrm{T}=\frac{2 \pi r^{3 / 2}}{\sqrt{G M}}\)

Periods are equal,

⇒ \({ P.E. }=-\frac{G M m}{r}\)

Potential energies are unequal,

V=\(\sqrt{\frac{G M}{r}}\)

orbital speeds are equal.

Question 69. For a satellite in an orbit around the Earth, the ratio of kinetic, energy to potential energy is:

1. \(\frac{1}{2}\)
2. \(\frac{1}{\sqrt{2}}\)
3. 2
4. \(\sqrt{2}\)

Answer: 1. \(\frac{1}{2}\)

⇒ \(-\frac{\mathrm{GMm}}{R^2}+m \omega^2 R \)=0

⇒ \(\frac{\mathrm{GMm}}{R^2} =m \omega^2 R\)

K.E. \( =\frac{1}{2} I \omega^2\)

=\(\frac{1}{2} m R^2 \omega^2\)

=\(\frac{\mathrm{GMm}}{2 R}\)

P.E =\(-\frac{\mathrm{GMm}}{R}\)

K.E =\(\frac{|P \cdot E|}{2}\)

∴ \(\frac{K . E}{P. E} =\frac{1}{2}\)

Question 70. The satellite of mass m is orbiting around the Earth in a circular orbit with a velocity v. What will be its total energy?

1. \((3 / 4) m v^2\)
2. \((1 / 2) m v^2\)
3. \(m v^2\)
4. \(-(1 / 2) m v^2\)

Answer: 4. \(-(1 / 2) m v^2\)

⇒ \({ K.E. of sattelite }=\frac{1}{2} m v^2\)

⇒ \({ (where, } v=\sqrt{\frac{G M}{r^2}}\)

P.E. of sattelite =\(\frac{G M m}{r}=-m v^2 \)

Total Energy = K. E. + P. E.

=\(\frac{1}{2} m v^2-m v^2=-\frac{1}{2} m v^2\)

Question 71. The mean radius of Earth is R, its angular speed on its axis is o and the acceleration due to gravity at Earth’s surface is g. What will be the radius of the orbit of a geostationary satellite?

1. \(\left(R^2 g / \omega^2\right)^{1 / 3}\)
2. \(\left(R g / \omega^2\right)^{1 / 3}\)
3. \(\left(R^2 \omega^2 / g\right)^{1 / 3}\)
4. \(\left(R^2 g / \omega\right)^{1 / 3}\)

Answer: 1. \(\left(R^2 g / \omega^2\right)^{1 / 3}\)

⇒ \(\frac{G M m}{r^2} =m \omega^2 r\)

\(r^3 =\frac{G M}{\omega^2}=\frac{g R^2}{\omega^2}\)

r =\(\left(g R^2 / \omega^2\right)^{1 / 3}\)

System Of Particles And Rotational Motion

Question 1. Three identical spheres, each of mass M, are placed at the comers of a right-angled triangle with mutually perpendicular sides equal to 2 m (see figure). Taking the point of intersection of the two mutually perpendicular sides as the origin, the position vector of the centre of mass is

1. \(2(\hat{i}+\hat{j})\)
2. \((\hat{i}+\hat{j})\)
3. \(\frac{2}{3}(\hat{i}+\hat{j})\)
4. \(\frac{4}{3}(\hat{i}+\hat{j})\)

Answer: 2. \((\hat{i}+\hat{j})\)

See The Figure,

⇒ \(\mathrm{OA}=2 \hat{i}\)

⇒ \(\mathrm{OB}=2 \hat{j}\)

Position vector of centre of mass, \(\mathrm{R}_{\mathrm{CM}} =\frac{m_1 r_1+m_2 r_2}{m_1+m_2}\)

=\(\frac{M \cdot \mathrm{OA}+M \cdot \mathrm{OB}}{M+M}\)

=\(\frac{M \times 2 \hat{i}+M \times 2 \hat{j}}{2 M}=\hat{i}+\hat{j}\)

Question 2. Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass. The centre of mass of the system from the 5 kg particle is nearly at a distance of:

1. 50 cm
2. 67 cm
3. 80 cm
4. 33 cm

Answer: 2. 67 cm

Given That,

⇒ \(\frac{m_1 r_1-m_2 r_2}{m_1+m_2}\) =0

⇒ \(5 r_1-10 r_2\) =0

⇒ \(r_2 =\frac{r_1}{2}\)

⇒ \(r_1+r_2\) =100

⇒ \(r_1+\frac{r_1}{2}\) =100

⇒ \(3 r_1\) =200

∴ \(r_1 =\frac{200}{3}=67 \mathrm{~cm}\)

Read and Learn More NEET Physics MCQs

Question 3. Three masses are placed on the x-axis: 300 g at origin, 500 g at x = 40 cm and 400 g at x = 70 cm. The distance of the centre of mass of the origin is:

1. 40 cm
2. 45 cm
3. 50 cm
4. 30 cm

Answer: 1. 40 cm

⇒ Centre of mass\(r_{\mathrm{cm}} =\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3}\)

=\(\frac{300 \times 0+500 \times 40+400 \times 70}{300+500+400}\)

=40 cm

Question 4. Two bodies of mass 1 kg and 3 kg have position vectors \(\hat{i}+2 \hat{j}+\hat{k}\) and \(-3 \hat{i}+2 \hat{j}+\hat{k}\) respectively. The centre of mass of this system has a position vector:

1. \(-2 \hat{i}+2 \hat{k}\)
2. \(-2 \hat{i}-\hat{j}+\hat{k}\)
3. \(2 \hat{i}-\hat{j}-2 \hat{k}\)
4. \(-\hat{i}-\hat{j}+\hat{k}\)

Answer: 2. \(-2 \hat{i}-\hat{j}+\hat{k}\)

According to the question,

Position Vector\(\vec{r} =\frac{m_1 \vec{r}_1+m_2 \vec{r}_2}{m_1+m_2}\)

⇒ \(\vec{r} =\frac{1(\hat{i}+2 \hat{j}+k)+3(-3 \hat{i}-2 \hat{j}+\hat{k})}{1+3}\)

⇒ \(\vec{r} =\frac{1}{4}(-8 \hat{i}-2 \hat{j}+4 \hat{k})\)

⇒ \(\vec{r} =-2 \hat{i}-2 \hat{j}+\hat{k}\)

Question 5. A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in a horizontal position is released. The initial angular acceleration of the rod is (moment of inertia of the rod about A is \(\frac{m l^2}{3}\):

1. \(m g \frac{l}{2}\)
2. \(\frac{3 g}{2 l}\)
3. \(\frac{2 l}{3 g}\)
4. \(\frac{3 g}{2 l^2}\)

Answer: 2. \(\frac{3 g}{2 l}\)

Here torque,

The moment of Inertia of the rod about A is:

I=\(\frac{m l^2}{3}\)

Angular acceleration of the rod is \(\alpha=\frac{\tau}{I}\)

∴ \(\alpha=\frac{m g\left(\frac{1}{2}\right)}{\frac{m l^2}{3}}=\frac{3 g}{2 l}\)

Question 6. Consider a system of two particles having masses m₁ and m₂. If the particle of mass m₁ is pushed towards the mass centre of a particle through a distance d, by what distance would the particle of mass m₂ move to keep the mass centre of particles at the original position?

1. \(\frac{m_1}{m_1+m_2} d\)
2. \(\frac{m_1}{m_2} d\)
3. \(\frac{d\left(m_1+m_2\right)}{m_1}\)
4. \(\frac{m_2}{m_1}\)

Answer: 2. \(\frac{m_1}{m_2} d\)

We know that, \(\mathrm{CM}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)

After changing the position of m₁ and keeping the position of C.M. the same.

C.M. =\(\frac{m_1\left(x_1 d\right)+m_2\left(\mathrm{x}_2-d_2\right)}{m_1+m_2}\)

0 =\(\frac{m_1 d-m_2 d_2}{m_1+m_2}\)

⇒ \(d_2  =\frac{m_1}{m_2} d\)

∴ C.M. =\(\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)

Question 7. Consider a system of two particles having masses m₁ and m₂. If the particle of mass m1 is pushed towards the centre of mass of the particles through a distance d, by what distance would be particle of mass m₂ move to keep the centre of mass of the particles at the original position?

1. \(\frac{m_1}{m_1+m_2} d\)
2. \(\frac{m_1}{m_2} d\)
3. d
4. \(\frac{m_2}{m_1} d\)

Answer: 2. \(\frac{m_1}{m_2} d\)

After changing the position of m₁ and keeping the position of C.M. same

C.M.=\(\frac{m_1\left(x_1-d\right)+m_2\left(x_2+d_2\right)}{m_1+m_2}\)

0 =\(\frac{-m_1 d+m_2 d_2}{m_1+m_2}\)

[Substituting value of C.M. from (1)

∴ \(d_2=\frac{m_1}{m_2} d\)

Question 8. Three identical metal balls, each of radius r are placed touching each other on a horizontal surface such that an equilateral triangle is formed when the centres of the three balls are joined. The centre of the mass of the system is located at:

1. line joining centres of my two balls
2. centre of one of the balls
3. horizontal surface
4. point of intersection of the medians

Answer: 4. point of intersection of the medians

Centre of mass of each ball lies on the centre

The Centre of mass of the combined body will be at the centroid of an equilateral triangle.

Question 9. Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the centre of mass of the system from the 10 kg mass is :

1. \(\frac{20}{3} m\)
2. \(10 \mathrm{~m}\)
3. \(5 \mathrm{~m}\)
4. \(\frac{10}{3} m\)

Answer: 1. \(\frac{20}{3} m\)

Given, M₁ = 10kg, M₂ = 20kg,

Length of rod = 10 m

Let M₁ be at the origin, and then X₁ and X₂ are 0 and 10 m respectively.

Centre of mass, X=\(\frac{M_1 X_1+M_2 X_2}{M_1+M_2}\)

Putting the values in above equation we get, X=\(\frac{10 \times 0+20 \times 10}{10+20}=\frac{200}{30}=\frac{20}{3}\)

The distance of the centre of mass of the system from the 10 kg mass is: 20/3 m

Question 10. Two persons of masses 55 kg and 65 kg respectively, are at the opposite ends of a boat. The length of the boat is 3.0 m and weighs 100 kg. The 55 kg man walks up to the 65 kg man and sits with him. If the boat is in still water. The centre of mass of the system shifts by:

1. 3.0 m
2. 2.3 m
3. zero
4. 0.75 m

Answer: 3. zero

Since no external force is applied to the system. And centre of mass (CM) be at rest. So our answer is option (C) zero

Question 11. The resultant of \(\vec{A}\) x 0 will be equal to:

1. zero
2. A
3. zero vector
4. unit vector

Answer: 3. zero vector

∴ \(\vec{A} \times 0\) is a zero vector

Question 12. Vectors, \(\vec{A}, \vec{B} \text { and } \vec{C}\) are such that \(\vec{A} \cdot \vec{B}\) = 0 and \(\vec{A} \cdot \vec{C}\)= 0. Then the vector parallel to A is:

1. \(\vec{A} \times \vec{B}\)
2. \(\vec{B}+\vec{C}\)
3. \(\vec{B} \times \vec{C}\)
4. \(\vec{B}and \vec{C}\)

Answer: 3. \(\vec{B} \times \vec{C}\)

Vector triple product of three vectors \(\vec{A}, \vec{B}\) and \(\vec{C}\) is:

Given : \(\vec{A} \times(\vec{B} \times \vec{C})=(\vec{A} \cdot \vec{C}) \vec{B}-(\vec{A} \cdot \vec{B}) \vec{C}\)

⇒ \(\vec{A} \cdot \vec{B}=0, \vec{A} \cdot \vec{C}\)=0

⇒ \(\vec{A} \times(\vec{B} \times \vec{C})\)=0

∴ Thus the vector \(\vec{A}\) is parallel to vector \(\vec{B} \times \vec{C}\).

Question 13.\(\vec{A} \text { and } \vec{B}\) are two vectors and θ is the angle between them, if \(|\vec{A} \times \vec{B}|=\sqrt{3}(\vec{A} \cdot \vec{B})\), the value of θ is:

1. 45°
2. 30°
3. 90°
4. 60°

Answer: 4. 60°

⇒ \(|\vec{A} \times \vec{B}| \) = \(\sqrt{3}(\vec{A} \cdot \vec{B})\)

⇒ \(A B \sin \theta =\sqrt{3} A B \cos \theta\)

or \(\tan \theta =\sqrt{3}\)

or \(\theta =\tan ^{-1} \sqrt{3}=60^{\circ}\)

Question 14. If the angle between the vectors\(\vec{A} \text { and } \vec{B}\) is θ, the value of the product \((\vec{B} \times \vec{A}) \cdot \vec{A}\) A is equal to:

1. \(B A^2 \sin \theta\)
2. \(B A^2 \cos \theta\)
3. \(B A^2 \sin \theta \cos \theta\)
4. zero

Answer: 4. zero

Let, \(|\vec{A} \times \vec{B}|=\vec{C}\)

The cross product of ,\(\vec{B}\) and \(\vec{A}\) is perpendicular to the plane containing \(\vec{A} \)and \(\vec{B}\) i.e. perpendicular to \(\vec{A}\)

Therefore Product Of \((\vec{B} \times \vec{A}) \cdot \vec{A}\)=0

Question 15.If \(|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B}\) then the value of \(|\vec{A}+\vec{B}|\) is:

1. \(\left(A^2+B^2+A B\right)^{1 / 2}\)
2. \(\left(A^2+B^2+\frac{A B}{\sqrt{3}}\right)^{1 / 2}\)
3. A+B
4. \(\left(A^2+B^2+\sqrt{3} A B\right)^{1 / 2}\)

Answer: 1. \(\left(A^2+B^2+A B\right)^{1 / 2}\)

⇒ \(|\vec{A} \times \vec{B}| =\sqrt{3} \vec{A} \cdot \vec{B}\)

⇒ \(|\vec{A} \| \vec{B}| \sin \theta =\sqrt{3}|\vec{A} \| \vec{B}| \cos \theta\)

⇒ \(\tan \theta =\sqrt{3}\)

⇒ \(\theta =60^{\circ}\)

⇒ \(|\vec{A}+\vec{B}| =\sqrt{|\vec{A}|^2+|\vec{B}|^2+2|\vec{A} \| \vec{B}| \cos \theta}\)

=\(\left(A^2+B^2+A B\right)^{1 / 2}\)

Question 16. A disc is rotating with angular velocity about its axis (without any translation push) on a smooth surface: Find the direction and magnitude of velocity at points B and A

1. \( -\omega {R} \) towards left
2. \(-\omega {R} \) towards right
3. \( +\omega {R} \) towards left
4. \( +\omega R \) towards left

Answer: 1. \( -\omega \mathrm{R} towards left \)

In the given question

Velocity at A, \( v_A=\omega\left(\frac{\mathrm{R}}{2}\right)\) towards right

∴ Velocity at B, \(v_B=-\omega\) R towards left

Question 17. What is the value of linear velocity, \(\vec{r}=3 \hat{i}-4 \hat{j}+\hat{k}\) and \(\vec{\omega}=5 \hat{i}-6 \hat{j}+6 \hat{k}\)?

1. \(4 \hat{i}-13 \hat{j}+6 \hat{k}\)
2. \(18 \hat{i}+13 \hat{j}-2 \hat{k}\)
3. \(6 \hat{i}+2 \hat{j}-3 \hat{k}\)
4. \(6 \hat{i}-2 \hat{j}+8 \hat{k}\)

Answer: 2. \(18 \hat{i}+13 \hat{j}-2 \hat{k}\)

Question 18. A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at the 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass ‘m’ is suspended from the rod at the 160 cm mark as shown in the figure. Find the value of ‘m’ such that the rod is in equilibrium, (g = 10 m/s²):

1. \(\frac{1}{2} \mathrm{~kg}\)
2. \(\frac{1}{3} \mathrm{~kg}\)
3. \(\frac{1}{6} \mathrm{~kg}\)
4. \(\frac{1}{12} \mathrm{~kg}\)

Answer: 4. \(\frac{1}{12} \mathrm{~kg}\)

Applying the principle of momentums, LHM = 2 x 20

RHM = 0.5 x 60 + m x 120

2 x 20=30+ 120m

10= 120m

∴ \(m=\frac{10}{120}=\frac{1}{12} \mathrm{~kg}\)

Question 19. Find the torque about the origin, when a force of 3\(\hat{\boldsymbol{j}}\)N acts on the particle whose position vector is \(\hat{k}\)

1. \(6 \hat{j} \mathrm{~N}-\mathrm{m}\)
2. \(-6 \hat{i} \mathrm{~N}-\mathrm{m}\)
3. \(-6 \hat{k} \mathrm{~N}-\mathrm{m}\)
4. \(6 \hat{i} \mathrm{~N}-\mathrm{m}\)

Answer: 2. \(-6 \hat{i} \mathrm{~N}-\mathrm{m}\)

According to question, Position vector \(\hat{r} =2 \hat{k} m\)

Force, \({\mathrm{F}} =3 \hat{j} \mathrm{~N}\)

Torque, \(\vec{\tau} =\vec{r} \times \overrightarrow{\mathrm{F}}=2 \hat{k} \times 3 \hat{j}\)

= \(-6 \hat{i}\)

= \(-6 \hat{i} \mathrm{~N}-\mathrm{m}\)

Question 20. A force F = \(\vec{F}=\alpha \hat{i}+3 \hat{j}+6 \hat{k}\) is acting at a point r = \(\vec{r}=2 \hat{i}-6 \hat{j}-12 \hat{k}\). The value of a for which angular momentum about the origin is conserved is:

1. – 1
2. 2
3. zero
4. 1

Answer: 1. – 1

From the question, Force, \(\vec{F} =\alpha \hat{i}+3 \hat{j}+6 \hat{k}\)

and Position, \(\vec{r}  =2 \hat{i}-6 \hat{j}-12 \hat{k}\)

Using the law of conservation of angular moments \(\tau\) =constant

Question 21. When a mass is rotating in a plane about a fixed point, its angular momentum is directed along :

1. a line perpendicular to the plane of rotation
2. the line at an angle of 45° to the plane of rotation
3. the radius
4. the tangent to the orbit

Answer: 1. a line perpendicular to the plane of rotation

The formula is L = m (r x v). This shows that when a mass is rotating in a plane about a fixed point, its angular momentum is directed along a line perpendicular to the plane of rotation

Question 22. A magnetic needle suspended parallel to a magnetic field requires \(\sqrt{3} \mathrm{~J}\) J of work to turn it through 60°. The torque needed to maintain the needle in the position will be:

1. \(2 \sqrt{3} \mathrm{~J}\)
2. \(3 \mathrm{~J}\)
3. \(\sqrt{3} \mathrm{~J}\)
4. \(\frac{3}{2} \mathrm{~J}\)

Answer: 2. \(3 \mathrm{~J}\)

Work done, W = \(M B \left(\cos \theta_1-\cos \theta_2\right)\)

⇒ \(\sqrt{3} =M B\left(\cos 0^{\circ}-\cos 60^{\circ}\right)\)

⇒ \(\sqrt{3}  =\frac{M B}{2}\)

And Torque, \(\tau =M B \sin \theta\)

⇒ \(\tau =M B \sin 60^{\circ}=\sqrt{3} \frac{M B}{2}\)

From Equations 1 And 2

∴ \(\tau=(\sqrt{3})(\sqrt{3})=3 \mathrm{~J}\)

Question 23. If \(\vec{F}\) is the force acting on a particle having position  vector \(\vec{r} \text { and } \vec{\tau}\) be the torque of this force about the origin, then:

1. \(\vec{r} \cdot \vec{\tau} \neq 0\) and \(\vec{F} \cdot \vec{\tau}\)=0
2. \(\vec{r} \cdot \vec{\tau}>0\) and \(\vec{F} \cdot \vec{\tau}<0\)
3. \(\vec{r} \cdot \vec{\tau}=0\) and \(\vec{F} \cdot \vec{\tau}\)=0
4. \(\vec{r} \cdot \vec{\tau}=0\) and \(\vec{F} \cdot \vec{\tau} \neq 0\)

Answer: 3. \(\vec{r} \cdot \vec{\tau}=0\) and \(\vec{F} \cdot \vec{\tau}\)=0

Torque is an axial vector i.e., its direction is always perpendicular to the plane containing vectors \(\vec{r}\) and \(\vec{F}\).

Therefore, \(\vec{\tau}=\vec{r} \times \vec{F}\)

Torque is perpendicular to both \(\vec{r}\) and \(\vec{F}\).

⇒ \(\vec{\tau} \cdot \vec{r}\)=0

∴ \(\vec{F} \cdot \vec{r}\)=0

Question 24. A particle of mass m moves in the XY plane with a velocity v along the straight line AB. If the angular momentum of the particle concerning origin O is L_A, when it is at A and L_B when it is at B, then:

1. L_A = L_B
2. the relationship between L_A and L_B depends upon the slope of the line AB
3. L_A<L_B
4. L_A>L_B

Answer: 1. L_A = L_B

The moment of Momentum is angular momentum. OP is the same whether the mass is at A or B

⇒ \(L_A=L_B\)

Question 25. Which of the following statements is correct?

1. The centre of mass of a body always coincides with the centre of gravity of the body.

2. The centre of mass of a body is the point at which the total gravitational torque of the body is zero.

3. A couple on a body produces both translation and rotational motion in a body.

4. Mechanical advantage greater than one means that a small effort can be used to lift a large load.

1. (2) and (4)
2. (1) and (2)
3. (2) and (3)
4. (3) and (4)

Answer: 1. (2) and (4)

The Centre of mass may or may not coincide with the centre of gravity.

Question 26. A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is:

1. \(\frac{w x}{d}\)
2. \(\frac{w d}{x}\)
3. \(\frac{w(d-x)}{x}\)
4. \(\frac{w(d-x)}{d}\)

Answer: 4. \(\frac{w(d-x)}{d}\)

So, W=\(N_1+N_2\)  →  Equation  1

For equilibrium, \(N_1 x=N_2(d-\boldsymbol{x})\)

From 1 \(N_1 x =\left(W-N_1\right)(d-x)\)

⇒ \(N_1 x =W_d-W_x-N_1 d+N_1 x\)

⇒ \(N_1 d =W(d-x)\)

∴ \(N_1 =\frac{W(d-x)}{d}\)

Question 27. ABC is an equilateral triangle with O as its centre. F₁, F₂ and F₃ represent three forces acting along the sides AB, BC and AC respectively. If the total torque about O is zero then the magnitude of FT is:

1. \(F_1+F_2\)
2. \(F_1-F_2\)
3. \(\frac{F_1+F_2}{2}\)
4. \(2\left(F_1+F_2\right)\)

Answer: 1. \(F_1+F_2\)

Using Torque, \(\tau=F r\)

Clockwise torque, \(\tau =\tau F_1+\tau F_2+\tau F_3\)

0 =\(F_1 r+F_2 r+F_3 r\)

∴ \(F_3  =F_1+F_2\)

Question 28.

1. The centre of gravity (CG) of a body is the point at which the weight of the body acts.
2. The centre of mass coincides with the centre of gravity if the earth is assumed to have an infinitely large radius.
3. To evaluate the gravitational field intensity due to anybody at an external point, the entire mass of the body can be considered to be concentrated at its CG.
4. The radius of gyration of anybody rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.

Which one of the following pairs of statements is correct?

1. (4) and (1)
2. (1) and (2)
3. (2) and (3)
4. (3) and (4)

Answer: 1. (4) and (1)

The CG of a body is the point at which the weight of the body acts.

The radius of gyration of anybody rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis

Question 29. The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the die about its diameter is ;

1. \(\sqrt{2}: 1\)
2. 4: 1
3. \(1: \sqrt{2}\)
4. 2:1

Answer: 1. \(\sqrt{2}: 1\)

The figure of a thin uniform disc and disc is shown below :

The radius of gyration of a thin uniform disc, k=\(\sqrt{\frac{I}{m}}\)  →   Equation  1

and moment of inertia of the disc, I=\(\frac{m R^2}{4}\)

Here we have R as the radius, and m as the mass Now, on putting the value in equation (1) we have The radius of gyration of the uniform disc;

⇒ \(k_1=\sqrt{\frac{\frac{m R^2}{2}}{m}}\)

⇒ \(k_1=\sqrt{\frac{R^2}{2}}\)  →   Equation  2

and the radius of gyration of the uniform disc;

⇒ \(k_2=\sqrt{\frac{m R^2}{4}}\)

⇒ \(k_2=\sqrt{\frac{R^2}{4}}\)  →  Equation  3

Now, on dividing equation (2) by equation (3) we have;

⇒ \(\frac{k_1}{k_2}  =\frac{\sqrt{\frac{R^2}{2}}}{\sqrt{\frac{R^2}{4}}}\)

∴ \(k_1: k_2 =\sqrt{2}: 1\)

Question 30. From a circular ring of mass ‘M’ and radius ‘R’, an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times ‘MR₂’. Then the value of  K is:

1. \(\frac{3}{4}\)
2. \(\frac{7}{8}\)
3. \(\frac{1}{4}\)
4. \(\frac{1}{8}\)

Answer: 1. \(\frac{3}{4}\)

Given,

Mass = M

Radius = R

⇒ \(\text { arc } \theta  =90^{\circ}\)

⇒ \(\mathrm{I}_{\text {remaining }} =\mathrm{KMR}^2\)

⇒ \(\mathrm{~K}\) =?

⇒ \(\mathrm{I}  =r^2 \int d m\)

⇒ \(\frac{d m}{d r} =\frac{\mathrm{M}}{2 \pi \mathrm{R}} d r \)

d m =\(\frac{\mathrm{M}}{2 \pi \mathrm{R}} d r\)

⇒ \(\mathrm{I} =r^2 \int_0^{\frac{3}{2} \pi \mathrm{R}} \frac{M}{2 \pi \mathrm{R}} d r\)

⇒ \(\mathrm{I} =\frac{\mathrm{MR}^2}{2 \pi \mathrm{R}} \cdot \frac{3}{2} \pi \mathrm{R}\)

=\(\frac{3}{4} \mathrm{MR}^2\)

∴ \(\mathrm{~K} =\frac{3}{4}\)

Question 31. A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation (E_sphere/E_Cylinder) will be:

1. 2 : 3
2. 1: 5
3. 1: 4
4. 3: 1

Answer: 2. 1: 5

We know That,KE=\(\frac{1}{2} I \omega^2\)

KE of sphere,\(K_S =\frac{1}{2} I_s \omega^2\)  Equation  1

= \(\frac{1}{2} \times \frac{2}{5} M R^2 \times \omega^2\)

In KE of cylinder,\(\mathrm{K}_{\mathrm{C}} =\frac{1}{2} I_e \omega^2\)   Equation  2

= \(\frac{1}{2} \times \frac{M R^2}{2} \times 4 \omega^2 \).

From (1) and (2),

∴ \(\frac{K_S}{K_C}=\frac{1}{5}\)

Question 32. A light rod of length / has two masses m₁ and m₂ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is:

1. \(\frac{m_1 m_2}{m_1+m_2} l^2\)
2. \(\frac{m_1+m_2}{m_1 m_2} l^2\)
3. \(\left(m_1+m_2\right)^2\)
4. \(\sqrt{m_1 m_2 l^2}\)

Answer: 1. \(\frac{m_1 m_2}{m_1+m_2} l^2\)

centre of mass of \(m_1 and m_2\) lies at

r =\(\frac{m_1 r_1+m_2 r_2}{m_1+m_2}\)

⇒ \(r_1 =\frac{m_2 l}{m_1+m_2}, r_2=\frac{m_1 l}{m_1+m_2}\)

\(I_{C M} =m_1 r_1^2+m_2 r_2^2 \)

=\(m_1\left[\frac{m_2 l}{m_1+m_2}\right]^2+m_2\left[\frac{m_1 l}{m_1+m_2}\right]^2\)

=\(\frac{m_1 m_2 l^2}{\left(m_1+m_2\right)^2}\left(m_1+m_2\right)\)

=\(\frac{m_1 m_2 l^2}{m_1+m_2}\)

Question 33. Two spherical bodies of masses M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before the collision is:

1. 2.5 R
2. 4.5 R
3. 7.5/2
4. 1.5 R

Answer: 3. 7.5/2

Mass of one spherical body = M its radii = R

Mass of second body = 5M,

Its radii = 2R

Then we have, M x =5 M(9 R-x)

x =45 R-5 x

x =\(\frac{45 R}{6}=7.5 \mathrm{R}\)

Question 34. From a circular disc of radius R and mass 9m, a small disc of mass M and radius \(\frac{R}{3}\)is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is:

1. \(\frac{40}{9} M R^2\)
2. \(M R^2\)
3. \(4 M R^2\)
4. \(\frac{4}{9} M R^2\)

Answer: 1. \(\frac{40}{9} M R^2\)

The Moment of Inertia of a disc about an axis perpendicular to the plane and passing through the centre,

I=\(I_1-I_2 \)

I=\(\frac{9 M R^2}{2}-\frac{M R^2}{18}\)

I=\(\frac{81 M R^2-M R^2}{18}\)

I=\(\frac{40 M R^2}{9}\)

Question 35. Three identical rings of mass m and radius r are placed as shown in the figure. Find the moment of inertia of the system about axis XX’.

1. \(\frac{7}{2} m r^2\)
2. \(\frac{5}{2} m r^2\)
3. \(\frac{3}{2} m r^2\)
4. \(\frac{1}{2} m r^2\)

Answer: 1. \(\frac{7}{2} m r^2\)

Moment of Inertia Of System About XX’

I=\(I_1+I_2+I_3\)

Here, \(I_1\)= M.I. of ring about its diameter \({1}{2} m r^2 \)

⇒ \(I_2=\frac{1}{2} m r^2+m r^2=\frac{3}{2} m r^2\)

Similarly, \(I_3=\frac{3}{2} m r^2\)

I =\(\frac{1}{2} m r^2+\frac{3}{2} m r^2+\frac{3}{2} m r^2\)

=\({7}{2} m r^2\)

Question 36. The ratio of the radii of gyration of a circular disc to that of a circular ring, each of the same mass and radius around their respective axes is:

1. \(\sqrt{3}: \sqrt{2}\)
2. \(1: \sqrt{2}\)
3. \(\sqrt{2}: 1\)
4. \(\sqrt{2}: \sqrt{3}\)

Answer: 2. \(1: \sqrt{2}\)

We know that, Radius of gyration is K=\(\sqrt{\frac{I}{M}}\)

Using This Formula We Have, \(M K_1 =M R^2\)

⇒ \(K_1\) =R

And \( M K_2^2 =\frac{M R^2}{2}\)

⇒ \( K_2  =\frac{R}{\sqrt{2}}\)

∴ \(\frac{K_2}{K_1}  =\frac{R}{\sqrt{2} R}=\frac{1}{\sqrt{2}}\)

Question 37. Two bodies have their moments of inertia 1 and 2  respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio:

1. 2: 1
2. 1: 2
3. \(\sqrt{2}: 1\)
4. \(1: \sqrt{2}\)

Answer: 4. \(1: \sqrt{2}\)

K.E. rotation,K=\(\frac{1}{2} I \omega^2\)

Angular momenta, L=100

K=\(\frac{L^2}{2}\)

Here, \(K_1=K_2, I_1=I_1, I_2=2 \mathrm{I}\)

⇒ \(\frac{L_1}{L_2}=\sqrt{\frac{2 I_1 K_1}{2 I_2 K_2}}\)

=\(\sqrt{\frac{I_1}{I_2}}=\sqrt{\frac{I}{2 I}}\)

=\(\frac{1}{\sqrt{2}}\)

Question 38. Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC side/cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram cm² units will be:

1. \(\frac{3}{4} m l^2\)
2. \(2 m l^2\)
3. \(\frac{5}{4} m l^2\)
4. \(\frac{3}{2} m l^2\)

Answer: 3. \(\frac{5}{4} m l^2\)

According to the question

⇒ \(I_{A X} =m l^2+m\left(\frac{l}{2}\right)^2\)

=\(m l^2+\frac{m l^2}{4}\)

=\(\frac{5}{4} m l^2\)

Question 39. A circular disc is to be made by using iron and aluminium so that it acquires a moment of inertia about the geometrical axis. It is possible with:

1. aluminium at the interior and iron surrounding it
2. iron at the interior and aluminium surrounding it
3. using iron and aluminium lives in an alternate order
4. sheet of iron is used at the external surface and aluminium sheet as internal layers

Answer: 1. aluminium in the interior and iron surrounding it

The density of Iron is more than Aluminum.

Question 40. The moment of inertia of a disc of mass M and radius R about a tangent to its rim in its plane is :

1. \(\frac{2}{3} M R^2\)
2. \(\frac{3}{2} M R^2\)
3. \(\frac{4}{5} M R^2\)
4. \(\frac{5}{4} M R^2\)

Answer: 4. \(\frac{5}{4} M R^2\)

Moment of inertia of a disc about its diameter

Now, according to the perpendicular axis theorem, a moment of inertia of a disc about a tangent passing through rim in the plane of the disc,

I= \(I_d+M R^2\)

I= \(\frac{1}{4} M R^2+M R^2\)

I= \(\frac{5}{4} M R^2\)

Question 41. From a disc of radius R and mass M, a circular hose of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the Centre?

1. \(\frac{13 \mathrm{MR}^2}{32}\)
2. \(\frac{11 \mathrm{MR}^2}{32}\)
3. \(\frac{9 \mathrm{MR}^2}{32}\)
4. \(\frac{15 \mathrm{MR}^2}{32}\)

Answer: 1. \(\frac{13 \mathrm{MR}^2}{32}\)

This Diagram is

Moment of Inertia of disc,

I =\(I_1+I_2+I_3\)

⇒ \(I_1 =\frac{2}{3} m r_2\)

⇒ \(I_2 =I_3=\frac{2}{3} m r^2+m r^2\)

(using parallel axis theorem)

⇒ \(I_3=I_2 =\frac{5}{3} m r^2\)

I =\(\frac{2}{3} m r^2+2 \times \frac{5}{3} m r^2\)

= \(m r^2\left(\frac{2}{3}+\frac{10}{3}\right)=4 m r^2\)

Question 42. Three identical spherical shells, each of mass m and radius r are placed as shown in the figure. Consider an axis XX, which is touching two shells and passing through the diameter of the third shell: The moment of inertia of the system consisting of these three spherical shells about XX’s axis is:

1. \(\frac{11}{5} m r^2\)
2. \(3 m r^2\)
3. \(\frac{16}{5} m r^2\)
4. \(4 m r^2\)

Answer: 4. \(4 m r^2\)

Let I₁, I₂ and I₃ be the Movement of Inertia of three spheres

So, the Moment of Inertia of the spheres about the axis passing through xx’ is,

I=\(I_1+I_2+I_3\)   Equation 1

⇒ \(I_1=\frac{2}{3} m r_2\)

⇒ \(I_2=I_3=\frac{2}{3} m r^2+m r^2\) (using parallel axis theorem)

⇒ \(I_3=I_2 =\frac{5}{3} m r^2\)

⇒ \(I =\frac{2}{3} m r^2+2 \times \frac{5}{3} m r^2\)

=\(m r^2\left(\frac{2}{3}+\frac{10}{3}\right)=4 m r^2\)

Question 43. The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through:

1. 2
2. 3
3. 4
4. 1

Answer: 1. 2

According to the parallel axis theorem,\(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+m d^2\) d is maximum to B

Question 44. The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its mid¬point and perpendicular to its length is \(l_0\)– Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is:

1. \(l_0+M L^2 / 4\)
2. \(l_0+2 M L^2\)
3. \(l_0+M L^2\)
4. \(l_0+M L^2 / 2\)

Answer: 1. \(l_0+M L^2 / 4\)

Moment of inertia about an axis passing t}rough one end

⇒ \(\mathrm{I}=I_{C M}+m d^2 \)

∴ \(\mathrm{I}=I_0+M\left(\frac{L}{2}\right)^2=I_0+\frac{M L^2}{4}\)

Question 45. Four identical thin rods each of mass M and length l, form a square frame. The moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is:

1. \(\frac{4}{3} \mathrm{Ml}^2\)
2. \(\frac{2}{3} \mathrm{Ml}^2\)
3. \(\frac{13}{3} \mathrm{Ml}^2\)
4. \(\frac{1}{3} \mathrm{Ml}^2\)

Answer: 1. \(\frac{4}{3} \mathrm{Ml}^2\)

From the parallel axis theorem,

Moment of Inertia =\(\frac{M l^2}{2}+M\left(\frac{1}{2}\right)^2=\frac{M l^2}{3}\)

Moment of Inertia of the system =\(\frac{M l^2}{3} \times 4\)

= \(\frac{4}{3} M l^2\)

Question 46. A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is 90°. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is:

1. \(\frac{M L^2}{24}\)
2. \(\frac{M L^2}{12}\)
3. \(\frac{M L^2}{6}\)
4. \(\frac{\sqrt{2} M L^2}{26}\)

Answer: 3. \(\frac{M L^2}{6}\)

Moment of Inertia of the system, I =\(\frac{M(L / 2)^2}{3}+\frac{M(L / 2)^2}{3}\)

=\(\frac{M L^2}{12}+\frac{M L^2}{12}\)

=\(\frac{M L^2}{6}\)

Question 47. The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is:

1. \(M R^2\)
2. \(\frac{1}{2} M R^2\)
3. \(\frac{3}{2} M R^2\)
4. \(\frac{7}{2} M R^2\)

Answer: 3. \(\frac{3}{2} M R^2\)

Moment of Inertia disc about its normal =\(\frac{1}{2} M R^2\)

Moment of Inertia about its one edge =\(M R^2+\frac{M R^2}{2}\)

Moment of Inertia = \(\frac{3}{2} M R^2\)

Question 48. The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is:

1. 2: 3
2. 2: 1
3. \(\sqrt{5}: \sqrt{6}\)
4. \(1: \sqrt{2}\)

Answer: 3. \(\sqrt{5}: \sqrt{6}\)

The radius of gyration of the disc about a tangential axis in the plane of the disc is \(k_1\),

⇒ \(k_1=\frac{5}{4} M R^2=\frac{\sqrt{5}}{2} R\)

(From parallel Axis Theorem) And radius of gyration of a circular ring of the same radius about a tangential axis is given by:

⇒ \(k_2=\frac{3}{2} M R^2=\frac{\sqrt{3}}{\sqrt{2}} R\)

(From parallel Axis Theorem)

⇒ \(\frac{k_1}{k_2} =\frac{\sqrt{5}}{2} R / \frac{\sqrt{3}}{\sqrt{2}} R\)

=\(\frac{\sqrt{5}}{2} \times \frac{\sqrt{2}}{\sqrt{3}}\)

=\(\frac{\sqrt{5}}{\sqrt{6}}\)

Question 49. The angular speed of the wheel of a vehicle is increased from 360 rpm to 1200 rpm in 14 s. Its angular acceleration is :

1. \(2 \pi \mathrm{rad} / \mathrm{s}^2\)
2. \(28 \pi \mathrm{rad} / \mathrm{s}^2\)
3. \(120 \pi \mathrm{rad} / \mathrm{s}^2\)
4. \(1 \mathrm{rad} / \mathrm{s}^2\)

Answer: 1. \(2 \pi \mathrm{rad} / \mathrm{s}^2\)

Initial angular velocity of the wheel,

⇒ \(\omega_i =2 \pi f_o=2 \pi \times \frac{360}{60} \mathrm{rad} / \mathrm{s}\)

=\(12 \pi \mathrm{rad} / \mathrm{s}\)

final angular velocity of wheel, \(\omega_f  =2 \pi f=2 \pi \times \frac{1200}{60} \mathrm{rad} / \mathrm{s}\)

=\(40 \pi \mathrm{rad} / \mathrm{s}\)

⇒ \(\Delta t=14 \mathrm{~s}\)

From the equation of rotational motion

⇒ \(\omega_f  =\omega_i+\alpha \Delta t\)

⇒ \(\alpha  =\frac{\omega_f-\omega_i}{\Delta t}\)

=\( 2 \pi \mathrm{rad} / \mathrm{s}\)

Question 50. A wheel has an angular acceleration of 3.0 rad/sec² and an initial angular speed of 2.00 rad sec. In a time of 2 seconds it has rotated through an angle (in radians) of:

1. 10
2. 12
3. 4
4. 6

Answer: 1. 10

⇒ \(\theta =\omega_0 t+\frac{1}{2} \alpha^2\)

=\((2)(2)+\frac{1}{2}(3)(2)^2\)

=4+6=10

Question 51. A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after the 2π revolution is :

1. \(2 \times 10^{-3} \mathrm{~N}-\mathrm{m}\)
2. \(2 \times 10^{-4} \mathrm{~N}-\mathrm{m}\)
3. \(2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
4. \(2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)

Answer: 4. \(2 \times 10^{-3} \mathrm{~N}-\mathrm{m}\)

Mass, M=2 kg, Radius R=4 cm

Initial angular speed, \(\omega_0=3 \times \frac{2 \pi}{60} \mathrm{rad} / \mathrm{s}\)

=\(\frac{\pi}{10} \mathrm{rad} / \mathrm{s}\)

We know that, \(\omega_2  =\omega_0^2+2 \alpha \theta\)

0 =\(\left(\frac{\pi}{10}\right)^2+2 \cdot \alpha \times 2 \pi \times 2 \pi\)

⇒ \(\alpha  =-\frac{1}{800} \mathrm{rad} / \mathrm{s}^2\)

Now moment of inertia of a solid cylinder,

I=\(\frac{M R^2}{2}=\frac{2 \times\left(\frac{4}{100}\right)^2}{2}=\frac{16}{10^4}\)

Now we know that, Torque,

⇒ \(\tau =\mathrm{I} \times \alpha=\frac{16}{10^4} \times\left(-\frac{1}{800}\right)\)

=\(-2 \times 10^{-6}(\mathrm{Nm})\) .

Question 52. Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed co about their symmetry axis. The amount of work (W) required to bring them to rest, would satisfy the relation :

1. \(W_B>W_A>W_C\)
2. \(W_A>W_B>W_C\)
3. \(W_C>W_B>W_A\)
4. \(W_A>W_C>W_B\)

Answer: 3. \(W_C>W_B>W_A\)

Work done required to bring them rest,

⇒ \(\Delta W  =\Delta K E\)

⇒ \(W_A  : W_B: W_C \)

=\(\frac{2}{5} M R^2: \frac{1}{2} M R^2: M R^2 \)

=\(\frac{2}{5}: \frac{1}{2}: 1\)

=4: 5: 10

∴ \(W_C >W_B>W_A\)

Question 53. A rope is around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder, if the rope is pulled with a force of 30 N?

1. 25 m/s²
2. 0.25 rad/s²
3. 25 rad/s²
4. 5 m/s²

Answer: 3. 25 rad/s²

Torque on the cylinder due to force F  \(\tau=F R\)

and we know that, \(\tau=\mathrm{I} \alpha\)

Where, I= Moment of Inertia of cylinder about axis = \(M R^2\)

and \(\alpha\)= angular acceleration on comparing both the equations,

⇒ \(\alpha=\frac{\tau}{I}=\frac{F R}{M R^2}=\frac{F}{M R}\)

∴ \(\alpha=\frac{30}{3 \times 40 \times 10^{-2}}=25 \mathrm{rad} / \mathrm{s}^2\)

Question 54. A uniform circular disc of radius 50 cm is at rest and is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s^-2. Its net acceleration in ms^-2 at the end of 2.0 is approximately:

1. 7.0
2. 6.0
3. 3.0
4. 8.0

Answer: 4. 8.0

According to the question,

Angular acceleration, \(\alpha=2 \mathrm{rad} \mathrm{s}^{-2}\)

Angular speed, \(\omega=\alpha t=4 \mathrm{rad} \mathrm{s}^{-1}\)

because centripetal acceleration at the end of 2.0

⇒ \(a_t=\alpha r=2 \times 0.5=1 \mathrm{~m} / \mathrm{s}^2\)

Net acceleration,\(\alpha=\sqrt{a_c^2+a_t^2}\)

=\(\sqrt{(8)^2+(1)^1}\)

=\(\sqrt{65} \approx 8 \mathrm{~m} / \mathrm{s}^2\)

Question 55. Point masses m₁ and m₂ are placed at the opposite ends of a rigid rod of length L and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass, so that the work required to set the rod rotating with angular velocity COQ is minimum, is given by:

1. \(x=\frac{m_1 \mathrm{~L}}{m_1+m_2}\)
2. \(x=\frac{m_1}{m_2} \mathrm{~L}\)
3. \(x=\frac{m_2}{m_1} \mathrm{~L}\)
4. \(x=\frac{m_2 \mathrm{~L}}{m_1+m_2}\)

Answer: 4. \(x=\frac{m_2 \mathrm{~L}}{m_1+m_2}\)

Here m₁ and m₂ masses are placed at opposite ends of the rigid rod whose length is L

Moment of Inertia of rod, \(\mathrm{I}=m_1 x^2+m_2(L-x)^2\)

=\(m_1 x^2+m_2 L^2+m_2 x^2-2 m_2 L^2\)

⇒ \(\mathrm{I}_a is\) minima then,

⇒ \(\frac{d \mathrm{I}}{d x}=2 m_1 x+0+2 x m_2-m_2 L\) =0

⇒ \(x\left(2 m_1+2 m_2\right)=2 m_2 L\)

x=\(\frac{m_2 L}{m_1+m_2}\)

Minimum work = Minimum rotational K.E.

Maximum angular moments = Minimum moment of inertia its rotation should be about CM

∴ x=\(\frac{m_2 L}{m_1+m_2}\)

Question 56. An automobile moves on a road with a speed of 50 km hr^-2. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m². If the vehicle is brought to rest in 15 s, the magnitude to average torque transmitted by its breaks to the wheel is:

1. 6.66 kg m2 s^-2
2. 8.58 kg m2 s^-2
3. 10.86 kg m2s^-2
4. 2.86 kg m2s^-2

Answer: 1. 6.66 kg m2 s^-2

According to question ,Velocity of automobile vehicle is v=\(54 \mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18}\)

And angular velocity, v=\(\omega_0 r\)

⇒ \(\omega_0=\frac{v}{R}=\frac{15}{0.45}=\frac{100}{3} \mathrm{rad} / \mathrm{s}\)

and then angular acceleration is \(\alpha  =\frac{\Delta \omega}{t}=\frac{\omega t-\omega_0}{t}\)

=\(\frac{0-\frac{100}{3}}{15}=-\frac{100}{15}\)

=\(-\frac{100}{15} \mathrm{rad} / \mathrm{s}^2\)

Torque, \(\tau =\mathrm{I} \alpha\)

=\(3 \times \frac{100}{45}=6.66 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}^2\)

In Short, \(\tau=\mathrm{I} \frac{v}{t r}=\frac{3 \times 54 \times \frac{5}{18}}{15 \times 0.45}\)

∴ \(\tau=6.66 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}^2\)

Question 57. A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound around the cylinder with one end attached to it and the other hanging freely. Tension in the string required to produce an angular acceleration of 2 rev/s² is:

1. 25 N
2. 50 N
3. 78.5 N
4. 157 N

Answer: 4. 157 N

According to the question, the Mass of the cylinder, M= 50 kg.

The radius of the cylinder,

R = 0.5 m

Angular acceleration = \(\alpha=2 \mathrm{rev} / \mathrm{s}^2\)

We know that torque, \(\tau=\mathrm{I} \alpha\)  →   Equation   1

Now torque produced in the string is \(\alpha=T \times R=T \times 0.5=\frac{T}{2}\) →   Equation   2

From equation 1 And 2

α = \(\frac{T}{2}\)

⇒ \(\frac{T}{2} =\left(\frac{M R^2}{2}\right) \times 2 \times 2 \pi\)

I =\(\frac{M R^2}{2}\)

⇒ \(\frac{T}{2} =\frac{50 \times(0.5)^2}{2} \times 4 \pi\)

I = \(50 \times \frac{1}{4} \times 4 \pi=50 \pi\)

=157 \(\mathrm{~N}\)

Question 58. A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When a string is cut, the initial angular acceleration of the rod is:

1. \(\frac{3 g}{2 \mathrm{~L}}\)
2. \(\frac{g}{\mathrm{~L}}\)
3. \(\frac{2 g}{\mathrm{~L}}\)
4. \(\frac{2 g}{3 \mathrm{~L}}\)

Answer: 1. \(\frac{3 g}{2 \mathrm{~L}}\)

Torque on the rod = Moment of the rod P

T= \(m g_{\frac{1}{2}}\)    →   Equation 1

The moment of inertia of the rod about P,

P= \(\frac{m L^2}{3}\)  →   Equation  2

From equation. (1) and (2), Since \(\mathrm{T}=\mathrm{I}\)α

⇒ \(m g \frac{1}{2} =\frac{m \mathrm{~L}^{\not}}{3} \alpha\)

∴ \(\alpha =\frac{3 g}{2 L}\)

Question 59. The instantaneous angular position of a point on a rotating wheel is given by the equation, Q(t) = 2/3 – 6/3. The torque on the wheel becomes zero at:

1. t = 0.5 x
2. t = 0.25 x
3. t = 2s
4. t = 1s

Answer: 4. t = 1 ,v

Given \(\theta(t)=2 t^3 -6 t^2\)

⇒ \(\frac{d \theta}{d t} =6 t^2-12 t\)

⇒ \(\frac{d^2 \theta}{d t^2}\) =12 t-12

It is given that, torque \( \tau\)=0 means \(\alpha \)=0 and

⇒ \(\alpha =\frac{d^2 \theta}{d t^2}\)=0

12 t-12 =0

t =1 sec

Question 60. A uniform rod AB of length L and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of. the rod about A is mP²/3, the initial angular acceleration of the rod will be :

1. \(\frac{m g l}{2}\)
2. \(\frac{3}{2} g l\)
3. \(\frac{3 g}{2 l}\)
4. \(\frac{2 g}{3 l}\)

Answer: 1. \(\frac{m g l}{2}\)

Torque About A,

⇒ \(\tau=m g \times \frac{l}{2}=\frac{m g l}{2}\)

⇒ \(\tau=I \alpha\)

∴ Also, Angular Acceleration  \(\alpha=m g \times \frac{l}{2}=\frac{m g l}{2}\)

Question 61. A wheel having moment of inertia 2 kg m² about its vertical axis, rotates at the rate of 60 rpm about the axis. The torque which can stop the wheel’s rotation in one minute would be:

1. \(\frac{2 \pi}{15} \mathrm{~N}-\mathrm{m}\)
2. \(\frac{\pi}{12} \mathrm{~N}-\mathrm{m}\)
3. \(\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)
4. \(\frac{\pi}{18} \mathrm{~N}-\mathrm{m}\)

Answer: 3. \(\frac{\pi}{12} \mathrm{~N}-\mathrm{m}\)

Given, \(\mathrm{I}  =2 \mathrm{~kg} m^2,\)

⇒ \(\omega_0 =\frac{60}{60} \times 2 \pi \mathrm{rad} / \mathrm{s}\)

t =60s

The torque required to stop the wheel’s rotation is,

⇒ \(\tau =\mathrm{I} \alpha l\left(\frac{\omega_0-\omega}{t}\right)\)

⇒ \(\tau  =\frac{2 \times 2 \pi \times 60}{60 \times 60}\)

=\(\frac{\pi}{15} \mathrm{~N}-\mathrm{m}\)

Question 62. A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass the same. Which of the following physical quantities would remain constant for the sphere?

1. Rotational kinetic energy
2. Moment of inertia
3. Angular velocity
4. Angular momentum

Answer: 4. Angular momentum

From question,

External torque, \(\tau\) =0

⇒ \(\frac{d \mathrm{~L}}{d t}\) =0

L = Constant

So, angular momentum remains constant.

Question 63. Two discs of the same moment of inertia rotate about their regular axis passing through the centre and perpendicular to the plane of the disc with angular velocities ω₁ and ω₂. They are brought into contact face to face coinciding with the axis of rotation. The expression for loss of energy during this process is:

1. \(\frac{1}{2}\left(\omega_1+\omega_2\right)^2\)
2. \(\frac{1}{4}\left(\omega_1-\omega_2\right)^2\)
3. \(l\left(\omega_1-\omega_2\right)^2\)
4. \(\frac{1}{8}\left(\omega_1+\omega_2\right)^2\)

Answer: 2. \(\frac{1}{4}\left(\omega_1-\omega_2\right)^2\)

Using the law of conservation of angular momentum,

Angular moments before contact = \(I_1 \omega_1+I_2 \omega_2\)

Angular momentum after contact = \(I_\omega\)

⇒ \(\left(I_1+I_2\right) \omega\)

⇒ \(\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}\)

Loss of energy =\(\frac{1}{2} I \omega_1^2+\frac{1}{2} I \omega_2^2-\frac{1}{2}(2 I) \omega^2\)

= \(\frac{1}{4} I\left(\omega_1-\omega_2\right)^2\)

Question 64. The rotating bodies A and B of masses m and 2m with moments of inertia and I_B(I_B > I_A) have equal kinetic energy of rotation. If L_A and L_B are their angular momenta respectively, then:

1. \(L_A=\frac{L_B}{2}\)
2. \(L_A=2 L_B\)
3. \(L_B>L_A\)
4. \(L_A>L_B\)

Answer: 3. \(L_B>L_A\)

The K.E. of a rotating body is, \(K E=\frac{1}{2} I \omega^2=\frac{1}{2} \frac{I^2 \omega^2}{I}=\frac{L^2}{2 I}\)

⇒ \((Since L=I \omega )\)

From question \(K_A=K_B\)

⇒ \(\frac{L_A^2}{2 I_A}=\frac{L_B^2}{2 I_B}\)

As, \(I_B>I_A\)

So,\(L_A^2<L_B^2\)

∴ \(L_A<L_B\)

Question 65. The 2 discs are rotating about their axis, normal to the plane of the discs and passing through the centre of the discs. Disc D, has a 2 kg mass and 0.2 m radius and an initial angular velocity of 50 rad s^-1. Disc D₁ has a 4 kg mass, 0.1 m radius and an initial angular velocity of 200 rad s^-1. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in rad,s^-1) of the system is:

1. 60
2. 100
3. 120
4. 40

Answer: 2. 100

Moment of Inertia of\(I_1=\frac{\mathrm{MR}^2}{2}=\frac{2(0.2)^2}{2}=0.04\) kg-m² initial angular velocity of \(I_2=\frac{4 \times(0.1)^2}{2}\)rad S^-1 M.J of D² is \(\omega_2=200\)= 0.02 kg m² initial angular velocity of D² fs co² = 200 and S^-1.

According to the law of conservation of angular momentum \(L_1 =L_2\)

⇒ \(\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2 =\left(\mathrm{I}_1+\mathrm{I}_2\right)\)

⇒ \(\omega =\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}\)

⇒ \(\omega =\frac{(0.04)^2 \times 50+(0.02)^2 \times 200}{0.04+0.02}\)

=\(\frac{2+4}{0.06}=100 \mathrm{rads}^{-1}\)

Question 66. A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity. If two objects each of mass m are attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

1. \(\frac{\omega(W-2 m)}{M+2 m}\)
2. \(\frac{\omega M}{M+2 m}\)
3. \(\frac{\omega(M+2 m)}{M}\)
4. \(\frac{\omega M}{M+m}\)

Answer: 2. \(\frac{\omega M}{M+2 m}\)

Applying the law of conservation of angular momentum 7icoi = I2to2

It is given that: I = MR₂

⇒ \(I_2 =M R^2+2 m R^2\)

⇒ \(\omega_1 =\omega_2\)

⇒ \(\omega_2 =\frac{I_1}{I_2} \omega\)

=\(\frac{M}{M+2 m} \omega\)

Question 67. A circular disc of the moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed to, Another disk of moment of inertia, is dropped coaxially onto the rotating disk. Initially, the second disk has zero angular speed. Eventually, both the disks rotate with a constant angular speed toy. The energy lost by the initially rotating disc due to friction is:

1. \(\frac{1}{2} \frac{I_b^2}{\left(I_t+I_b\right)} \omega_i^2\)
2. \(\frac{1}{2} \frac{I_t^2}{\left(I_t+I_b\right)} \omega_i^2\)
3. \(\frac{1}{2} \frac{I_b-I_t}{\left(I_t+I_b\right)} \omega_i^2\)
4. \(\frac{1}{2} \frac{I_b I_t}{\left(I_t+I_b\right)} \omega_i^2\)

Answer: 4. \(\frac{1}{2} \frac{I_b I_t}{\left(I_t+I_b\right)} \omega_i^2\)

Loss of energy is \(\Delta\)E

⇒ \(\Delta E =\Delta E_i-\Delta E_f \)

=\(\frac{1}{2} I_t \omega_i^2-\frac{1}{2} \frac{I_i^2 \omega_i^2}{\left(r_e+I_B\right)}\)

=\(\frac{1}{2} \frac{I_b I_t \omega_i^2}{\left(I_t+I_b\right)}\)

Question 68. A round disc of the moment of inertia/2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of the moment of inertia I rotating with an angular velocity ω about the same axis. The final angular velocity of the combination of discs is:

1. \(\frac{I_2 \omega_1}{I_2+I_2}\)
2. \(\frac{\omega\left(I_1+I_2\right)}{I_2}\)
3. \(\frac{I_1 \omega_1}{I_1+I_2}\)
4. \(\frac{\left(I_1+I_2\right) \omega_1}{I_1}\)

Answer: 3. \(\frac{I_1 \omega_1}{I_1+I_2}\)

From the law of conservation of angular momentum we have :

⇒ \(I_1 \omega_1 \equiv\left(I_1+I_2\right) \omega_2\)

∴ \(\omega_2 =\frac{I_1 \omega_1}{\left(I_1+I_2\right)}=\frac{I_1 \omega}{I_1+I_2}\)

Question 69. A thin circular ring of Mass M and radius r is rotating about its axis with a constant angular velocity co. Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:

1. \(\frac{M \omega}{4 m}\)
2. \(\frac{M \omega}{M+4 m}\)
3. \(\frac{(M+4 m) \omega}{M}\)
4. \(\frac{(M-4 m) \omega}{M+4 m}\)

Answer: 2. \(\frac{M \omega}{M+4 m}\)

According to the law of conservation of angular momentum

⇒ \(M r^2 \omega =\left(M r^2+4 m r^2\right) \omega^{\prime}\)

∴ \(\omega^{\prime} =\frac{M \omega}{M+4 m}\)

Question 70. A disc is rotating with angular speed co. If a child sits on it, what is conserved?

1. Linear momentum
2. Angular momentum
3. Kinetic energy
4. Potential energy

Answer: 2. Angular momentum

\(\tau =\frac{d L}{d t}, \text { so }\)=0

L =\(\frac{d L}{d l}\)constant

Law of conservation of angular momentum.

Question 71. At any instant, a rolling body may be considered to be in pure rotation about an axis through the point of contact. This axis is translating forward with speed :

1. equal to the centre of mass
2. zero
3. twice of centre of mass
4. None of the above

Answer: 1. equal to the centre of mass

Since the instantaneous axis of rotation is always below the centre of mass in this case. This is only possible if the point of contact moves at the same speed as the centre of mass

Question 72. A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle inclination 30°. The centre of mass of the cylinder has a speed of 4 m/s. The distance travelled by the cylinder on the inclined surface will be : (Take g = 10 m/s²)

1. 2.2 m
2. 1.6 m
3. 1.2 m
4. 2.4 m

Answer: 4. 2.4 m

Since, v= \(\sqrt{\frac{2 g h}{1+\frac{\mathrm{K}^2}{\mathrm{R}^2}}}\)

⇒ \(v^2= \frac{2 g h}{1+\frac{1}{2}}\)

⇒ \(\left[\frac{\mathrm{K}^2}{\mathrm{R}^2}=\frac{1}{2} \text { for solid cylinder }\right]\)

2 g h =\(4^2 \times \frac{3}{2}\)

h =\(\frac{12}{10}=1.2 \mathrm{~m}\)

∴ Now, X =\(\frac{h}{\sin 30^{\circ}}=\frac{1.2}{\frac{1}{2}}=2.4 \mathrm{~m}\)

Question 73. A solid sphere is in a rolling motion. In rolling motion, a body possesses translational kinetic energy (K,) as well as rotational kinetic energy (Kr) simultaneously. The ratio K,: (K, + Kr) for the sphere is :

1. 10: 7
2. 5: 7
3. 7: 10
4. 2: 5

Answer: 3. 7: 10

The translational kinetic energy of the rolling body,

Total kinetic energy = Rotational KE + Translational KE = latex]k_t+k_r[/latex]

(Solid sphere I=\(\frac{2}{5} M R^2)\)

= \(k_t+k_r\)

=\( \frac{1}{2}\left(\frac{2}{5} M R^2\right) \omega^2+\frac{1}{2} m v^2\)

=\( \frac{1}{2}\left(\frac{2}{5} M R^2\right)\left(\frac{V}{R}\right)^2+\frac{1}{2} m v^2\)

=\( \frac{1}{5} m v^2+\frac{1}{2} m v^2=\frac{7}{10} m v^2\)

Ratio = \(\frac{k_t}{k_t+k_2}=\frac{\frac{1}{2} m v^2}{\frac{7}{10} m v^2}\)

=\(\frac{1}{2} \times \frac{10}{7}=\frac{5}{7}\)=5: 7

Question 74. A disc and a sphere of the same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?

1. Both reach at the same time
2. Depends on their masses
3. Disc
4. Sphere

Answer: 4. Sphere

The time taken by the body to reach the bottom when it rolls down on an inclined plane without slipping is given by

⇒ \(t=\sqrt{\frac{2 l\left(1+\frac{k^2}{R^2}\right)}{g \sin \theta}}\)

Question 75. The ratio of the acceleration for a solid sphere (mass m and radius R) rolling down an incline of angle θ without slipping and slipping down the incline without rolling is :

1. 5: 7
2. 2 : 3
3. 2: 5
4. 7: 5

Answer: 1. 5: 7

In this question, there are two cases. First when a solid sphere rolls without slipping down an inclined plane and second a sphere slips down in the inclined plane.

First Case

\(a_1 =\frac{g \sin \theta}{1+\frac{k^2}{\mathrm{R}^2}}\)

=\(\frac{g \sin \theta}{1+\frac{(215) R^2}{R^2}}\) →  Equation  1

⇒ \(a_1 =\frac{5 g \theta}{7}-6\) (Since, solid sphere \(k^2=\frac{2}{5} R^2)\)

Second case

\(a^2=g \sin \theta\)  →   Equation  2

From (1) and (2),

⇒ \(\frac{a_1}{a_2}=\frac{\frac{5}{7} g \sin \theta}{g \sin \theta}=\frac{5}{7}\)

∴ \(\frac{a_1}{a_2}=\frac{5}{7}\)

Question 76. A small object of uniform density rolls up a curved surface with an initial velocity of V. It reaches up to a maximum height of\(\frac{3 v^2}{4 g}\) concerning the initial position. The object is:

1. hollow sphere
2. disc
3. ring
4. solid sphere

Answer: 2. disc

The Kinetic energy of the rolling object is converted into potential energy at height

So by the law of conservation of mechanical energy, we have

⇒ \(\frac{1}{2} M v^2+\frac{1}{2} I \omega^2\) =M g h

⇒ \(\frac{1}{2} M v^2+\frac{1}{2}\) I\((\frac{v}{R})^2\) =Mg \((\frac{3 v^2}{4 g})\)

⇒ \(\frac{1}{2} I \frac{v^2}{R^2} =\frac{3}{4} M v^2-\frac{1}{2} M v^2\)

⇒ \(\frac{1}{2} I \frac{v^2}{R^2} =\frac{1}{4} M v^2\)

or I =\(\frac{1}{2} M R^2\)

Question 77. A solid cylinder of mass 3 kg is rolling on a horizontal surface with a velocity of 4 ms^-1. It collides with a horizontal spring of force constant 200 Nm^-1. The maximum compression produced in the spring will be the:

1. 0.5 m
2. 0.6 m
3. 0.7 m
4. 0.2 m

Answer: 1. 0.5 m

At maximum compression, the solid cylinder will stop.

According to the law of conservation of mechanical energy Loss in kinetic energy of the cylinder = Gain in the potential of the spring

⇒ \(\frac{1}{2} m v^2+\frac{1}{2} I \omega^2=\frac{1}{2} k x^2\)

⇒ \(\frac{1}{2} m v^2+\frac{1}{2}\left(\frac{m R^2}{2}\right)\left(\frac{v}{R}\right)^2=\frac{1}{2} k x^2\)

v\(=R \omega\) and for solid cylinders. \(\left.I=\frac{1}{2} m R^2\right)\)

⇒ \(\frac{1}{2} m v^2+\frac{1}{4} m v^2 =\frac{1}{2} k x^2 \)  or,

⇒ \(\frac{3}{4} m v^2  =\frac{1}{2} k x^2 \)  or

⇒ \(x^2 =\frac{3}{2}=\frac{m v^2}{k}\)

Here, m^-3 kg, v = 4ms^-1, A: = 200 Nm^-1 Substituting the given values, we get

⇒ \(x^2=\frac{3 \times 3 \times 4 \times 4}{2 \times 200}\)

∴ \(x^2=\frac{36}{60} \text { or } x=0.6 \mathrm{~m}\)

Question 78. The solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

1. Both together only when the angle inclination of the plane is 45°
2. Both together
3. Hollow cylinder
4. Solid cylinder

Answer: 4. Solid cylinder

T =\(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}\left(1+\frac{K^2}{R^2}\right)}\)

⇒ \(\text { ce, }\left(\frac{K^2}{R^2}\right)_S <\left(\frac{K^2}{R^2}\right)_H\)

Solid-body will reach the bottom first with greater velocity

Question 79. A roller coaster is designed such that riders experience “weightlessness” as they go around the top of a hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is between:

1. 14 m/s and 15 m/s
2. 15 m/s and 16 m/s
3. 16 m/s and 17 m/s
4. 13 m/s and 14 m/s

Answer: 1. 14 m/s and 15 m/s

Balancing the force,M g-N= \(\frac{M v^2}{r}\)

When N=0 for weightiness

⇒ \(\frac{M v^2}{r} =\mathrm{Mg}\)

⇒ \(v^2 =g r=20 \times 10=200 \)

v =\(\sqrt{200}=14.14 \mathrm{~m} / \mathrm{s}\)

Question 80. A drum of radius R and mass M rolls down without slipping along an inclined plane of angle θ. The frictional force:

1. dissipates energy as heat
2. decreases the rotational motion’
3. decreases the rotational and translational motion
4. converts translational energy

Answer: 4. converts translational energy

Required frictional force converts translational energy into rotational energy.

Question 81. A drum of radius R and mass M rolls down without slipping along an inclined plane of angle θ. The frictional force:

1. dissipates energy as heat
2. decreases the rotational motion
3. decreases the rotational and translational motion
4. converts translational energy to rotational energy.

Answer: 4. converts translational energy to rotational energy.

Required frictional force converts some part of translational energy into rotational energy

Question 82. A solid cylinder of mass M and radius R rolls without slipping an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom?

1. \(\sqrt{2 g h}\)
2. \(\sqrt{\frac{3}{4} g h}\)
3. \(\sqrt{\frac{4}{3} g h}\)
4. \(\sqrt{4 g h}\)

Answer: 2. \(\sqrt{\frac{3}{4} g h}\)

Here, K.E. of the centre of mass when the cylinder reached the bottom\(\frac{1}{2} m v^2+\frac{1}{2} \mathrm{I} \omega^2\)  → Equation 1

Here, I =\(\frac{1}{2} M K^2\)

And \(\omega =\frac{v}{R}\)

⇒ \(\mathrm{KE} =\frac{1}{2} m v^2+\frac{1}{2} M K^2 \cdot \frac{v^2}{r^2}\)

= \(\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right]\)

But for solid Cylinder,\(\frac{R}{\sqrt{2}}\)

Or \(\frac{K^2}{R^2}=\frac{1}{2}\)

K E=\(\frac{3}{4} m v^2\)  →  Equation 2

PE of the solid cylinder at height h is PE = mgh  →  Equation  3

From Equation (2) and (3)

mg =\(\frac{3}{4} m v^2 \)

v = \(\sqrt{\frac{3}{4} g h}\)

Question 83. A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If the radius of the ball is R, then the fraction of total energy associated with its rotational energy will be:

1. \(\frac{K^2+R^2}{R^2}\)
2. \(\frac{K^2}{R^2}\)
3. \(\frac{K^2}{K^2+R^2}\)
4. \(\frac{R^2}{K^2+R^2}\)

Answer: 3. \(\frac{K^2}{K^2+R^2}\)

Total energy = \(\frac{1}{2} I \omega^2+\frac{1}{2} m v^2\)

= \(\frac{1}{2} m v^2\left(1+\frac{K^2}{R^2}\right)\)

Rotational energy = \(\frac{1}{2} I \omega^2\)

= \(\frac{K^2+R^2}{1+\frac{K^2}{R^2}}\)

Required fraction = \(\frac{K^2 / R^2}{1+K^2 / R^2}\)

= \(\frac{K^2}{R^2+K^2}\)

Question 84. A solid sphere of radius R is placed on a smooth horizontal surface. A horizontal force F is applied at height ‘h’ from the lowest point. For the maximum acceleration of the centre of mass, which is correct?

1. h = R
2. h = 2R
3. h = 0
4. No relation between h and R

Answer: 4. No relation between h and R

A smooth surface is given so rolling motion is not possible. Sphere will perform linear motion

Question 85. Consider a point P at the contact point of a wheel on the ground which rolls on the ground without slipping, then the value of the displacement of point P when the wheel completes half of the rotation (if the radius of the wheel is 1 m) is:

1. \(2 \mathrm{~m}\)
2. \(\sqrt{\pi^2+4} \mathrm{~m}\)
3. \(\pi \mathrm{m}\)
4. \(\sqrt{\pi^2+2} \mathrm{~m}\)

Answer: 2. \(\sqrt{\pi^2+4} \mathrm{~m}\)

Displacement PQ=\(\sqrt{(PM)^2+(Q M)^2}\)

⇒ \(\sqrt{(\pi R)^2+(2 R)^2}\)

∴ \(\sqrt{\pi^2+4}\)

Question 86. A disc is rolling, and the velocity of its centre of mass is V_cm. Which one will be correct?

1. The velocity of the highest point is 2v_cm and at the point of contact is zero.
2. The velocity of the highest point is v_cm and at the point of contact is v_cm.
3. The velocity of the highest point is 2v_cm and the point of contact is v_cm.
4. The velocity of the highest point is 2v_cm and the point of contact is 2v_cm.

Answer: 1. The velocity of the highest point is 2v_cm and at the point of contact is zero

Question 87. A solid spherical ball rolls on a table. The ratio of its rotational kinetic energy to total kinetic energy is:

1. 1/2
2. 1/6
3. 7/10
4. 2/7

Answer: 4. 2/7

Linear K.E. of ball=\(\frac{1}{2} m v^2\) and

rotational K.E. of ball =\(\frac{1}{2} I \omega^2\)

=\(\frac{1}{2}\left(\frac{2}{5} m r^2\right) \omega^2=\frac{1}{5} m v^2\)

Total K. E. =\(\frac{1}{2} m v^2+\frac{1}{5} m v^2=\frac{7}{10} m v^2\)

Ratio of rotational K.E. and total K.E. =\(\frac{(1 / 5) m v^2}{(7 / 10) m v^2}=\frac{2}{7}\)

Question 88. A solid sphere, disc and solid cylinder all of the same mass and radius are allowed to roll down (from rest) on the inclined plane, then:

1. solid sphere reaches the bottom first
2. solid sphere reaches the bottom last
3. the disc will reach the bottom first
4. all reach the bottom at the same time.

Answer: 1. solid sphere reaches the bottom first

For solid sphere\(\frac{K^2}{R^2}=\frac{2}{5}\)

For disc and solid cylinder,\(\frac{K^2}{R^2}=\frac{1}{2}\)

As for the solid sphere \(, K^2 / R^2\) is the smallest, it takes minimum time to reach the bottom of the incline, disc and cylinder reach together later.

Question 89. The speed of a homogenous solid sphere after rolling down an inclined plane of vertical height h from rest without sliding is:

1. \(\sqrt{\frac{10}{7} g h}\)
2. \(\sqrt{g h}\)
3. \(\sqrt{\frac{6}{5} g h}\)
4. \(\sqrt{\frac{4}{3} g h}\)

Answer: 1. \(\sqrt{\frac{10}{7} g h}\)

PE. = total K.E.

∴ \(m g h=\frac{7}{10} m v^2, v=\sqrt{\frac{10 g h}{7}}\)

Question 90. A solid homogenous sphere of mass M and radius is moving on a rough horizontal surface, partly rolling and partly. During this kind of motion of the sphere:

1. total kinetic energy is conserved
2. the angular momentum of the sphere about the point of contact with the plane is conserved
3. only the rotational kinetic energy about the centre of mass is conserved
4. angular momentum about the centre of mass is conserved.

Answer: 2. the angular momentum of the sphere about the point of contact with the plane is conserved

Angular momentum about the point of contact with the surface includes the angular momentum about the centre. Because of friction, linear momentum will not be conserved.

Work, Energy And Power

Question 1. A particle moves so that its position vector is given by \(\vec{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}\) where to is a constant. Which of the following is true?

1. Velocity is perpendicular to \(\vec{r}\) and acceleration is directed towards the origin.
2. Velocity is perpendicular to \(\vec{r}\) and acceleration is directed away from the origin.
3. Velocity and acceleration both are perpendicular to \(\vec{r}\).
4. Velocity and acceleration both are perpendicular to \(\vec{r}\).

Answer: 2. Velocity is perpendicular to \(\vec{r}\) and acceleration is directed away from the origin.

Given,\(\vec{r}= \cos \omega t \hat{x}+\sin \omega t \hat{y}\)

∴ \(\vec{v} =\frac{d \vec{r}}{d t}=-\omega \sin \omega t \hat{x}+\omega \cos \omega t \hat{y}\)

⇒ \(\vec{a} =\frac{d \vec{v}}{d t}=-\omega^2 \cos \omega t \hat{x}-\omega^2 \sin \omega t \hat{y}=-\omega^2 \vec{r}\)

Since position vector \((\vec{r})\) is directed away from the origin, so, acceleration \(\left(-\omega^2 \vec{r}\right)\) is directed towards the origin.

Also, \(vec{r} . \vec{v}=(\cos \omega t \hat{x}+\sin \omega t \hat{y}) .(-\omega \sin \omega t \hat{x}+\omega \cos \omega t \hat{y})\)

=\(-\omega \sin \omega t \cos \omega t+\omega \sin \omega t \cos \omega t=0\)

∴ \(\vec{r} \perp \vec{v}\)

Question 2. If vectors \(\vec{A}=\cos \omega t \hat{i}+\sin \omega t \hat{j} \text { and } \vec{B}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}\) are functions of time, then the value of t at which they are orthogonal to each other is:

1. \(\mathrm{t}=\frac{\pi}{\omega}\)
2. t=0
3. t=\(\frac{\pi}{4 \omega}\)
4. \(\mathrm{t}=\frac{\pi}{2 \omega}\)

Answer: 1. \(\mathrm{t}=\frac{\pi}{\omega}\)

Two vectors\(\vec{A} \text { and } \vec{B}\) are orthogonal to each other, if their scalar product is zero i.e. \(\vec{A} \cdot \vec{B}=0\)

Here, \(\vec{A}=\cos \omega \mathrm{t} \vec{i}+\sin \omega \mathrm{t} \vec{j}\)

and \(\vec{B}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}\)

⇒ \(\vec{A} \cdot \vec{B} =(\cos \omega t \hat{i}+\sin \omega t \hat{j}) \cdot\left(\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}\right)\)

=\(\cos \omega t \cos \frac{\omega t}{2}+\sin \omega t \sin \frac{\omega t}{2} \cos\)

=\(\cos \left(\omega t-\frac{\omega t}{2}\right)\)

Read and Learn More NEET Physics MCQs

But \(\vec{A} \cdot \vec{B}=0\) as \(\vec{A}\) and \(\vec{B}\) are orthogonal to each other

⇒ \(\cos \left(\omega t-\frac{\omega t}{2}\right)\)=0

⇒ \(\left(\omega t-\frac{\omega t}{2}\right)=\cos \frac{\pi}{2}\)

{ or } \(\omega t-\frac{\omega t}{2}=\frac{\pi}{2}\)

⇒ \(\frac{\omega t}{2}=\frac{\pi}{2}\)

∴ \(\text { or } \quad t=\frac{\pi}{\omega}\)

Question 3. If a vector \(2 \hat{i}+3 \hat{j}+8 \widehat{k}\) is perpendicular to the vector \(4 \hat{j}-4 \hat{i}+\alpha \hat{k}\) , then the value of a is:

1. 1/2
2. -1/2
3. 1
4. -1

Answer: 2. -1/2

⇒ \(\vec{a}=2 \hat{i}+3 \hat{j}+8 \hat{k}, \hat{b}=4 \hat{j}-4 \hat{i}+\alpha \hat{k}\)

⇒ \(\vec{a} \cdot \vec{b}=0 \text { if } \vec{a} \perp \vec{b}\)

⇒ \((2 \hat{i}+3 \hat{j}+8 \hat{k}) \cdot(-4 \hat{i}+4 \hat{j}+\alpha \hat{k})\)=0

⇒ \(\text { or, }-8+12+8 \alpha\)=0

⇒ \(4+8 \alpha=0 \)

∴ \(\alpha=-1 / 2\)

Question 4. The vector sum of two forces is perpendicular to their vector differences. In that case, the forces:

1. are equal to each other
2. are equal to each other in magnitude
3. are not equal to each other in magnitude
4. cannot be predicted.

Answer: 2. are equal to each other in magnitude

Given \(\left(\vec{F}_1+\vec{F}_2\right) \perp\left(\vec{F}_1-\vec{F}_2\right)\)

⇒ \(\left(\vec{F}_1+\vec{F}_2\right) \cdot\left(\vec{F}_1-\vec{F}_2\right)\)=0

⇒ \(F_1^2-F_2^2-\vec{F}_1 \cdot \vec{F}_2+\vec{F}_2 \vec{F}_1\)=0

⇒ \(F_1^2=F_2^2\)

∴ F₁, F₂ are equal to each other in magnitude

Question 5. The position vector of a particle is \(\vec{r}=(a \cos \omega t) \hat{i}-(a \sin \omega t) \hat{j}\) . The velocity of the particle is:

1. directed towards the origin
2. directed away from the origin
3. parallel to the position vector
4. perpendicular to the position vector.

Answer: 2. directed away from origin

Position vector of the particle,

⇒ \(\vec{r}=(\mathrm{a} \cos \omega t) \vec{i}+(\mathrm{a} \sin \omega t) \vec{j}\)

velocity vector \(\vec{v}=\frac{d \vec{r}}{d t}=(-\mathrm{a} \omega \sin \omega t) \hat{i}+(\mathrm{a} \omega \cos \omega t) \hat{j}\)

=\(\omega[(-\mathrm{a} \sin \omega t) \hat{i}+(\mathrm{a} \cos \omega t) \hat{j}]\)

⇒ \(\vec{v} \cdot \vec{r}=\omega a[-\sin \omega t \hat{i}+\cos \omega t \hat{j}] \cdot[\mathrm{a} \cos \omega t \vec{i}+\mathrm{a} \sin \omega t \hat{j}]\)

=\(\omega\left[-\mathrm{a}^2 \sin \omega t \cos \omega t+a^2 \cos \omega t \sin \omega t\right]\)=0

Therefore velocity vector is perpendicular to the position vector.

Question 6. The angle between the two vectors \(\vec{A}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{B}=3 \hat{i}+4 \hat{j}-5 \hat{k}\) will be:

1. 90°
2. 180°
3. zero
4. 45°

Answer: 1. 90°

⇒ \(\vec{A}=3 \hat{i}+4 \hat{j}+5 \hat{k} \text { and } \hat{B}=3 \hat{i}+4 \hat{j}-5 \hat{k}\)

⇒ \(\cos \theta=\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}\)

=\(\frac{(3 \hat{i}+4 \hat{j}+5 \hat{k}) \cdot(3 \hat{i}+4 \hat{j}-5 \hat{k})}{\left[\sqrt{(3)^2+(4)^2+(5)^2}\right]}\)\(\times\left[\sqrt{(3)^2+(4)^2+(5)^2}\right]\)

=\(\frac{9+16-25}{50}=0 \text { or } \theta=90^{\circ}\)

Question 7. Consider a drop of rainwater having a mass of 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take g, constant with a value of 10 m/s². The work done by the [1] gravitational force and the [2] resistive force of air is:

1. (1) – 10J,     (2) – 8.25 J
2. (1) – 1.25J,  (2) – 8.25 J
3. (1) – 100J,   (2) – 8.75 J
4. (1) – 10J,     (2) – 8.75 J

Answer: 4. (1) – 10J, (2) – 8.75 J

Work done by gravitational force, W = mgh

W= 10^-3x 10 x 1 x 10³= [10]

From the work-energy theorem,

Work done = change in KE

⇒ \(\Delta K=W_{\text {gravity }}+W_{\text {air resistance }}\)

⇒ \(\frac{1}{2}=m g h+W_{\text {air resist }}\)

⇒ \(W_{\text {air resist }}=\frac{1}{2}-m g h=10^{-3}\)

= \(10^{-3}\left(\frac{1}{2} \times 50 \times 50-10 \times 10^3\right)\)

= -8.75J

Question 8. A particle of mass l0g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 x 10^-4 by the end of the second revolution after the beginning of the motion?

1. 0.18 m/s²
2. 0.2 m/s²
3. 0.1 m/s²
4. 0.15 m/s²

Answer: 3. 0.1 m/s²

Here, m = 10 g = 10^-2 kg, R = 6.4 cm = 6.4 x 10^-2 m,

K_f = 8 x 10^-4 J, K, = 0, a, = 7

Using the work-energy theorem, Work done by all the forces = Change in KE

⇒ \(\mathrm{W}_{\text {tangential force }}+\mathrm{W}_{\text {centripetal force }}=\mathrm{K}_f-\mathrm{K}_i\)

⇒ \(a_t=\frac{K_f}{4 \pi R m}\)

=\(\frac{8 \times 10^{-4}}{4 \times \frac{22}{7} \times 6.4 \times 10^{-2} \times 10^{-2}}\)

=\(0.099=0.1 \mathrm{~m} \mathrm{~s}^{-2}\)

Question 9. A bullet of mass 10 g leaves a rifle at an initial velocity of 1000 m/s and strikes the earth at the same level with a velocity of 500 m/s. The work done in joule to overcome the resistance of air will be:

1. 375
2. 3750
3. 5000
4. 500

Answer: 2. 3750

According to the work-energy theorem, work done by a force in displacing a body determines the change in kinetic energy of the body or W=\(\Delta K E\)

= final \(\mathrm{KE}-\text { initial } \mathrm{KE}\)

=\(\frac{1}{2} m v^2-\frac{1}{2} m u^2\)

Given, v=1000\(\mathrm{~m} / \mathrm{s}, m=10 \mathrm{~g}\)

Putting the values of \(m_1, u_1\) and \(v_1,\) we get

⇒ \(\quad W =\frac{1}{2} \times 0.01\left[(1000)^2-(500)^2\right]\)

=\(3750 \mathrm{~J}\)

Question 10. A force F = 20 + 10y acts on a particle in y-direction, where F is in Newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is :

1. 5 J
2. 25 J
3. 20 J
4. 30 J

Answer: 2. 25 J

Work done by the variable force is,

⇒ \(\mathrm{W}=\int_{y_i}^{y_f} \mathrm{~F} d y\)

Here, \(y_i=0 and y_f=1 \mathrm{~m}\)

⇒ \(W =\int_{y_i}^{y_f}(20+10 y) d y\)

=\(\left[20 y+\frac{10 y^2}{2}\right]_0^1=25 \mathrm{~J}\)

Question 11. The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is:

1. 2mgR
2. \(\frac{1}{2}\) mgR
3. \(\frac{3}{2}\)mgR
4. mgR

Answer: 2. \(\frac{1}{2}\) mgR

The initial potential energy at the earth’s surface is,

⇒ \(U_i=-\frac{G M m}{R}\)

The final potential energy at height, h =R

⇒ \(U_f =-\frac{G M m}{2 R}\)

From the work-energy theorem,

Work done \(=\text { Change in } \mathrm{KE}\)

W=\(U_f-V_i\)

=\(\frac{G u m}{2 R}=\frac{g R^2 m}{2 R}\)

=\(\frac{m g R}{2}\)

Question 12. A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its center of mass has a speed of 20 cm/s. How much work is needed to stop it?

1. 33 kJ
2. 2 J
3. 1 J
4. 3 J

Answer: 4. 3 J

We know that, Work required = Change in kinetic energy

So, \(\mathrm{K}. \mathrm{E}=\frac{1}{2} m v^2+\frac{1}{2} \mathrm{I} \omega^2\)

=\(\frac{3}{4} m v^2\)

=\(\frac{3}{4} \times 100 \times\left(20 \times 10^{-2}\right)^2\)

∴ \(\Delta \mathrm{KE} =3 \mathrm{~J}\)

Question 13. A particle moves from a point \((-2 \hat{i}+5 \hat{j}) \text { to }(4 \hat{j}+3 \hat{k})\) when a force of \((4 \hat{i}+3 \hat{j}) \mathrm{N}\) is applied. How much work has been done by the force?

1. 8 J
2. 11 J
3. 5 J
4. 2 J

Answer: 3. 5 J

Here \(\vec{F}=(-2 \hat{i}+5 \hat{j}) \mathrm{m}\)

⇒ \(\vec{r}=(4 \hat{j}+3 \hat{k}) \mathrm{m}\)

⇒ \(\vec{F}=(4 \hat{i}+3 \hat{j}) \mathrm{N}, \mathrm{W}\)=?

Work done by force in moving from \(\overrightarrow{r_1}\) to \(\overrightarrow{r_1}\)

⇒ \(\mathrm{W} =\vec{F} \cdot\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right)\)

⇒ \(\mathrm{W} =(4 \hat{i}+3 \hat{j}) \cdot(4 \hat{j}+3 \hat{k}+2 \hat{i}-5 \hat{j})\)

= \((4 \hat{i}+3 \hat{j}) \cdot(2 \hat{i}-\hat{j}+3 \hat{k})\)

= \(8+(-3)=5 \mathrm{~J}\)

Question 14. A uniform force of \((3 \hat{i}+\hat{j})\) N acts on a particle of mass 2 kg. Hence the particle is displaced from position \((2 \hat{i}+\hat{k})\) m to position \((4 \hat{i}+3 \hat{j}-\hat{k})\) m. The work done by the force on the particle is:

1. 9 J
2. 6 J
3. 13 J
4. 15 J

Answer: 1. 9 J

Work done = F.S

From the question, \(F=(3 \hat{i}+\hat{j}) N\)

⇒ \(\mathrm{~S}=r_2-r_1\)

⇒ \(\left\{r_1\right.=2 \hat{i}+\hat{k} ; r_2=4 \hat{i}+3 \hat{j}-\hat{k}\)

= \(4 \hat{i}+3 \hat{j}-\hat{k}-(2 \hat{i}+\hat{k})\)

= \((2 \hat{i}+3 \hat{j}-2 \hat{k}) m\)

⇒ \(\mathrm{~W}=\mathrm{F} . \mathrm{S} .=(3 \hat{i}+\hat{j})(2 \hat{i}+3 \hat{j}-2 \hat{k})\)

⇒ \(3 \times 2+3+0=09\)

⇒ \(\{\hat{i} . \hat{i}=1 \hat{i} \cdot \hat{j}=0\} \mathrm{J} \hat{j} \cdot \hat{j}=1 \hat{j} \cdot \hat{k}\)=0

∴ \(\hat{k} \cdot \hat{k}=1 \hat{k} \cdot \hat{i}=0\}\)

Question 15. A body of mass 10 kg is released from a wall of height 20 m and acquires a velocity of 10 ms^-1 after falling through the distance 20 m then the work done by the push of the air on the body is:

1. 1456 J
2. 1500 J
3. -1500 J
4. -1456 J

Answer: 3. -1500 J

From the Work Energy Theorem, Work done = change in K.E.

⇒ \(W_{\text {gravity }}+W_{\text {air }}\) =K E

⇒ \(m g h +W_{\text {air }}=\frac{1}{2} m v^2-\frac{1}{2} m u^2\)

Putting the values in the above equation, \(10 \times 10 \times 20+W_{\text {air }} =\frac{1}{2}(10)(10)^2-0\)

∴ \(W_{\text {air }}=-1500 \mathrm{~J}\)

Question 16. A chain of mass m and length L is placed, on a table in such a way that its \(\frac{1}{n}\) the part is hanging below the edge of

1. \(\frac{\mathrm{MgL}}{2 \mathrm{n}^2}\)
2. \(\frac{\mathrm{MgL}}{\mathrm{n}^2}\)
3. \(\frac{\mathrm{MgL}}{2 \mathrm{n}}\)
4. \(\frac{\mathrm{MgL}}{\mathrm{n}}\)

Answer: 1. \(\frac{\mathrm{MgL}}{2 \mathrm{n}^2}\)

The potential energy of chain at table level, \(U_f =0 \text { and } v_i=-m g h\)

⇒ \(v_i =-\left(\frac{\mathrm{M}}{\mathrm{L}}\right)\left(\frac{\mathrm{L}}{n} g\right)\left(\frac{\mathrm{L}}{2 n}\right)=-\frac{\mathrm{MgL}}{2 n^2}\)

⇒ \(\text { Work done } =v_f-v_i\)

=\(\left[-\frac{M g L}{2 n^2}\right]=\frac{M g L}{2 n^2}\)

Question 17. A body moves a distance of 10 m in a straight line under the action of a 5 N force. If the work done is 25 J, then the angle between the force and direction of motion of the body is:

1. 60°
2. 75°
3. 30°
4. 45°

Answer: 1. 60°

Distance(s) = 10 m;

Force(F)=5N

and work done (W) = 25 J Work done (W) = Fs cosθ

25 = 5 x 10 cosθ = 50 cosθ  or  cos θ = 25/50 = 0.5   or  θ = 60°

Question 18. A body, constrained to move in the y-direction, is subjected to a force given by F=\((-2 \hat{i}+15 \hat{j}+6 \hat{k})\) N. The work done by this force in moving the body through a distance of 10 \(\hat{j}\) m along the y-axis, is:

1. 150 J
2. 20 J
3. 190 J
4. 160 J

Answer: 1. 150 J

Force \(\vec{F}=(-2 \hat{i}+15 \hat{j}+6 \hat{k}) \mathrm{N}\), and distance, \(\mathrm{d}=10 \vec{j} \mathrm{~m}\)

Work done \(\mathrm{W} =\vec{F} \vec{d}=(-2 \hat{i}+15 \hat{j}+6 \hat{k}) \cdot(10 \hat{j})\)

=\(150 \mathrm{Nm}=150 \mathrm{~J}\)

Question 19. When a body moves with a constant speed along a circle:

1. no work is done on it
2. no acceleration is produced in it
3. its velocity remains constant
4. no force acts on it.

Answer: 1. no work is done on it

Question 20. A particle is released from height S from the surface of the Earth. At a certain height, its kinetic energy is three times its potential energy. The light from the surface of the earth and the speed of the period. At that instant are respectively:

1. \(\frac{\mathrm{S}}{4}, \frac{3 g \mathrm{~S}}{2}\)
2. \(\frac{\mathrm{S}}{4}, \frac{\sqrt{3 g S}}{2}\)
3. \(\frac{\mathrm{S}}{2}, \frac{\sqrt{3 g \mathrm{~S}}}{2}\)
4. \(\frac{\mathrm{S}}{4}, \frac{\sqrt{3 g S}}{2}\)

Answer: 2. \(\frac{\mathrm{S}}{4}, \frac{\sqrt{3 g S}}{2}\)

Given, Height=S

Arbitrary hieight=h

where KE_h=3 PE_h

h=? v_h=?

At the height ‘h’, TE = mg S

At arbitrary height ‘h’, (h< S)

PE= mgh

⇒ \(\mathrm{KE} =\frac{1}{2} \mathrm{mv}^2\)

⇒  \(\mathrm{v}^2=2 \mathrm{~g}(\mathrm{~S}-h)\)

Where \(\mathrm{v}^2=2 \mathrm{~g}(\mathrm{~S}-h)\)

⇒ \(\mathrm{KE}=\frac{1}{2} \mathrm{mv}^2\)

⇒ \(\mathrm{v}^2 =2 \mathrm{~g}(\mathrm{~S}-h)\)

⇒  \(\mathrm{KE} =\frac{1}{2} m 2 g(\mathrm{~S}-h)\)

= \(m g(\mathrm{~S}-h)\)

Given,\(\mathrm{KE}_n =3 \mathrm{PE}_n\)

⇒  \((\mathrm{~S}-h) =3 m g h\)

⇒ \(\mathrm{~S}-h =3 h\)

⇒  \(\mathrm{~S} =4 h\)

∴ h =\(\frac{\mathrm{S}}{4}\)

∴ \(v^2 =2 g\left(\mathrm{~S}-\frac{\mathrm{S}}{4}\right)\)

⇒ \(v^2 =2 g \frac{3 \mathrm{~S}}{4}\)

∴ \(v =\sqrt{\frac{3 g s}{2}}\)

Question 21. A particle of mass 5 m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular directions with each speed v. The energy released during the process is:

1. \(\frac{3}{5} m v^2\)
2. \(\frac{5}{3} m v^2\)
3. \(\frac{3}{2} m v^2\)
4. \(\frac{4}{3} m v^2\)

Answer: 4. \(\frac{4}{3} m v^2\)

From the law of conservation of linear moments,

0 =\(m v \hat{j}+m v \hat{i}+3 m \vec{v}_{\mathrm{t}}\)

⇒  \(\vec{v}_1 =\frac{v}{3}(\hat{i}+\hat{j})\)

⇒  \(v_1 =\frac{\sqrt{2}}{3} v\)

\(\mathrm{KE}_i\)=0

⇒  \(\mathrm{KE}_f =\frac{1}{2}+\frac{1}{2}+\frac{1}{2}(3 \mathrm{~m})\left(\frac{\sqrt{2}}{3}\right)^2 v^2\)

=\(m v^2+\frac{m v^2}{3}=\frac{4}{3} m v^2 \)

∴ \(\Delta \mathrm{KE} =\mathrm{KE}_f-\mathrm{KE}_i=\frac{4}{3} m v^2\)

Question 22. A block of mass 10 kg moving in the x-direction with a constant speed of 10 ms^-1, is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20 m to 30 m. Its final KE will be:

1. 475 J
2. 450 J
3. 275 J
4. 250 J

Answer: 1. 475 J

According to the question, Mass of the block, m = 10 kg

Speed of block, V = 10 ms^-1

Retarding force, f = 0.1 J/m

Using the work-energy theorem, Work done = change in kinetic energy

⇒ \(W =K E_f-K E_i\)

⇒ \(K E_f =W+K E_i\)

=\(\int_{20}^{30}-0.1 \times x d x+\frac{1}{2} \times 10 \times 10^2\)

=\(\left[-0.1 \frac{x^2}{2}\right]_{20}^{30}+500\)

=\(-0.1 \times \frac{500}{2}+500\)

=\(-25+500=475 \mathrm{~J}\)

Question 23. A body of mass (4 m) is lying in the XY plane at rest. It explodes into three pieces. Two pieces each of mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to the explosion is:

1. \(m v^2\)
2. \(\frac{3}{2} m v^2\)
3. \(2 m v^2\)
4. \(4 m v^2\)

Answer: 2. \(\frac{3}{2} m v^2\)

⇒ \(\sqrt{2} m v\)= Resultant momentum of two small mass masses

Using law of conversation of momentum \(\sqrt{2} =2 m v^{\prime}\)

⇒ \(v^{\prime} =\frac{1}{\sqrt{2} m}\)

Now total kinetic energy by the explosion is \(\mathrm{k}=\frac{1}{2} m v^2+\frac{1}{2} m v^2+\frac{1}{2}(2 \mathrm{~m}) v^{\prime 2}\)

= \(m v^2+m\left(\frac{\mathrm{v}}{\sqrt{2}}\right)^2=m v^2+\frac{m v^2}{2}\)

= \(\frac{3}{2} m v^2\)

Question 24. An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?

1. \(\frac{1}{2} m v^3\)
2. \(m v^2\)
3. \(\frac{1}{2} m v^2\)
4. \(\frac{1}{2} m^2 v^2\)

Answer: 1. \(\frac{1}{2} m v^3\)

Let m = mass per unit length

Rate of mass per sec =\(\frac{m x}{t}\)

=\(m \frac{x}{t}=m v\)

Rate Of KE =\(\frac{1}{2}(m v) V^2\)

=\(\frac{1}{2} m v^3\)

Question 25. A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 ms^-1. The kinetic energy of the other mass is:

1. 324 J
2. 486 J
3. 256 J
4. 524 J

Answer: 2. 486 J

According to the law conservation of angular momentum

30 x 0 = 18 x 6 + 12 x v

v = – 9 m/s

The negative sign indicates that both fragments move in opposite directions.

⇒ \(\text { K.E. of } 12 \mathrm{~kg} =\frac{1}{2} \mathrm{~m} v^2\)

=\(\frac{1}{2} \times 12 \times 81\)

=\(486 \mathrm{~J}\)

Question 26. A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60-foot-tall building. After a fall of 30 feet each toward Earth, their respective kinetic energies will be in the ratio of:

1. \(\sqrt{2}+1\)
2. 1:4
3. 1:2
4. \(1:\sqrt{2}\)

Answer: 3.

Let E₁ and E₂ be the K.E. of two bodies.

Then, \(E_1=\frac{1}{2} m_1 v_1^2\)

⇒ \(E_2=\frac{1}{2} m_2 v_2^2\)

⇒ \(\frac{E_1}{E_2}=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v^2}=\frac{m_1 v_1^2}{m_2 v_2^2}\)

The initial velocity of both bodies is zero. v²= 2gh is the same for both bodies

∴ \(\frac{E_1}{E_2}=\frac{m_1}{m_2}=\frac{2}{4}=\frac{1}{2}\)

Question 27. A particle of mass m₁ is moving with a velocity vj and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum but their different kinetic energies are E₁ and E₂ respectively. If m₁ > m₂ then:

1. \(E_1<E_2\)
2. \(\frac{E_1}{E_2}=\frac{m_1}{m_1}\)
3. \(E_1>E_2\)
4. \(E_1=E_2\)

Answer: 1. \(E_1<E_2\)

We know that, K.E = \(\frac{p^2}{2 m}\)

⇒ \(\text { K.E. }=\frac{p^2}{2 m}\)

⇒ \(\frac{E_1}{E_2}=\frac{p_1^2 / 2 m_1}{p_2^2 / 2 m_2}\)

⇒ \(\frac{E_1}{E_2}=\frac{m_2}{m_1}\)

Here, \(m_1=m_2\)

∴ \(E_1<E_2\)

Question 28. A stationary particle explodes into two particles of masses m₁ and m₂ which are in opposite directions with velocities v₁ and v₂. The ratio of their kinetic energies E₁/ E₂ is:

1. \(\frac{m_2}{m_1}\)
2. \(\frac{m_1}{m_2}\)
3. 1
4. \(\frac{m_1+m_2}{m_2 m_1}\)

Answer: 1. \(\frac{m_2}{m_1}\)

According to the law of conservation of linear momentum, \(m_1 v_1 =m_2 v_2\)

⇒ \(E_1=\frac{1}{2} m_1 v_1^2, E_2 =\frac{1}{2} m_2 v_2^2\)

⇒ \(\frac{E_1}{E_2}=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v_1^2} =\frac{\left(m_1 v_1\right)^2}{\left(m_2 v_2\right)^2} \times \frac{m_2}{m_1}\)

∴ \(\frac{E_1}{E_2} =\frac{m_2}{m_1}\)

Question 29. If the kinetic energy of a body is increased by 300%, then the percentage change in momentum will be:

1. 100%
2. 150%
3. 265%
4. 73.2%

Answer: 1. 100%

Here, \(p_1=\sqrt{2 m E_1} and p_2=\sqrt{2 m E_2}\)

⇒ \(p^{\prime} =\sqrt{2 m\left(E+\frac{300}{100} E_1\right)}\)

=\(\sqrt{2 m 4 E_1}\)

=\(2 n_1\)

So, the momentum will change by 100%

Question 30. A particle is projected making an angle of 45° with horizontal having kinetic energy K. The kinetic energy at the highest point will be:

1. \(\frac{K}{\sqrt{2}}\)
2. \(\frac{K}{2}\)
3. \(2 \mathrm{~K}\)
4. \(\mathrm{K}\)

Answer: 2. \(\frac{K}{2}\)

Kinetic energy of ball = K and angle of projection (9) = 45°

The velocity of the ball at the highest point = v cos θ

⇒ \(v \cos 45^{\circ}=\frac{v}{\sqrt{2}}\)

Therefore the kinetic energy of the ball \(\frac{1}{2} \mathrm{mx}\left(\frac{v}{\sqrt{2}}\right)^2=\frac{1}{4} \mathrm{mv}^2=\frac{k}{2}\)

Question 31. If the momentum of a body is increased by 50%, then the percentage increase in its kinetic energy is:

1. 50%
2. 100%
3. 125%
4. 200%

Answer: 3. 125%

Let p₁, and p₂ be the initial momentum and inversed momentum respectively

⇒ \(So, p_2 =\frac{150}{100} p_1\)

⇒ \(m v_2 =\frac{15}{10} m v_1\)

⇒ \(m v_2=\frac{15}{10} m v_1\)

⇒ \(\left(p_1=m v_1, p_2=m v_2\right)\) or

⇒ \(v_2=\frac{15}{10} v_1\)

Now \(,\frac{E_2}{E_1} =\frac{\frac{1}{2} m v_2^2}{\frac{1}{2} m v_1^2}=\left(\frac{v_2}{v_1}\right)^2\)

=\(\left(\frac{15}{10}\right)^2=\frac{225}{100}\)

⇒ Clearly, \(\quad E_2>E_1\)

So, percentage increase in K.E. =\(\frac{\left(E_2-E_1\right)}{E_1} \times 100\)

=\(\left(\frac{225}{100}-1\right) \times 100=125 \%\)

Question 32. The kinetic energy acquired by a mass m in traveling distance d, starting from rest, under the action of a constant force is directly proportional to:

1. \(\mathrm{m}\)
2. \(\mathrm{m}^0\)
3. \(\sqrt{m}\)
4. \(1 / \sqrt{m}\)

Answer: 2. \(\mathrm{m}^0\)

⇒ \(v^2=u^2+2 a s \text { or } v^2-u^2=2 a s\) or

⇒ \(v^2-(0)^2=2 \times \frac{F}{m} \times \mathrm{s} \text { or } v^2=\frac{2 F s}{m}\)

⇒ \(\text { K.E. }=\frac{1}{2} m v^2=\frac{1}{2} m \times \frac{2 F s}{m}=\text { Fs. }\) and

∴ Thus K.E. is independent of m or directly proportional to\( \mathrm{m}^0\).

Question 33. Two masses of 1 g and 9 g are moving with equal kinetic energies. The ratio of the magnitudes of their respective linear momenta is:

1. 1: 9
2. 9: 1
3. 1 :3
4. 3: 1

Answer: 3. 1 :3

⇒ \(\frac{K_1}{K_2}=\frac{p_1^2}{p_2^2} \times \frac{M_1^2}{M_2^2}\)

⇒ \(\text { Here } \quad \mathrm{K}_1=\mathrm{K}_2\)

⇒ \(\frac{p_1}{p_2}=\sqrt{\frac{M_1}{M_2}}=\sqrt{\frac{1}{9}}=\frac{1}{3}\)

or,\(P_1: P_2=1: 3\)

Question 34. A particle of mass M is moving in a horizontal circle of radius R with uniform speed v. When it moves from one point to a diametrically opposite point, it:

1. kinetic energy change by Mv^2/4
2. momentum does not change
3. momentum change by 2 Mv
4. kinetic energy changes by mv²

Answer: 3. Momentum change by 2 Mv

∴ On the diametrically opposite points, the velocities have the same magnitude but opposite directions. Therefore change in momentum is Mv -(- Mv)= 2 Mv

Question 35. Force F on a particle moving in a straight line varies with distance as shown in the figure. The work done on the particle during its displacement of 12 m is:

1. 21 J
2. 26 J
3. 13 J
4. 18 J

Answer: 3. 13 J

From Question,Work done = Area under (F – x) graph

=\(2 \times(7-3)+\frac{1}{2} \times 2 \times(12-7)\)

=\(8+\frac{1}{2} \times 10\)

=\(8+5=13 \mathrm{~J}\)

Question 36. A body of mass 3 kg is under a constant force which causes a displacement s in meters in it, given by the 1, relation s = \(\frac{1}{3} t^2\), where t is in seconds. Work done by the force in 2 seconds is:

1. \(\frac{19}{5} \mathrm{~J}\)
2. \(\frac{5}{19} \mathrm{~J}\)
3. \(\frac{3}{8} \mathrm{~J}\)
4. \(\frac{8}{3} \mathrm{~J}\)

Answer: 4. \(\frac{8}{3} \mathrm{~J}\)

It is given that:\(\mathrm{s}=\frac{t^2}{3}\)

Differentiating w.r.t. t, we have,

U=\(\frac{d s}{d t}=\frac{2 t}{3}\)

Again differentiating w.r.t. t, we have,

a=\(\frac{d^2 s}{d t^2}=\frac{2}{3}\)

Now, \(\mathrm{W}=\int \mathrm{F} d s =\int \mathrm{mads}\)

=\(\int_0^2 m \times \frac{2}{3} \times \frac{2 t}{3} d t\)

=\(\frac{4}{9} m \int^2 t d t\)

W =\(\frac{4}{9} m \times\left[\frac{t^2}{2}\right]_0^2\)

=\(\frac{4}{9} \times 3 \times 2=\frac{8}{3} \mathrm{~J}\)

Question 37. A force F acting on an object varies with distance x as shown here. The force is in N and x in m. The work done by the force in moving the object x = 0 to x = 6m is:

1. 18.0 J
2. 13.5 J
3. 9.0 J
4. 4.5 J

Answer: 2. 13.5 J

Work done = Area under f – x curve.

= area of trapezium = \(\frac{1}{2} \times(6+3) \times 3=\frac{27}{2}=13.5 \mathrm{~J}\)

Question 38. A force acts on a 3.0 g particle in such a way that the position of the particle as a function of time is given by, x = 3t – 4t²- + t³, where x is in meter and nd t in second. The work done during the first 4s is:

1. 570 mJ
2. 450 mJ
3. 490 mJ
4. 528 mJ

Answer: 4. 528 mJ

Given,\( x=3 t-4 t^2+t^3\)

So, velocity v =\(\frac{d x}{d t}=3-8 t+3 t^2\)

⇒ \(\text { At } t=0 \mathrm{~s}\),

⇒ \(v_1 =3-0+0=3 \mathrm{~m} / \mathrm{s}\)

At \(t =4 \mathrm{~s}\),

⇒ \(v_2 =3-8 \times 4+3 \times 4^2\)

=\(3-32+48=19 \mathrm{~m} / \mathrm{s}\)

Now, from t=0 to t=4s

work done = gain in kinetic energy

=\(\frac{1}{2} m v_2^2-\frac{1}{2} m v_1^2\)

=\(\frac{1}{2} m\left(v_2^2-v_1^2\right)\)

=\(\frac{1}{2} \times 3 \times 10^{-3}\left[(19)^2-(3)^2\right]\)

⇒ [Using, \(a^2-b^2=(a+b)(a-b) ]\)

=\(1.5 \times 10^{-3} \times[(19+3)(19-3)]\)

=\(1.5 \times 10^{-3} \times 22 \times 16\)

=\(528 \times 10^{-3} \mathrm{~J}=528 \mathrm{~mJ}\)

Question 39. A position-dependent force F = (7 – 2x + 3x²) N, acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m. Work done in joule is:

1. 35
2. 70
3. 135
4. 270

Answer: 3. 135

Work done by a variable force F in displacement from x = x¹ to x = x² = x is given by

⇒ \(\mathrm{W} =\int_{x_1}^{x_2} F(d x)\)

⇒ \(x_1 =0, x_2=5 \)

F =\(\left(7-2 x+3 x^2\right) \mathrm{N}\)

W =\(\int_0^5\left(7-2 x+3 x^2\right) d x\)

=\(\left[7 x-x^2+x^3\right]_0^5\)

=\(\left[7 \times 5-(5)^2+(5)^3\right]\)

=\([35-25+125]=135 \mathrm{~J}\)

Question 40. A body of mass m taken from the earth’s surface to the height equal to twice the radius (R) of the earth. The change in potential energy of the body will be:

1. 2mgR
2. mgR
3. 3 mgR
4. \(\frac{1}{3}\)mgR

Answer: 2. Mgr

According to the question, Change in potential energy

⇒ \(\Delta \mathrm{P} . \mathrm{E}  =-\frac{G M m}{R+2 R}-\left(-\frac{G M m}{R}\right)\)

= \(-\frac{G M m}{3 R}-\frac{G M m}{R}=\frac{2 G M m}{3 R}\)

⇒ \(\Delta \mathrm{P} . \mathrm{E} =\frac{2}{3} m g R\)

Where, g =\(\frac{G M}{R^2}\)

Question 41. The potential energy of a particle in a force field is U =\(\mathrm{U}=\frac{A}{r^2}-\frac{B}{r}\) where A and B are positive constants and r is the distance of the particle form the center of the field. For stable equilibrium, the distance of the particle is:

1. \(\frac{B}{2 A}\)
2. \(\frac{2 A}{B}\)
3. \(\frac{A}{B}\)
4. \(\frac{B}{A}\)

Answer: 2. \(\frac{2 A}{B}\)

Here,\(\mathrm{U} =\frac{A}{r^2}-\frac{B}{r}\)

⇒ \(\frac{d U}{d r}\) =0

⇒ \(-\frac{2 A}{r^3}+\frac{B}{r^2}\) =0

or \(\frac{2 A}{r^3} =\frac{B}{r^2}\)

or r =\(\frac{2 A}{B}\)

Question 42. The potential energy of a system increases if work is done:

1. by the system against a conservative force
2. for the system against a conservative force
3. upon the system by a conservative force
4. against the system by a conservative force

Answer: 1. By the system against is a conservative force

The potential energy of a system increases. If work is done by the system against a conservative force.

⇒ \(\mathrm{W}_{\text {int }} =-\Delta \mathrm{U}\)

=\(-\left(\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}\right)\)

Question 43. The potential energy of a system increases if work is done:

1. upon the system by nonconservative force
2. by the system against a conservative force
3. by the system against a nonconservative force
4. upon the system by a conservative force.

Answer: 2. by the system against a conservative force

The work done by the system against a conservative force. Because we know that, \(-\Delta \mathrm{U}=\mathrm{W}_{\text {conservative force }}\)

Question 44. The potential energy between two atoms, in a molecule, is given by U (x) = \(\frac{a}{x^{12}}-\frac{b}{x^6}\)where a and b are positive and x is the distance between the atoms. The atom is in stable equilibrium, when:

1. x=\(\left(\frac{2 a}{b}\right)^{1 / 6}\)
2. \(x=\left(\frac{11 a}{5 b}\right)^{1 / 6}\)
3. x=0
4. x=\(\left(\frac{a}{2 b}\right)^{1 / 6}\)

Answer: 1. x=\(\left(\frac{2 a}{b}\right)^{1 / 6}\)

U(x)=\(\frac{a}{x^{12}}-\frac{b}{x^6}\)

or \(\frac{12 a}{x^{13}}-\frac{-6 b}{x^7}\)=0

or \( x^6=\frac{2 a}{b}\)

Therefore x=\(\left(\frac{2 a}{b}\right)^{1 / 6}\)

Question 45. Two similar springs P and Q have spring constant K_p and K_q, such that K_p > K_q. They are stretched, first by the same amount (case 1), then by the same force (case 2). The work done by the springs WP and WQ are related, in case and case (2), respectively:

1. W_P = W_Q; W_P > W_Q
2. W_P = W_Q; W_P = W_Q
3. W_P > W_Q; W_Q > W_P
4. W_P < W_Q; W_Q < W_P

Answer: 3. W_P > W_Q; W_Q > W_P

In case (1) elongation (x) is the same for both springs. So work done by springs P and Q is respectively.

⇒ \(W_P=\frac{1}{2} \mathrm{~K}_{\mathrm{P}} x^2 \text { and } W_Q=\frac{1}{2} \mathrm{~K}_{\mathrm{Q}} x^2\)

⇒ \(K_P>K_Q\)

⇒ \(W_P>W_Q\)

In case (2) force of elongation (F) is the same and so elongation is given by \(x_{\mathrm{P}}=\frac{F}{K_P} \text { and } x_{\mathrm{Q}}=\frac{F}{K_Q}[latex]\)

⇒ \(W_P=\frac{1}{2} k_{\mathrm{P}} x_P^2=\frac{1}{2} \frac{F^2}{k_P}\)

⇒ \(\mathrm{~W}_{\mathrm{Q}}=\frac{1}{2} k_{\mathrm{Q}} x_g^2=\frac{1}{2} \frac{\mathrm{F}^2}{k_g}\)

∴ \(W_P<W_Q\)

Question 46. A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be:

1. \(\frac{2 M g}{k}\)
2. \(\frac{4 M g}{k}\)
3. \(\frac{M g}{2 k}\)
4. \(\frac{M g}{k}\)

Answer: 1. \(\frac{2 M g}{k}\)

When the mass attached to a spring fixed at the other end is allowed to fall suddenly, it extends the spring by x. Potential energy lost by the mass is gained by the spring. \(m g x =\frac{1}{2} k x^2\)

x =\(-\frac{2 m g}{k}\)

Question 47. A vertical spring with force constant k is fixed on a table. A ball of mass m at a height h above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance d. The net work done in the process is:

1. \(m g(h+d)-\frac{1}{2} k d^2\)
2. \(m g(h-d)-\frac{1}{2} k d^2\)
3. \(m g(h-d)+\frac{1}{2} k d^2\)
4. \(m g(h+d)+\frac{1}{2} k d^2\)

Answer: 1. \(m g(h+d)-\frac{1}{2} k d^2\)

Net work done = work done by gravitational force + work done by spring force \(m g(h+d)-\frac{1}{2} k d^2\)

Question 48. The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is:

1. U/4
2. 4 U
3. W
4. 16 U

Answer: 4. 16 U

Potential energy of a spring=\(\frac{1}{2} \times \text { force constant } \times(\text { extension })^2\)

Potential energy\(\propto(\text { extension })^2\)

Hence, \(\frac{U_1}{v_2}=\left(\frac{x_1}{x_2}\right)^2\)

⇒\(\frac{v_1}{v_2}=\left(\frac{2}{5}\right)^2\)

⇒  \(\frac{U_1}{U_2}=\frac{1}{16}\)

∴ \(U_2=16 U_1=16 v\)

Question 49. A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be:

1. 0.15 m
2. 0.12 m
3. 1.5m
4. 0.5m

Answer: 1. 0.15 m

According To Question , \(\frac{1}{2} m v^2 =\frac{1}{2} k x^2\)

x =\(v \sqrt{\frac{m}{k}}=1.5 \sqrt{\frac{0.5}{50}}\)

=\(0.15 \mathrm{~m}\)

Question 50. Two springs of spring constant k₁ and k₂ are joined in series. The effective spring constant of the combination is given by:

1. \(\sqrt{k_1 k_2}\)
2. \(\frac{\left(k_1+k_2\right)}{2}\)
3. \(k_1+k_2\)
4. \(\frac{k_1 k_2}{k_1+k_2}\)

Answer: 4. \(\frac{k_1 k_2}{k_1+k_2}\)

When the spring joined in series the total extension of spring is y=y₁+y₂

=\(\frac{-F}{k_1}-\frac{F}{k_2}\)

y =\(-\mathrm{F}\left[\frac{1}{k_1}+\frac{1}{k_2}\right]\)

∴ Spring Constant k =\(\frac{k_1 k_2}{k_1+k_2}\)

Question 51. A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be:

1. 0.15 m
2. 0.12 m
3. 1.5 m
4. 0.5 m
5. Answer: 1. 0.15 m

The kinetic energy of Mass is converted into energy required to compress a spring which is given by

⇒ \(\frac{1}{2} m v^2 =\frac{1}{2} k x^2\)

x =\(v \sqrt{\frac{m}{k}}=1.5 \sqrt{\frac{0.5}{50}}\)

=0.15

Question 52. When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be:

1. \(\frac{U}{5}\)
2. 5U
3. 10U
4. 25U

Answer: 4. 25U

U=\(\frac{1}{2}\)

⇒ \(K(2)^2\)

⇒ \(U^{\prime}=\frac{1}{2} K(10)^2\)

From above equations (1) and (2) \(U^{\prime}=25 \mathrm{U}\)

Question 53. A mass is suspended separately by two different springs in successive order, and then the period is T₁ and T₂, respectively. If it is connected by both springs as shown in the figure, then the period T₀, so the correct equation is:

1. \(T_0^2=T_1^2+T_2^2\)
2. \(T_0^{-2}=T_1^{-2}+T_2^{-2}\)
3. \(T_0^{-1}=\mathrm{T}_1^{-1}+\mathrm{T}_2^{-1}\)
4. \(T_0=T_1+T_2\)

Answer: 2. \(T_0^{-2}=T_1^{-2}+T_2^{-2}\)

Time period of spring,T=\(2 \pi \sqrt{\frac{m}{\mathrm{~K}}}\)

K=\(4 \pi^2 \frac{m}{T^2}\)

⇒ \(K \propto \frac{1}{T^2}\)

For a parallel combination of spring \(K =K_1+K_2\)

⇒ \(\frac{1}{T_0^2} =\frac{1}{T_1^2}+\frac{1}{T_2^2}\)

∴ \(T_0^{-2} =T_1^{-2}+T_2^{-2}\)

Question 54. Two springs A and B having spring constant K_A and K_B (K_A = 2 K_B) are stretched by applying a force of equal magnitude. If energy stored in spring A is EA then energy stored in B will be:

1. 2 E_A
2. E_A/4
3. E_A/2
4. 4 E_A

Answer: 1. 2 E_A

⇒ \(\frac{1}{2} K x^2 =\frac{1}{2} \frac{F^2}{K}\)

⇒ \(\frac{K_A}{K_B}\) =2

∴ \(\frac{E_A}{E_B} =\frac{1}{2} \text { or } \mathrm{E}_{\mathrm{B}}=2 \mathrm{E}_{\mathrm{A}}\)

Question 55. The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is:

1. 36 x 10⁴ J
2. 36 x 10⁵ J
3. 1 x 10⁵ J
4. 36 x 10⁷ J

Answer: 4. 36 x 10⁷ J

Given, P= 100 kW

Time, t= 1 hour

E = Pxt= 100x 1 = 100 kWh

E = 100 x 3.6 x 10⁶ J = 36 x 10⁷J ( v₁ kWh = 3.6 x 10⁶J)

Question 56. Light with an average flux of 20 W/cm² falls on a non-reflecting surface at normal incidence having a surface area of 20 cm². The energy received on the surface during 1 min is:

1. 12 x 10³ J
2. 24 x 10³ J
3. 48 x 10³ J
4. 10 x 10³ J

Answer: 2. 24 x 10³ J

From the question, average flux = 20 W/cm², surface area 20 cm²

Time = 1 min = 60 s For non-reflecting surface

Energy received = average flux x surface area x time

= 20 x 20 x 60

= 24 x 10³ J.

Question 57. The energy required to break one bond in DNA is 10^-20 J. This value (in eV) is nearly:

1. 0.6
2. 0.062
3. 0.006
4. 6

Answer: 2. 0.062

The energy in electron volts is given as lev = 1.602 x 10^-19 J

The energy required to break the DNA bond is 10^-20J. So, the energy in terms of eV will be:

⇒ \(10^{-20} \mathrm{~J} =\frac{1}{1.602 \times 10^{-19}} \times 10^{-20} \mathrm{eV}\)

∴ \(\mathrm{E}_{e v} =0.062 \mathrm{eV}\)

Question 58. A particle of mass m is driven by a machine that delivers a constant power of k watts. If the particle starts from rest, the force on the particle at time t is:

1. \(\sqrt{\frac{m k}{2}} t^{-1 / 2}\)
2. \(\sqrt{m k} t^{-1 / 2}\)
3. \(\sqrt{2 m k} t^{-1 / 2}\)
4. \(\frac{1}{2} \sqrt{m k} t^{-1 / 2}\)

Answer: 1. \(\sqrt{\frac{m k}{2}} t^{-1 / 2}\)

According to the question. The machine delivers constant power

F. v = Constant

⇒ \(\frac{d v}{d t} \cdot v=\mathrm{K}\)

⇒ \(\int v d v=\frac{k}{m} \int d t\)

⇒ \(\frac{v^2}{2}=\frac{k}{m} t \Rightarrow v=\sqrt{\frac{2 k t}{m}}\)

force on the particle is F=\(m \frac{d v}{d t}=m \frac{d}{d t}\left(\sqrt{\frac{2 k t}{m}}\right)\)

=\(\sqrt{2 k m} \cdot\left(\frac{1}{2} t^{-1 / 2}\right)=\sqrt{\frac{m k}{2}} \cdot t^{-1 / 2}\)

Question 59. 300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking, g = 10 m/s², work done against friction is:

1. 1000 J
2. 200 J
3. 100J
4. Zero

Answer: 3. 100J

Potential energy, P.E. = 2 x 10 x 10 = 200 J

and workdone w = 300 J.

Work done against friction = 300-200= 100 J.

Question 60. An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms^-1. The minimum power delivered by the motor to the lift in watts is (g = 10 ms^-2):

1. 20000
2. 34500
3. 23500
4. 23000

Answer: 2. 34500

Given: v = 1.5 m/s

m = 2000 kg F = 3000 N

Here we are given that the velocity is constant, and acceleration is defined as the rate of change of velocity concerning time, therefore for the constant velocity the acceleration (a) is zero, a = 0

The figure of an electric lift is shown below:

Therefore, The tension T = W + F ⇒  T = 2000 x 10 + 3000

T = 20000 + 3000

T = 23000 N

Now, using equation (1) we have; P = Tv

P = 23000 x 1.5

P = 34500 watts.

Question 61. Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? (g = 10 m/s²)

1. 10.2 kW
2. 8.1 kW
3. 12.3 kW
4. 7.0 kW

Answer: 2. 8.1 kW

Given, Height (h) = 60 m

Mase/time = 15 kg/s

Loss = 10% ri = 90%

P = ? if g = 10 ms²

⇒ \(\text { Power }=\frac{\text { work }}{\text { time }}\)

=\(\frac{m g h}{t} \eta\)

⇒ \(\mathrm{P}=15 \times 16 \times 60 \times \frac{9}{16}\)

= 900 x 9 = 8100 J/S P = 8.1 kW

Question 62. A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time T is V, the power delivered to the particle is:

1. \(\frac{M v^2}{T}\)
2. \(\frac{1}{2} \frac{M v^2}{T^2}\)
3. \(\frac{M v^2}{T^2}\)
4. \(\frac{1}{2} \frac{M v^2}{T}\)

Answer: 4. \(\frac{1}{2} \frac{M v^2}{T}\)

K.E. of particle=\(\frac{1}{2} \mathrm{Mv}^2\)

⇒ \(\text { Power } =\frac{\text { Energy }}{\text { Time }}\)

∴ \(\mathrm{P} =\frac{1}{2} \frac{M v^2}{t}\)

Question 63. A body of mass 1 kg begins to move under the action of a time-dependent force F =\(\left(2 t \hat{i} \times 3 t^2 \hat{j}\right)\) N, where \(\hat{i}\) and \(\hat{j}\) are unit vectors along X and Y-axis. What power will be developed by the force at the time (t)?

1. (2t² + 4t⁴) W
2. (2t³ + 3t⁴) W
3. (2t³ + 3t⁴) W
4. (2t + 3t³) W

Answer: 3. (2t³ + 3t⁴) W

From the question, A body of mass 1 kg begins to move under the action of a time-dependent force isF=\(\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{N}\)

F=ma

a=\(\frac{F}{m} \text { and } \mathrm{m}=1 \mathrm{~kg}\)

a=\(\left(2 \hat{i}+3 t^2 \hat{j}\right) \mathrm{m} / \mathrm{s}^2\)

a=\(\frac{d v}{d t}\)

DV = adt

⇒ \(\int d v=\int a d t\)

v=\(\int\left(2 t \hat{i}+3 t^2 \hat{j}\right) d t=t^2 \hat{i}+t^3 \hat{j}\)

⇒ \(\text { Power, } P=F \cdot v\)

= \(\left(2 \hat{t} \hat{i}+3 t^2 \hat{j}\right) \cdot\left(t^2 \hat{t}+t_3 \hat{j}\right)\)

P=\(\left(2 t^3+3 t^5\right) \mathrm{W}\)

Question 64. One coolie takes 1 minute to raise a suitcase through a height of 2m but the second coolie takes 30 s to raise the same suitcase to the same height. The power two coolies are in the ratio of

1. 1 : 3
2. 2: 1
3. 3 :1
4. 1: 2

Answer: 4. 1: 2

We know that,\(\text { Power }=\frac{\text { Work done }}{\text { Time }}\)

Work done = mgh,

In same in both the case \(\frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{f_2}{f_1}\)

=\(\frac{30 \mathrm{sec}}{1 \mathrm{~min}}=\frac{1}{2}\)

Question 65. An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 ms^-1. The mass per unit length of water in the pipe is 100 km-1. What is the power of the engine?

1. 400 W
2. 200 W
3. 100 W
4. 800 W

Answer: 4. 800 W

We know that P=FV=maV

= \(m \times \frac{v}{t} \times v=\left(\frac{m}{l}\right)_t^l . v . v\)

⇒ \(\frac{m}{l}= \text { mass per unit }\)

= \(100 \text { (given) }\)

= \((100)\left(2^3\right)=800 \mathrm{w}\)

Question 66. Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10% of energy. How much power is generated by the turbine? (g= 10 m/s²)

1. 8.1 kW
2. 10.2 kW
3. 12.3 kW
4. 7.0 kW

Answer: 1. 8.1 kW

Power generated by the turbine is, \(P_{\text {generated }}=P_{\text {input }} \times \frac{90}{100}=\frac{M g h}{t} \times \frac{90}{100}\)

Putting the given value, \(\frac{M}{t}=15 \mathrm{~kg} / \mathrm{s}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\)

and h=60

P=\((15 \times 10 \times 60) \times \frac{90}{100}=8.1 \mathrm{kw}\)

Question 67. A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2: 2: 1. If the fragments having equal mass fly off in mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is:

1. \(\sqrt{2} v\)
2. \(2 \sqrt{2} v\)
3. \(3 \sqrt{2} v\)
4. v

Answer: 2. \(2 \sqrt{2} v\)

The total mass of the shell is m.

The ratio of the masses of the fragments is 2 : 2: 1

2x + 2x + x = m

x = 0.2m

Let the velocity of the lighter fragment be v ‘

⇒ \(\mathrm{P}^{\prime} =\sqrt{P_1^2+P_2^2}\)

⇒ \(0.2 m v^{\prime} =\sqrt{(0.4 m v)^2+(0.4 m v)^2}\)

⇒ \(0.2 m v^{\prime} =\sqrt{2} \times 0.4 m v\)

∴ \(v^{\prime} =2 \sqrt{2} v\)

Question 68. A mass of m moving horizontally (along the x-axis) with velocity v collides and strikes to mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is:

1. \(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)
2. \(\frac{1}{3} v \hat{i}+\frac{2}{3} v \hat{j}\)
3. \(\frac{2}{3} v \hat{i}+\frac{1}{3} v \hat{j}\)
4. \(\frac{3}{2} v \hat{i}+\frac{1}{4} v \hat{j}\)

Answer: 1. \(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)

⇒ \(m v+(3 m)(2 v) =(4 m) v^{\prime}\)

⇒ \(m v \hat{i}+6 m v \hat{j} =4 m v^{\prime}\)

⇒ \(v^{\prime} =\frac{1}{4} v \hat{i}+\frac{6}{4} v \hat{j}\)

=\(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)

Question 69. Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head-on-end elastic. After the collision, the fraction of energy lost by the colliding body A is:

1. \(\frac{8}{9}\)
2. \(\frac{4}{9}\)
3. \(\frac{5}{9}\)
4. \(\frac{1}{9}\)

Answer: 1. \(\frac{8}{9}\)

Fractional loss of KE. of the colliding body is \(\frac{4 \mathrm{KE}}{\mathrm{KE}}=\frac{4\left(m_1 m_2\right)}{\left(m_1+m_2\right)^2} =\frac{4(m)(2 m)}{(4 m+2 m)^2}\)

=\(\frac{32 m^2}{36 m^2}=\frac{8}{9} .\)

Question 70. Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head-on-end elastic. After the collision, the fraction of energy lost by the colliding body A is:

1. \(\frac{5}{9}\)
2. \(\frac{1}{9}\)
3. \(\frac{8}{9}\)
4. \(\frac{4}{9}\)

Answer: 3. \(\frac{8}{9}\)

According to the conservation of momentum, \(4 m u_1=4 m v_1+2 m v_2\)

⇒ \(2\left(u_1-v_1\right)=v_2\)  →   Equation – 1

From conservation of energy, \(\frac{1}{2}(4 m) u_1^2=\frac{1}{2}(4 m) v_1^2+\frac{1}{2}(2 \mathrm{~m}) v_2^2\)

⇒  \(2\left(u_1^2-v_1^2\right)=v_2^2\)  → Equation – 2

From 1 and 2,

⇒  \(2\left(u_1^2-v_1^2\right)=4\left(u_1-v_1\right)^2\)

⇒  \(3 v_1=u_1\) Equation – 3

Now, a fraction of loss in kinetic energy for mass 4 m, \(\frac{\Delta k}{k_i}=\frac{k_i-k_f}{k_i}\)

=\(\frac{\frac{1}{2}(4 m) u_1^2-\frac{1}{2}(4 m) v_1^2}{\frac{1}{2}(4 m) u_1^2}\) Equation – 4

Substituting 3 in 4, we get \(\frac{\Delta K}{K_i}=\frac{8}{9}\)

Question 71. A moving block having a mass of m collides with another stationary block having a mass of 4 m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of the coefficient of restitution (e) will be:

1. 0.8
2. 0.25
3. 0.5
4. 0.4

Answer: 2. 0.25

According to the law of conservation of linear momentum, we have, m₁U₁+ m₂u₂ = m₁V₁ + m₂v_2. Equation -1

Where m₁ and m₂ are the masses of two bodies, their initial velocities are u₁ and u₂, and final velocities are v₁ and v₂ respectively in the equation m₁ = m₂ = 4m

u₁ = V₁ u₂ = 0 and v₁ = 0

In equation. 1,

mv + 4m x 0 = 0 + 4 mv₂

⇒ \(\quad v_2=\frac{v}{4}\)

Now coefficient of restitution e,

e=\(\frac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}=\frac{\frac{v}{4}}{v}\)

e=\(\frac{1}{4}=0.25\)

Question 72. Two identical balls A and B having velocities of 0.5 m/s and – 0.3 m/s respectively, collide elastically in one dimension. The velocities of B and A after the collision respectively will be:

1. – 0.5 m/s and 0.3 m/s
2. 0.5 m/s and 0.3 m/s
3. – 0.3 m/s and 0.5 m/s
4. 0.3 m/s and 0.5 m/s

Answer: 3. – 0.3 m/s and 0.5 m/s

Since both bodies are identical and collision is elastic. Therefore velocities will be interchanged after collision.

v_A = – 0.3 m/s and v_B = 0.5 m/s

Question 73. A bullet of mass 10 g moving horizontally with a velocity of 400 m s^-1 strikes a wooden block of mass 2 kg which is suspended by light in an extensible string of length 5 m. As a result, the center of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges horizontally from the block will be:

1. 100 ms^-1
2. 80 ms^-1
3. 120 ms^-1
4. 160 ms^-1

Answer: 3. 120 ms^-1

Mass of bullet,m= 10 g=0.01 kg

The initial speed of a bullet, u=400 ms^-1

Mass of block, M=2 kg

Length of string, l= 5 m

Speed of the block after collision = v₁

Speed of the bullet on emerging from block, v=?

Using the energy conservation principle from the block,

⇒ \((\mathrm{KE}+\mathrm{PE})_{\text {Reference }}=(\mathrm{KE}+\mathrm{PE})_{\mathrm{h}}\)

⇒ \(\frac{1}{2} \mathrm{M} v_1^2=\mathrm{Mgh}\) or

⇒ \(v_1=\sqrt{2 g h}\)

⇒ \(v_1=\sqrt{2 \times 10 \times 0.1}=\sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\)

Using the momentum conservation principle for block and bullet systems,

⇒ \((M \times 0+m u)_{\text {Before collision }} =\left(M \times v_1+m v\right)_{\text {after collision }}\)

⇒ \(0.01 \times 400 =2 \frac{1}{2}+0.01 \times v\)

v =\(\frac{4-2 \sqrt{2}}{0.01}\)

=\(117.15 \mathrm{~m} \mathrm{~s}^{-1} \approx 120 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 74. Two particles of masses m₁, and m₂ move with initial velocities u₁ and u₂. On collision, one of the particles gets excited to a higher level, after absorbing energy If the final velocities of particles are v₁ and v₂ then we must have:

1. \(m_1^2 u_1+m_2^2 u_2-\mathrm{E}=m_1^2 v_1+m_2^2 v_2\)
2. \(\frac{1}{2} m_1^2 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m^2 v_2^2-\mathrm{E}\)
3. \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2-\mathrm{E}=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)
4. \(\frac{1}{2} m_1^2 u_1^2+\frac{1}{2} m_2^2 u_2^2+\mathrm{E}=\frac{1}{2} m_1^2 v_1^2+\frac{1}{2} m_2^2 v_2^2\)

Answer: 3. \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2-\mathrm{E}=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)

Let, m₁, m₂ be the masses of particles u₁, u₂ be the initial velocities of particles and v₁, V₂ be the final velocities of particles

from the law of conservation of energy

Initial total energy = final total energy

∴ \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2+\mathrm{E}\)

Question 75. A ball is thrown vertically downwards from a height of 20 m with an initial velocity of V_Q. It collides with the ground, loses 50% of its energy in a collision, and rebounds to the same height. The initial velocity v₀ is: (Take, g = 10 ms^-2)

1. 14 ms^-1
2. 20 ms^-1
3. 28 ms^-1
4. 10 ms^-1

Answer: 2. 20 ms^-1

Let v be the speed of the ball then \(v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 20}=20 \mathrm{~m} / \mathrm{s}\)

Now the energy of a ball just after rebound \(E=\frac{1}{2} m v^2=200 \mathrm{~m}\)

According to the question, 50% of energy cases in a collision means just before collision energy is 400 m.

Using law of conservation of energy,\(\frac{1}{2} m v_0^2+m g h=400\)

⇒ \(\frac{1}{2} m v_0^2+m \times 10 \times 20\)=400

∴ \(v_0 =20 \mathrm{~m} / \mathrm{s}\)

Question 76. On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of the same mass M which is initially at rest. After the collision, the first block moves at an angle θ to its initial direction and has a speed \(\frac{v}{3}\). The second block’s speed after the collision is:

1. \(\frac{2 \sqrt{2}}{3} v\)
2. \(\frac{3}{4} v\)
3. \(\frac{3}{\sqrt{2}} v\)
4. \(\frac{\sqrt{3}}{2} v\)

Answer: 1. \(\frac{2 \sqrt{2}}{3} v\)

According to the question, Using Law Of Conservation of Kinetic Energy

⇒ \(\frac{1}{2} M v^2+0=\frac{1}{2} M\left(\frac{v}{3}\right)^2+\frac{1}{2} M v_2^2\)

⇒  \(v^2=\frac{v^2}{9}+v_2^2\)

⇒  \(v_2^2=\frac{8 v^2}{9}\)

∴ \(v_2=\frac{2 \sqrt{2}}{3} v\)

Question 77. Two particles A and B, move with constant velocities \(\vec{r}_1 \text { and } \overrightarrow{r_2}\). At the initial moment their position vectors are \(\vec{r}_1 \text { and } \overrightarrow{r_2}\) respectively. The condition for particles A and B for their collision is:

1. \(\overrightarrow{r_1} \times \overrightarrow{v_1}=\overrightarrow{r_2} \times \overrightarrow{v_2}\)
2. \(\overrightarrow{r_1}-\overrightarrow{r_2}=\overrightarrow{v_1}-\overrightarrow{v_2}\)
3. \(\frac{\overrightarrow{r_1}-\overrightarrow{r_2}}{\left|\overrightarrow{r_1}-\overrightarrow{r_2}\right|}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{\left|\overrightarrow{v_2}-\overrightarrow{v_1}\right|}\)
4. \(\overrightarrow{r_1}, \overrightarrow{v_1}=\overrightarrow{r_2}, \overrightarrow{v_1}\)

Answer: 3. \(\frac{\overrightarrow{r_1}-\overrightarrow{r_2}}{\left|\overrightarrow{r_1}-\overrightarrow{r_2}\right|}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{\left|\overrightarrow{v_2}-\overrightarrow{v_1}\right|}\)

Let the particles A and B collide at time t. For their collision, the position vectors of both particles should be the same at time t,\(\vec{r}_1+\vec{v}_1 t=\vec{r}_2+\vec{v}_2 t\)

⇒ \(\vec{r}_1-\vec{r}_2 \approx \vec{v}_2 t-\vec{v}_1 t =\left(\vec{v}_2-\vec{v}_1\right) t\) Equation – 1

⇒ \(\left|\vec{r}_1-\vec{r}_2\right| =\left|\vec{v}_2-\vec{v}_1\right| t\)

or t =\(\frac{\left|\vec{r}_1-\vec{r}_2\right|}{\left|\vec{v}_2-\vec{v}_1\right|}\)

Substituting this value of t in eqn. 1, we get,

∴ \(\vec{r}_1-\vec{r}_2=\left(\vec{v}_2-\vec{v}_1\right) \frac{\left|\vec{r}_1-\vec{r}_2\right|}{\left|\vec{v}_2-\vec{v}_1\right|}\)

Question 78. A ball is thrown vertically downwards from a height of 20 m with an initial velocity of v₀. It collides with the ground, loses 50 percent of its energy in a collision, and rebounds to the same height. The initial velocity v0 is (Take g = 10 m s²):

1. 28 m s^-2
2. 10 m s^-2
3. 14 ms^-2
4. 20 m s^-2

Answer: 4. 20 m s^-2

The situation is shown in the figure.

Let v be the velocity of the ball with which it collides with the ground. Then according to the law of conservation of energy,

The gain in kinetic energy = loss in potential energy

⇒ \(\frac{1}{2} m v^2-m v_0^2=m g h\) (where m is the mass of the ball) → Equation 1

Now, when the ball collides with the ground, 50% of its energy is lost and it rebounds to the same height h

⇒ \(\frac{50}{100}\left(\frac{1}{2} m v^2\right) =m g h\)

⇒ \(\frac{1}{4} v^2 =g h\)

or \(v^2 =4 g h\)

Substituting this value of v² in eqn. (1), we get \(4 g h-v_0^2=2 g h\)

Here, \(\mathrm{g}=10 \mathrm{~ms}^{-1} and h=20 \mathrm{~m}\)

∴ \(v_0 =\sqrt{2\left(10 \mathrm{~ms}^{-1}\right)(20 m)} 20 \mathrm{~ms}^{-1} =20 \mathrm{~ms}^{-1}\)

Question 79. On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of the same mass M which is initially at rest. After the collision, the first block moves at an angle θ to its initial direction and has a speed of v/3. The second block’s speed after the collision is:

1. \(\frac{3}{\sqrt{2}} v\)
2. \(\frac{\sqrt{3}}{2} v\)
3. \(\frac{2 \sqrt{2}}{3} v\)
4. \(\frac{3}{4} v\)

Answer: 3. \(\frac{2 \sqrt{2}}{3} v\)

The situation is shown in the figure.

Let v’ be the speed of the second block after the collision. As the collision is elastic, so kinetic energy is conserved. According to the conservation of kinetic energy,

⇒ \(\frac{1}{2} M v^2+0 =\frac{1}{2} M\left(\frac{v}{3}\right)^2+\frac{1}{2} M v^{\prime 2}\)

∴ \(v^2 =\frac{v^2}{9}+v^{\prime} 0^2\)

Question 80. Two spheres A and B of masses my and m2 respectively collide. A is at rest initially and after collision B has a velocity \(\frac{v}{2}\) in a direction perpendicular to the original direction. The mass A moves after collision in the direction:

1. same as that of B
2. opposite to that of B
3. a = tan^-1 \(\left(\frac{1}{2}\right)\) to the x-axis
4. a = tan^-1 \(\left(\frac{-1}{2}\right)\)to the x-axis

Answer: 1. same as that of B

There is no external force acting on the spheres. So linear momentum will be conserved.

Before the collision, In direction x, Linear momentum = m₂v  → Equation 1

In direction y, Linear momentum = 0

After the collision, the spheres move as shown in the figure, let the velocity of sphere A be v₁

In direction x, Linear momentum = m₁v₁cos(θ) →  Equation …(2)

In direction y, Linear momentum = \(\frac{m_2 v}{2} m_1 v_1 \sin (\theta)\)  → Equation 3

Linear momentum will be conserved, From equations 1 and 2, m₁v₁cosθ=0

From the equation 2,\(\frac{m_2 v}{2} m_1 v_1 \sin (\theta)\)=0

⇒ \(=m_1 v_1 \sin (\theta)=\frac{m_2 v}{2}\) Equation 5

Dividing equation 5 by 4,

⇒ \(\tan (\theta) =\frac{1}{2}\)

∴ \(\theta =\tan ^{-1}\left(\frac{1}{2}\right)\) Equation 5

Question 81. A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to a mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is:

1. \(\frac{3}{4} v \hat{i}+\frac{1}{4} v \hat{j}\)
2. \(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)
3. \(\frac{1}{3} v \hat{i}+\frac{2}{3} v \hat{j}\)
4. \(\frac{2}{3} v \hat{i}+\frac{1}{3} v \hat{j}\)

Answer: 2. \(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)

According to conservation of momentum, we get\(m v^{\frac{2 a}{b}}+(3 m) 2 v^{\left(\frac{2a}{b}\right)^{\prime 1/6}}=(m+3 m) \overrightarrow{v^{\prime}}\)

where v ‘ is the final velocity after collision\(\overrightarrow{v^{\prime}}=\frac{1}{4} v \hat{i}+\frac{6}{4} v \hat{j}=\frac{1}{4} v \hat{i}+\frac{3}{4} v \hat{j}\)

Question 82. A ball moving with velocity ms^-1 collides head-on with another stationery ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in ms^-1) after the collision will be:

1. 0, 1
2. 1, 1
3. 1,0.5
4. 0,2

Answer: 1. 0, 1

The situation is shown as :\(v_1=\frac{\left(m_1-e m_2\right)}{m_1+m_2} u_1+0 \times u_2\)

∴ \(v_1=\left[\frac{m-(0.5)(2 m)}{3 m}\right]\)=0

Question 83. Two equal masses m₁ and m₂ moving along the same straight line with velocities + 3 m/s and -5 m/s respectively collide elastically. Their velocities after the collision will be respectively:

1. – 4 m/s and + 4 m/s
2. + 4 m/s for both
3. – 3 m/s and + 5 m/s
4. – 5 m/s and + 3 m/s

Answer: 4. – 5 m/s and + 3 m/s

Equal masses after elastic collision interchanges their velocities. – 5 m/s and + 3 m/s.

Question 84. A rubber ball is dropped from a height of 5 m on a plane. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of:

1. \(\frac{3}{5}\)
2. \(\frac{2}{5}\)
3. \(\frac{16}{25}\)
4. \(\frac{9}{25}\)

Answer: 1. \(\frac{3}{5}\)

Initial energy equation \(m g h=\frac{1}{2} m v^2 \text { i.e. } 10 \times 5=\frac{1}{2} v_1^2\)

V₁=10

After one Bounce,\(10 \times 1.8=\frac{1}{2} v_2^2\)

v₂=6

∴ Loss in velocity On bouncing \(\frac{6}{10}=\frac{3}{5}\)

Question 85. A metal ball of mass 2 kg moving at the speed of 36 km/h has a head-on collision with a stationary ball of mass 3 kg. If after collision, both the balls move as a single mass, then the loss in K.E. due to collision is:

1. 100 J
2. 140 J
3. 40 J
4. 60 J.

Answer: 4. 60 J

Mass of metal ball = 2 kg;

Speed of metal ball (v₁) = 36 km/h = 10 m/s and

mass of stationary ball = 3 kg

Applying law of conservation of momentum, m₁v₁ + m₂v₂ = (m₁ + m₂)v

or \(\mathrm{v} =\frac{m_1 v_1+m_2 v_2}{m_1+m_2}\)

=\(\frac{(2 \times 10)+(3 \times 0)}{2+3}=\frac{20}{5}=4 \mathrm{~m} / \mathrm{s}\)

Question 86. A moving body of mass m and velocity 3 km/hour collides with a body at rest of mass 2m and sticks to it. Now the combined mass starts to move. What will be the combined velocity?

1. 3 km/hour
2. 4 km/hour
3. 1 km/hour
4. 2 km/hour

Answer: 3. 1 km/hour

Mass of body (m₁) = m;

Velocity of first body (u₁) = 3 km/hour;

Mass of second body at rest (m₂) = 2m and velocity of second body (u₂) = 0.

After combination, mass of the body M= m + 2m = 3m

From the law of conservation of momentum, we get Mv = m₁U₁ + m₂u₂

or 3mv = (m x3) + (2m x 0) = 3m

or v = 1 km/hour.

Question 87. The coefficient of restitution e for a perfectly elastic collision is:

1. 1
2. zero
3. infinite
4. -1

Answer: 1. 1

A quantity known as the coefficient of restitution or coefficient of resilience of the collision determines the degree of elasticity of the impact. It is defined as the ratio of relative separation velocity after a collision to relative approach velocity before a collision. It is denoted by the letter e. relative velocity of separation (after collision)

=\(\frac{\text { relative velocity of separation(after collision) }}{\text { relative velocity of approach (before collision) }}\)

⇒\(e=\frac{v_2-v_1}{u_1-u_2}\)

where, u₁ and u₂ are the velocities of two bodies before collision, and u₁, and u₂ are the velocities of the two bodies after the collision. The relative velocity of separation after a perfectly elastic collision is equal to the relative velocity of the approach before the collision.

∴ e = 1

Laws of Motion

Question 1. Physical independence of force is a consequence of:

1. third law of motion
2. second law of motion
3. first law of motion
4. all of these laws

Answer: 3. first law of motion

Newton’s first law of motion is related to the physical independence of force.

Question 2. A ball of mass 0.15 kg is dropped from a height of 10 m, strikes the ground, and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s²) nearly:

1. 0 kg m/s
2. 4.2 kg m/s
3. 2.1 kg m/s
4. 1.4 kg m/s

Answer: 2. 4.2 kg m/s

Given, height (h) = 10m

mass (m) = 0.15 kg

⇒ Velocity at ground, v =\(\sqrt{2 g h}\)

v=\(\sqrt{2 \times 10 \times 10}\)

=\(10 \sqrt{2} \mathrm{~m} / \mathrm{s}\)

⇒ \(\Delta \mathrm{P}=\mathrm{P}_f-\mathrm{P}_i\)

⇒ Since rebound takes place for the ball to reach the same height, hence, P_f =-P_i

=\((0.15 \times 10 \sqrt{2})\)

⇒ \(\Delta \mathrm{P} =3 \sqrt{2}\)

=4.2 kg m/s

Question 3. The moment of the force F =\(4 \hat{i}+5 \hat{j}-6 \hat{k}\) at (2, 0 – 3)m about the point (2,-2,-2), is given by:

1. \(-7 \hat{i}+8 \hat{j}-4 \hat{k}\)
2. \(4 \hat{i}+\hat{j}-8 \hat{k}\)
3. \(-8 \hat{i}-4 \hat{j}-7 \hat{k}\)
4. \(-7 \hat{i}-4 \hat{j}-8 \hat{k}\)

Answer: 4. \(-7 \hat{i}-4 \hat{j}-8 \hat{k}\)

Read and Learn More NEET Physics MCQs

Moment of force = Torque

⇒ \(\mathrm{F}=4 \hat{i}+5 \hat{j}-6 \hat{k}\)

⇒ \(r_0=2 \hat{i}+0 \hat{j}-3 \hat{k}\)

∴ \(\mathrm{r} =2 \hat{i}+2 \hat{j}-2 \hat{k}\)

Question 4. The force F acting on a particle of mass m is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from 0 s to 8 s is:

1. 14 Ns
2. 20 Ns
3. 12 Ns
4. 6 Ns

Answer: 3. 12 Ns

The f-t graph is

⇒ Now, Change of momentum,For 1^st → ΔP₁ = \(\frac{1}{2}\) x 2 x 6 = 6 kg-m/s

⇒ For 2^nd– ΔP₂ = 2 (- 3) = – 6 kg-m/s  ⇒ For 3^rd  ΔP₃ = 4X3 = 12 kg-m/s

∴ Total change in momentum,ΔP = ΔP₁ + ΔP₂ + ΔP₃ = 6-6+12 = 12N-s.

Question 5. Three blocks with masses m, 2m, and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m? (g is the acceleration due to gravity)

1. Zero
2. 2 mg
3. 3 mg
4. 6 mg

Answer: 1. Zero

As a block of mass 2 m moves with constant velocity the net force is 2 cm [Since, Velocity = constant

⇒ acceleration = 0

∴ F = m x a = m x 0 = 0

Question 6. A person holding a rifle (mass of person and rifle together is 100 kg) stands on a smooth surface and fires 10 shots horizontally, in 5 s. Each bullet has a mass of 10 g with a muzzle velocity of 800 ms^-1. The final velocity acquired by the person and the average force exerted on the person are:

1. – 0.08 ms^-4, 16 N
2. – 0.8 ms^-4,16 N
3. – 0.08 ms^-4, 16 N
4. – 0.08 ms^-4, 16 N

Answer: 2. – 0.8 ms^-4,16 N

Using the law of conservation of moments, Initial momentum = mu₁ + mu₂

⇒ final momentum = nmv₁+ (M- nm)v₂ mu₁ + mu₂ = nmv₁ + (M- nm)v₂

⇒ 0=\(\frac{10 \times 10}{1000} \times 800+\left(100-\frac{10 \times 10}{1000}\right) v_2\)

⇒ -80=\(\frac{999}{10} v_2\)

∴ v₂ = -0.8ms^-1

⇒ And average force exerted on the person\(\frac{n \times m \times u}{\Delta t}\)

=\(\frac{10 \times \frac{10}{1000} \times 800}{5}\)

=\(\frac{80}{5}=16 \mathrm{~N}\)

Question 7. A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 ms^-2, the velocity with which it hits the ground is:

1. 20 m/s
2. 40 m/s
3. 5 m/s
4. 10 m/s

Answer: 1. 20 m/s

∴ Since, \(v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 20}=20 \mathrm{~m} / \mathrm{sec}\)

Question 8. A body, under the action of a force\(\overrightarrow{\mathrm{F}}=6 \hat{i}-8 \hat{j}+10 \hat{k}\), acquires an acceleration of 1 ms^-2. The mass of this body must be:

1. \(2 \sqrt{10} \mathrm{~kg}\)
2. 10
3. 20
4. \(10 \sqrt{2} \mathrm{~kg}\)

Answer: 4. \(10 \sqrt{2} \mathrm{~kg}\)

We know that, force = mass x acceleration

⇒ \(\vec{F} =6 \hat{i}-8 \hat{j}+10 \hat{k}\)

⇒ |F| =\(\sqrt{6^2+(8)^2+10^2}\)

=\(\sqrt{36+64+100}\)

=\(10 \sqrt{2} \mathrm{~N}\)

⇒ a =\(1 \mathrm{~ms}^{-2}\)

∴ m =\(\frac{10 \sqrt{2}}{1}=10 \sqrt{2} \mathrm{~kg}\)

Question 9. An object of mass 3 kg is at rest. Now a force of \(\vec{F}=6 t^2 \hat{i}+4 \hat{j}\) is applied on the object then velocity of object at t = 3 s is:

1. \(18 \hat{i}+3 \hat{j}\)
2. \(18 \hat{i}+6 \hat{j}\)
3. \(3 \hat{i}+18 \hat{j}\)
4. \(18 \hat{i}+4 \hat{j}\)

Answer: 1. \(18 \hat{i}+3 \hat{j}\)

⇒ Mass, m = 3 kg, force, F =\(\mathrm{F}=6 \mathrm{t} \hat{i}+4 t \hat{j}\)

⇒ Acceleration a =\(\frac{F}{m}=\frac{6 t^2 \hat{i}+4 t \hat{j}}{3}\)

=\(2 t^2 \hat{i}+\frac{4}{3} t \hat{j}\)

⇒ \(\mathrm{a} =\frac{d v}{d t}=2 t^2 \hat{i}+\frac{4}{3} t \hat{j}\)

⇒ DV =\(\left(2 t^2 \hat{i}+\frac{4}{3} t \hat{j}\right) d t\)

⇒ v =\(\int_0^3\left(2 t^2 \hat{i}+\frac{4}{3} t \hat{j}\right) d t\)

=\(\frac{2}{3} t^3 \hat{i}+\left.\frac{4}{6} t^2 \hat{j}\right|_0 ^3=18 \hat{i}+6 \hat{j}\)

Question 10. A cricketer catches a ball of mass 150 gm in 0.1 sec moving with a speed of 20 m/s, then he experiences a force of

1. 300 N
2. 30 N
3. 3 N
4. 0.3 N

Answer: 2. 30 N

Impulse = Change in momentum \(\mathrm{F} \cdot \Delta \mathrm{t} =\mathrm{m} \mathrm{v}\)

⇒ \(\mathrm{F}=\frac{m \cdot v}{\Delta t}\)

=\(\frac{150 \times 10^{-3} \times 20}{0.1}=30 \mathrm{~N}\)

Question 11. If the force on a rocket, moving with a velocity of 300 m/s is 210 N, then the rate of combustion of the fuel is:

1. 0.07 kg/s
2. 1.4 kg/s
3. 0.7 kg/s
4. 10.7 kg/s

Answer: 2. 1.4 kg/s

⇒ \(\text { Force }=\frac{d}{d t}(\text { momentum })\)

=\(\frac{d}{d t}(m v)=v\left(\frac{d m}{d t}\right)\)

⇒ \(210=300\left(\frac{d m}{d t}\right)\)

⇒ \(\frac{d m}{d t}\)= rate of combustion

=\(\frac{210}{300}=0.7 \mathrm{~kg} / \mathrm{s}\)

Question 12. A bullet is fired from a gun. The force on the bullet is given by F = 600 – 2 x 10⁵ t. Where F is in Newton and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

1. 9 Ns
2. zero
3. 1.8 Ns
4. 0.9 Ns

Answer: 4. 0.9 Ns

F =0,600-2 x times 10⁵ t=0

t =\(\frac{600}{2 \times 10^5}=3 \times 10^{-3} \mathrm{~s}\)

⇒ Now, Impulse I =\(\int_0^t F d t=\int_0^t\left(600-2 \times 10^5 t\right) d t\)

=\(600 t-2 \times 10^5 \frac{t^2}{2}\)

Question 13. A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 m s^-1. To give an initial upward acceleration of 20 m s^-1. the amount of gas ejected per second to supply the needed thrust will be (g = 10 m s^-2):

1. 185. kgs^-1
2. 187.5 kgs^-1
3. 127.5 kg s^-1
4. 137.5 kg s^-1

Answer: 2. 187.5 kgs^-1

\(\text { Thrust }=\mathrm{M}(g+a)=u \frac{d m}{d t}\) \(\frac{d m}{d t} =\frac{M(g+a)}{u}=\frac{5000(10+20)}{800}\)

=\(187.5 \mathrm{~kg} / \mathrm{s}\)

Question 14. A force of 6 N acts on a body at rest and of mass 1 kg. During this time, the body attains a velocity of 30 m/s. The time for which the force acts on the body is:

1. 7 seconds
2. 5 seconds
3. 10 seconds
4. 8 seconds

Answer: 2. 5 seconds

Force (F) = 6 N;

Initial velocity (u) = 0;

Mass (m) = 1 kg and final velocity (v) = 30 m/s.

Acceleration \((a)=\frac{F}{m}=\frac{6}{1}=6 \mathrm{~m} / \mathrm{s}^2\)

velocity, (v) = 30 = u + at = 0 + 6 x t

Question 15. A 10 N force is applied to a body to produce an acceleration of 1 m/s². The mass of the body is:

1. 15 kg
2. 20 kg
3. 10 kg
4. 5 kg

Answer: 3. 10 kg

Force (F) = 10 N and acceleration (a) = 1 m/s².

⇒ Mass(m)=\(\frac{F}{a}=\frac{10}{1}\) = 10kg

Question 16. In a rocket, fuel bums at the rate of 1 kg/s. This fuel is ejected from the rocket with a velocity of 60 km/s. This exerts a force on the rocket equal to:

1. 6000 N
2. 60000
3. 60 N
4. 600 N

Answer: 2. 60000

Rate Of Burning Of Fuel\(\left(\frac{d m}{d t}\right)\)=1kg/s^-1

and velocity of ejected fuel, (v) = 60 km/s = 60 x 10³ m/s

⇒Force = Rate of change of momentum\(\frac{d p}{d t}=\frac{d(m v)}{d t}=v \frac{d m}{d t}\)

∴ (60 x 10³)x 1 =60000N

Question 17. A satellite in force-free space sweeps stationary interplanetary dust at a rate of \(\frac{d M}{d t}=\alpha v\), where M is mass and v is the speed of the satellite and a is a constant. The acceleration of the satellite is:

1. \(\frac{-\alpha v^2}{2 M}\)
2. \(-\alpha v\)
3. \(\frac{-2 \alpha v^2}{M}\)
4. \(\frac{-\alpha v^2}{M}\)

Answer: 4. \(\frac{-\alpha v^2}{M}\)

⇒ Rate of change of mass \(\frac{d M}{d t} \alpha \mathrm{v}\)

⇒ Retarding force = Rate of change of momentum

= Velocity x Rate of change in mass

⇒ \(-v \times \frac{d M}{d t}= -v \times \alpha v=\alpha v^2\)

∴ Therefore, acceleration=\(-\frac{\alpha v^2}{M}\)

Question 18. A particle of mass m is moving with a uniform velocity V₁. It is given an impulse such that its velocity becomes V₂– The impulse is equal to:

1. \(m\left[\left|\mathbf{v}_2\right| \times\left|\mathbf{v}_1\right|\right]\)
2. \(\frac{1}{2} m\left[v_2^2-v_1^2\right]\)
3. \(m\left[\mathrm{v}_1+\mathrm{v}_2\right]\)
4. \(m\left(\mathrm{v}_2-\mathrm{v}_1\right)\)

Answer: 4. \(m\left(\mathrm{v}_2-\mathrm{v}_1\right)\)

Impulse is a vector quantity and is equal to the change in momentum of the body thus,(same as F x / where t is short)

∴ Impulse = mv₂ – mv₁ = m(v₂ – v₁)

Question 19. A 600 kg rocket is set for a vertical firing. If the exhaust speed is 1000 m s^-1, the mass of the gas ejected per second to supply the thrust needed to overcome the weight of the rocket is:

1. 117.6kgs^-1
2. 58.6kgs^-1
3. 6 kg s^-1
4. 76.4 kg s^-1

Answer: 3. 6 kg s^-1

⇒ Thrust is the force with which the rocket moves upward given by \(\mathrm{F}=\mathrm{u} \frac{d m}{d t}\)

⇒ Thus the mass of the gas ejected per second to supply the thrust needed to overcome the weight of the rocket is \(\frac{d m}{d t}=\frac{F}{u}=\frac{m \times a}{u}\)

∴ or \(\frac{d m}{d t}=\frac{600 \times 10}{1000}=6 \mathrm{~kg} \mathrm{~s}^{-1}\)

Question 20. A rigid ball of mass m strikes a rigid wall at 60° and gets reflected without loss of speed as shown in the figure. The value of impulse imparted by the wall on the ball will be:

1. mv
2. 2mv
3. mv/2
4. mv/3

Answer: 1. mv

∴ Impulse,\(|\overrightarrow{\Delta P}|=m|\overrightarrow{\Delta V}|\)= m (2 x cos 60°) = mv

Question 21. A body of mass M hits normally a rigid wall with velocity v and bounces back with the same velocity. The impulse experiment by the body is:

1. 2 Mv
2. 1.5 Mv
3. zero
4. Mv

Answer: 1. 2 Mv

Impulse = Change in linear momentum = MV – (-MV) = 2 MV

Question 22. A man of 50 kg mass is standing in a gravity-free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed of 2 ms^-1. When the stone reaches the floor, the distance of the man above the floor will be:

1. 9.9 m
2. 10.1m
3. 10 m
4. 20 m

Answer: 1. 9.9 m

From the question, Mr = constant

m₁r₁ = m₂r₂

∴ \(r_2=\frac{m_1 r_1}{m_2}=\frac{0.5 \times 10}{50}=0.1\)

Question 23. Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v m/s will be:

1. Mv newton
2. 2 Mv newton
3. \(\frac{\mathrm{Mv}}{\mathrm{v}^2}\)newton
4. zero

Answer: 1. Mv newton

Force(F)=\(\frac{d(m v)}{d t}\)

⇒ \(v\left(\frac{d m}{d t}\right)\)=v M

as velocity v is constant

∴ Hence F = Mv Newton

Question 24. A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed at the same angle. If the ball is in contact with the wall for 0.25 seconds, the average force acting on the wall is:

1. 96 N
2. 48 N
3. 24 N
4. 12 N

Answer: 3. 24 N

Components of momentum parallel to the wall are in the same direction and components of momentum perpendicular to the wall are opposite to each other. Therefore change of momentum = 2mvsinθ

F x t = Change in momentum = 2 mv sinθ

⇒ F =\(\frac{2 m v \sin \theta}{t}\)

=\(\frac{2 \times 0.5 \times 12 \times \sin 30^{\circ}}{0.25}\)

∴ \(48 \times \frac{1}{2}=24 \mathrm{~N}\)

Question 25. A body of mass 3 kg hits a wall at an angle of 60° and returns at the same angle. The impact time was 0.2 s. The force exerted on the wall is:

1. \(150 \sqrt{3} \mathrm{~N}\)
2. \(50 \sqrt{3} \mathrm{~N}\)
3. 100
4. \(75 \sqrt{3} \mathrm{~N}\)

Answer: 1. \(150 \sqrt{3} \mathrm{~N}\)

⇒ Change in momentum- mv₂sin0- (mv₁sinθ)= 2 mvsinθ = 2 x 3 X 10 x sin60°

⇒ 60x \(\frac{\sqrt{3}}{2}\)

⇒ Force =\(\frac{\text { Change in momentum }}{\text { Impact time }}\)

=\(\frac{30 \sqrt{3}}{0.2}=150 \sqrt{3} \mathrm{~N}\)

Question 26. An object flying in air with velocity \((20 \hat{i}+25 \hat{j}-12 \hat{k})\) suddenly breaks into two pieces whose masses are in the ratio 1: 5. The smaller mass flies off with a velocity \((100 \hat{i}+35 \hat{j}-8 \hat{k})\). The velocity of the larger piece will be:

1. \((4 \hat{i}+23 \hat{j}-16 \hat{k})\)
2. \((-100 \hat{i}+35 \hat{j}-8 \hat{k})\)
3. \((20 \hat{i}+15 \hat{j}-80 \hat{k})\)
4. \((-20 \hat{i}-15 \hat{j}-80 \hat{k})\)

Answer: 1. \((4 \hat{i}+23 \hat{j}-16 \hat{k})\)

From the law of conservation of linear momentum\(m \vec{v}=m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}\)

⇒ \(6 \mathrm{k}(20 \hat{i}+25 \hat{j}-12 \hat{k})=(100 \hat{i}+35 \hat{j}+8 \hat{k})+5 k \hat{v}_2\)

⇒ \(5 \vec{v}_2=(120-100) \hat{i}+(150-35) \hat{j}+(-72-8) \hat{k}\)

⇒ \(5 \vec{v}_2=20 \hat{i}+115 \hat{j}-80 \hat{k}\)

∴ \(\vec{v}_2=4 \hat{i}+23 \hat{j}-16 \hat{k}\)

Question 27. What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

1. \(\sqrt{2 g \mathrm{R}}\)
2. \(\sqrt{3 g R}\)
3. \(\sqrt{5 g \mathrm{R}}\)
4. \(\sqrt{g \mathrm{R}}\)

Answer: 3. \(\sqrt{5 g \mathrm{R}}\)

When the minimum speed of the body is \(\sqrt{5 g R}\) then no matter from where it enters the loop, it will complete a full vertical loop.

Question 28. A bullet of mass 10 g moving horizontally with a velocity of 400 m/s strikes a wood block of inextensible string of length 5 m. As a result, the center of gravity of the block was found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges horizontally from the block will be:

1. 100 m/s
2. 80 m/s
3. 120 m/s
4. 160 m/s

Answer: 3. 120 m/s

Applying law of conservation of momentum\(\frac{10}{1000} \times 400+0\)

⇒ \(2 \times v_1+\frac{10}{1000} \times v_2\)

4 = 2v₁ + 0.01 v₂        →    Equation – 1

⇒from the work-energy theorem for block, W=AKE

⇒ \(2 \times 10 \times 0.1=\frac{1}{2} \times 2 \times v_1^2\)

⇒ \(v_1=\sqrt{2}=1.4 \mathrm{~m} / \mathrm{s}\)

Putting the value of V₁ in the equation. 1

⇒ 4 = 2 x 1.4 + 0.01 v₂

∴ v₂= 120 m/s

Question 29. A person holding a rifle (mass of person and rifle together is 100 kg) stands on a smooth surface and fires 10 shots horizontally, in 5 s. Each bullet has a mass of 10 g with a muzzle velocity of 800 m s^-1. The final velocity acquired by the person and the average force exerted on the person are:

1. -0.8 ms^-1, 16N
2. – 0.8 ms¹, 8N
3. -1.6ms^-1, 16N
4. -1.6ms^-1, 8N

Answer: 1. -0.8 ms^-1, 16N

P (initial) = P (final)

⇒ 0 = n x m x u + (M-n x m) x v where: n = 10, m = 10g = 0.01kg, u = 800m/s, M = 100kg

⇒ 0 = 10 x 0.01kg x 800m/s + (100kg – 10 x 0.01kg) x v

⇒ v=\(\frac{80 \mathrm{kgm} / \mathrm{s}}{99.9 \mathrm{kgm} / \mathrm{s}}\)

⇒ v=-0.8m/s

∴ Then F= \(\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{10 \times 0.01 \mathrm{~kg} \times 800 \mathrm{~m} / \mathrm{s}}{5 \mathrm{~s}}=16 \mathrm{~N}\)

Question 30. An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms^-1 and the second part of mass 2 kg moves with 8 ms 1 speed, if the third part flies off with 4 ms-1 speed, then its mass is:

1. 3 kg
2. 5 kg
3. 7 kg
4. 17 kg

Answer: 2. 5 kg

We know momentum, P = mv

⇒ Using the law of conservation of momentum\(1 \times 12 \hat{i}+2 \hat{i}+2 \times 8 \hat{j}+P_3  =0
P_3 =-(12 \hat{i}+16 \hat{j})\)

⇒ \(\left|P_3\right|=\sqrt{(12)^2+(16)^2} =\sqrt{144+256}\)

=\(20 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)

⇒ \(\text { Now, } \quad P_3 =m_3 v_3\)

∴ \(m_3=\frac{P_3}{v_3}  =\frac{20}{4}=5 \mathrm{~kg}\)

Question 31. A solid cylinder of mass 3 kg is rolling on a horizontal surface with a velocity of 4 ms^-1. It collides with a horizontal spring of force constant 200 Nm”1 The maximum compression produced in the spring will be:

1. 0.5 m
2. 0.6 m
3. 0.7 m
4. 0.2 m

Answer: 2. 0.6 m

It is given figure

⇒ mass = 3 kg, v = 4 ms 1 k = 200 Nm 1 for max compression.

⇒ P.E. of Spring = Less in K.E of cylinder

⇒ K.E. of cylinder =\(\frac{1}{2} m v^2+\frac{1}{2} I \omega^2\)

=\(\frac{1}{2} m v^2+\frac{1}{2}\left(\frac{1}{8} m l^2\right)\left(\frac{\mathrm{v}}{3}\right)^2\)

=\(\frac{1}{2} m v^2+\frac{1}{4} m v^2=\frac{3}{4} m v^2\)

⇒ And \(\quad \frac{1}{2} k x^2=\frac{3}{4} m v^2\)

⇒ \(\quad x^2  =\frac{\frac{3}{4}}{\frac{1}{2}} \times \frac{m v^2}{\mathrm{~K}}\)

⇒ \(x^2 =\frac{3}{4} \frac{m v^2}{\mathrm{~K}}\)

⇒ \(\quad x =\sqrt{\frac{3}{2} \times 3 \times \frac{11}{200}}\)

=\(\frac{3 \times 4}{20}=0.6 \mathrm{~m}\)

Question 32. A stone is dropped from a height of h. It hits the ground with a certain momentum p. If the same stone is dropped from a height 100% more than the previous height, the momentum, when it hits the ground, will change by:

1. 68%
2. 41%
3. 200%
4. 100%

Answer: 2. 41%

⇒ We know that, \(v=\sqrt{2 g h}\) and momentum,\(p=m v=p \propto \sqrt{h}\)

⇒ \(p \propto \sqrt{h}\)

⇒ \(\frac{p_2}{p_1}=\sqrt{\frac{h_2}{h_1}}=\sqrt{\frac{2 h}{h}}=\sqrt{2}\)

⇒ \(p_2=\sqrt{2} p_1\)

⇒ \(p_2=1.414 p_1\)

∴ \(\% \text { change }=\frac{p_2-p_1}{p_1} \times 100 \%=41 \%\)

Question 33. A shell, in flight, explodes into four unequal parts. Which of the following is conserved?

1. Potential energy
2. Momentum
3. Kinetic energy
4. Both 1 and 3.

Answer: 2. Momentum

Question 34. A man fires a bullet of mass 200 g at a speed of 5 m/s. The gun is of one kg mass. By what velocity does the gun rebound backward?

1. 1 m/s
2. 0.01 m/s
3. 0.1 m/s
4. 10 m/s.

Answer: 1. 1 m/s

⇒ Mass of bullet (m₁) = 200 g = 0.2 kg

⇒ speed of bullet (v₁) = 5 m/s

and mass of gun (m₂) = 1 kg. Before firing, the total momentum is zero. After firing total momentum is m₁V₁ + m₂V₂

From the law of conservation of momentum m₁V₁ + m₂v₂ = 0

Question 35. A 600 kg rocket is set for a vertical firing. If the exhaust speed is 1000 ms^-1, the mass of the gas ejected per second to supply the thrust needed to overcome the weight of the rocket is:

1. 117.6 kgs^-1
2. 58.6 kgs^-1
3. 6kg^-1
4. 76.4 kg s^-1

Answer: 3. 6kg^-1

The thrust of a rocket is the force that accelerates it higher.

Thrust on the rocket at time t is given by,\(F=-u \frac{d m}{d t}\)

where u is the relative velocity of exhaust gases concerning the rocket. \(\frac{d m}{d t}\) is the rate of combustion of dt fuel at that instant.

⇒ \(F=-u \frac{d m}{d t}=m g\)

⇒ \(-\frac{d m}{d t}=\frac{m g}{u}\)

⇒ Here, m = 600 kg, u = 1000 ms^-1

∴ \(-\frac{d m}{d t}=\frac{600 \times 10}{1000}=6 \mathrm{~kg} \mathrm{~s}^{-1}\)

Question 36. A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1:1:3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment in m/s will be:

1. \(7 \sqrt{2}\)
2. \(5 \sqrt{2}\)
3. \(3 \sqrt{2}\)
4. \(\sqrt{2}\)

Answer: 1. \(7 \sqrt{2}\)

Since the 5 kg body explodes into three fragments with masses in the ratio 1:1:3 thus, the masses of fragments will be 1 kg, 1 kg, and 3 kg respectively. The magnitude of the resultant momentum of two fragments each of mass 1 kg, moving with velocity 21 m/s, in perpendicular directions is

=\(\sqrt{\left(m_1 v_1\right)^2+\left(m_2 v_2\right)^2}\)

=\(\sqrt{(21)^2+(21)^2}=21 \sqrt{2} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)

⇒ According to the law of conservation of linear momentum.

∴ \(m_3 v_3=21 \sqrt{2} \text { or } 3 v_3=21 \sqrt{2}\)

Question 37. Two blocks A and B of masses 3 m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively:

1. \(\mathrm{g}, \frac{g}{3}\)
2. \(\frac{g}{3}, \mathrm{~g}\)
3. \(\mathrm{g}, \mathrm{g}\)
4. \(\frac{g}{3}, \frac{g}{3}\)

Answer: 2. \(\frac{g}{3}, \mathrm{~g}\)

This problem has two cases,

Case 1: The system is in equilibrium with a mass of 3m (weight = 4 mg) over the spring.

Case 2: When the string is cut according to the figure then:

1. free body diagram of mass m

Here, the force of mass m = mg Acceleration = g

2. free body diagram of mass 3 m Net force,

Ket = 4 mg-3 mg

3 m.a. = mg

⇒ \(a=\frac{g}{3}\)

From 1 and 2

∴acceleration of blocks A and B is a = \(\frac{g}{3}\) and a = g

Question 38. The displacement V (in meter) of a particle of mass ‘m’ (in kg) moving in one dimension under the action of a force is related to time V (in a sec) by i = \(\sqrt{x}\)+ 3 The displacement of the particle when its velocity is zero, will be:

1. 4 m
2. 0 m(zero)
3. 6 m
4. 2 m

Answer: 2. 0 m(zero)

According to the question t =\(\sqrt{x}+3\)

⇒ \(\sqrt{x} =t-3\)

⇒ Squaring both sides we get, x = (t- 3)² →  Equation… 1

⇒ Now Velocity v= \(\frac{d x}{d t}=\frac{d}{d t}(t-3)^2\)= 2(t -3)

⇒ velocity becomes zero means 2(t-3) = 0

t = 3 sec.

At t = 3s → from…1

∴ x = (3 – 3)² = 0

Question 39. Calculate the acceleration of the block and trolly system shown in the figure. The coefficient of kinetic friction between the trolly and the surface is 0.05. g = 10 m/s², the mass of the string is negligible and no other friction exists).

1. 1.25 m/s²
2. 1.50 m/s²
3. 1.66 m/s²
4. 1.00 m/s²

Answer: 1. 1.25 m/s²

The situation is shown in the figure

Here, a = acceleration of the system.

Here from free free-body diagram.

⇒ T- μR = 10a

T- 0.05 x log = 10a {tR = 0.05 x 100 = 5}

T-5 = 10a.    →  Equation.. 1

For a 2 kg block the equation of motion

⇒ 2g-T=20 2 x 10 – T = 2a

⇒ 20-T=2a  →  Equation  ..2

From equation 1 and 2, we get T-5 = 10a

⇒ 20-T=2a

20-5 = 12a

15 = 12a

∴  a=\(\frac{15}{12}=\frac{5}{4}=1.25 \mathrm{~ms}^{-2}\)

Question 40. Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity g is:

1. \(\frac{g}{2}\)
2. \(\frac{g}{5}\)
3. \(\frac{g}{4}\)
4. g

Answer: 2. \(\frac{g}{5}\)

Given that, m₁= 4 kg m₂ = 6 kg and a = ?

⇒ From the free free-body diagram,

m₂g- T = m₂ a

⇒ T-m₁g- m₁a

⇒ (m₂g- m₁g) = m₁a+ m₂a

⇒ (m₂– m₁)g = a(m₁+ m₂)

⇒ \(a=\left(\frac{m_2-m_1}{m_1+m_2}\right) g\) → Equation 1

Putting the values of m₁, m₂, and g in equation 1.

⇒ \(a=\left(\frac{6-4}{4+6}\right) g=\frac{2}{10} g=\frac{g}{5}\)

∴ a=\(\frac{g}{5}\)

Question 41. A body of mass m is kept on a rough horizontal surface (coefficient of friction = μ). Horizontal force is applied to the body, but it does not move. The resultant of the normal reaction and the frictional force acting on the object is given as F, where F is:

1. \(|F|=m g+\mu m g\)
2. \(|F|=\mu m g\)
3. \(|F| \leq m g \sqrt{1+\mu^2}\)
4. \(|F|=m g\)

Answer: 3. \(|F| \leq m g \sqrt{1+\mu^2}\)

Since the body does not move hence it is in equilibrium. Fr = Frictional force which is less than or equal to friction

⇒ \(\mathrm{N}=m g\)

⇒  \(\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{N}}+\vec{f}_r\)

⇒  \(|\overrightarrow{\mathrm{F}}| \leq(m g)^2+(\mu m g)^2\)

∴ \(|\overrightarrow{\mathrm{F}}| \leq m g \sqrt{1+\mu^2}\)

Question 42. Which one of the following statements is incorrect?

1. Frictional force opposes the relative motion.
2. The limiting value of static friction is directly proportional to normal reaction.
3. Rolling friction is smaller than sliding friction.
4. The coefficient of sliding friction has a dimension of length.

Answer: 4. The Coefficient of sliding friction has a dimension of length.

⇒  The coefficient of sliding friction is,\(\mu_{\mathrm{S}}=\frac{N}{F_{\text {sliding }}}\)

∴ The dimensions of N and F are the same. So, ^ is a dimensionless quantity.

Question 43. Three blocks A, B, and C of masses 4 kg. 2 kg, and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is:

1. 2 N
2. 6 N
3. 8 N
4. 18 N

Answer: 2. 6 N

From the question, it is given that, M_A = 4 kg, M_B = 2 kg, M_C = 1 kg

M = M_A + M_B + M_c = (4 + 2 + 1) = 7 kg

We know, that F = Ma

14 = 7a

a = 2 m/s²

⇒ Let f be the contact force between A and B then the net force

F-f= 4a

14-f= 4×2

∴ f= 6N

Question 44. A Block A of mass m₁ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of the table and from its other another block B of mass m₂ is suspended. The coefficient of kinetic friction between the block and the table is μ_k  When the, T block A is sliding on the table, the tension in the string is:

1. \(\frac{\left(m_2+\mu_k+m_1\right) g}{\left(m_1+m_2\right)}\)
2. \(\frac{\left(m_2-\mu_k m_1\right) g}{\left(m_1+m_2\right)}\)
3. \(\frac{m_1 m_2\left(1+\mu_g\right) g}{\left(m_1+m_2\right)}\)
4. \(\frac{m_1 m_2\left(1-\mu_k\right) g}{\left(m_1+m_2\right)}\)

Answer: 4. \(\frac{m_1 m_2\left(1-\mu_k\right) g}{\left(m_1+m_2\right)}\)

The situation is

Let T be the tension in the string a be the acceleration of the block. Then from the free body diagram of and B is

From both the diagram, we have,

m₂g- T = m₂a  → equation 1

T-μRm₁g = m₁a    →  Equation 2

From eq. 1 and 2,

m₂S – μRm₁g = (m₁+ m₂)a

⇒ \(a=\frac{\left(m_2-m_1\right) g}{m_1+m_2}\)

Putting The value of A in Equation 1

T=\(m_2 g-m_2 a\)

=\(m_2 g-m_2\left[\frac{\left(m_2-\mu_k m_1\right) g}{m_1+m_2}\right]\)

=\(m_2 g\left[1-\frac{\left(m_2-\mu_k m_1\right) g}{m_1+m_2}\right]\)

∴ T =\(\frac{m_1 m_2\left(1+\mu_k\right) g}{m_1+m_2}\)

Question 45. A plank with a box on one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30°, the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficient of static and kinetic friction between the box and the plank will be respectively:

1. 0.6 and 0.6
2. 0.6 and 0.5
3. 0.5 and 0.6
4. 0.4 and 0.3

Answer: 2. 0.6 and 0.5

The situation is shown in the figure

⇒ Here coefficient of static friction,\(\mu_s=\tan 30^{\circ}=\frac{1}{\sqrt{3}}=0.6\)

⇒  And distance covered by plank,\(S=u t+\frac{1}{2} a t^2\)

⇒  Here, u = 1 and a = g (sin θ – μ cos θ)=\(4=\frac{1}{2} g\left(\sin 30^{\circ}-\mu_k \cos 30^{\circ}\right)(4)^2\)

⇒  \(0.5 =10 \times \frac{1}{2}-\mu_k \times 10 \times \frac{\sqrt{3}}{2}\)

⇒  \(5 \sqrt{3} \mu_k =45\)

∴ \(\mu_k  =0.51\)

Question 46. A system consists of three masses m₁, m₂, and m₃ connected by a string passing over a pulley P. The m₁ hangs freely and m₂ and w3 are on a rough horizontal table (the coefficient of friction = μ). The pulley is frictionless and of negligible mass. The downward acceleration of mass m₁ is (Assume, m₁ = m₂ = m₃ = m):

1. \(\frac{g(1-g \mu)}{g}\)
2. \(\frac{(2 g \mu)}{3}\)
3. \(\frac{g(1-2 \mu)}{3}\)
4. \(\frac{g(1-2 \mu)}{2}\)

Answer: 3. \(\frac{g(1-2 \mu)}{3}\)

Consider free body diagram for block m₁V₁

⇒ Here m₁ = m₂ = m₂

mg-T₁ = ma  →  Equation 1

For block m₂ and m₃

T₁ -2μmg = 2ma  →   Equation  2

Solving equation 1 and 2

mg- T₁ = ma

T₁– 2μmg = 2 ma

mg- 2μmg = 3ma

mg(1- 2μ) = 3pa

⇒ \(a=\frac{g(1-2 \mu)}{3}\)

∴ This is a downward acceleration of mass

Question 47. The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom. The coefficient of friction between the block and the lower half of the plane is given by:

1. \(\mu=\frac{1}{\tan \theta}\)
2. \(\mu=\frac{2}{\tan \theta}\)
3. \(\mu=2 \tan \theta\)
4. \(\mu=\tan \theta\)

Answer: 3. \(\mu=2 \tan \theta\)

From work energy theorem

(mg sin θ) (2S)- (μmg cos θ) 5 = 0- 0

mg sin θ(2s)- (μ mg cos θ)S = 6

2s Sin θ- μs cos θ = 0

2 Sin θ- p cos θ = 0

μ cos θ = 2 sin θ

∴ \(\mu=2 \frac{\sin \theta}{\cos \theta}=2 \tan \theta\)

Question 48. A block of mass 5 kg is placed on a horizontal surface and a pushing force of 20 N is acting on the block as shown in the figure. If the coefficient of friction between the block and the surface is 0.2, then calculate the speed of the block after 15. (Given g = 10 m/s²):

1. 3.936 m/s
2. 6.396 m/s
3. 9.369 m/s
4. 3.215 m/s

Answer: 1. 3.936 m/s

The different forces acting on the block are shown in the given figure

Horizontal component = 20 cos 45°

Vertical component = 20 sin 45°

Now we can write, 20 cos 45°-f= 5a →  Equation 1

Where  = frictional force

= μN

= 0.2 x N  →  Equation  2

And  N = 5g + 20 sin 45° = 50 +10/2

Considering eq. 1 frictional force in eq 1

= 10V₂– 0.2(50 +10V₁)2624

⇒ speed ofblock after 15 sec. is v = u + at

= 0.2624 x 15 = 3.936 ms^-1

Question 49. Block B is pushed momentarily along a horizontal surface with an initial velocity v. If μ is the coefficient of sliding friction between B and the surface, block B will come to rest after a time.

1. \(\frac{g \mu}{v} s\)
2. \(\frac{v}{g} s\)
3. \(\frac{v}{g} s\)
4. \(\frac{v}{g \mu} s\)

Answer: 3. \(\frac{v}{g} s\)

By Using, v =u+at

0 =\(v-\mu g t\)

t =\(\frac{v}{\mu g}\)

Question 50. The coefficient of static friction μg, between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. (g = 10 m/s²):

1. 2.0 kg
2. 4.0 kg
3. 0.2 kg
4. 0.4 kg

Answer: 4. 0.4 kg

⇒ According To Question,

T=\(m_{\mathrm{B}} g=\mu\left(m_{\mathrm{A}} g\right)\)

∴ \(m_{\mathrm{B}} =\mu m_{\mathrm{A}}=0.02 \times 2 =0.4 \mathrm{~kg}\)

Question 51. A block of mass 10 kg placed on the rough horizontal surface having a coefficient of friction p = 0.5, if a horizontal force of 100 N acting on it then the acceleration of the block will be:

1. 10 ms^-1
2. 5 ms^-2
3. 15 ms^-2
4. 0.5 ms^-2

Answer: 2. 5 ms^-2

⇒ Here,\(f_{\max } =\mu N=\mu m g\)

=\(0.5 \times 10 \times 10=50 \mathrm{~N}\)

⇒ \(\quad a =\frac{\text { net force }}{\text { mass }}\)

=\(\frac{100-50}{10}=5 \mathrm{~m} / \mathrm{sec}^2\)

Question 52. A lift of mass 1000 kg which is moving with an acceleration of 1 ms^-2 in the upward direction, then the tension developed in the string which is connected to the lift is:

1. 9800 N
2. 10,800 N
3. 11000 N
4. 10,000 N

Answer: 2. 10,800 N

^∴ Using the equation,T=m(g + a) = 1000(9.8 + 1) = 10800 N

Question 53. A block has been placed on an inclined plane with a slope angle of θ, block slides down the plane at constant speed. The coefficient of kinetic friction is equal to:

1. sin θ
2. cos θ
3. g
4. tan θ

Answer: 4. tan θ

The angle of sliding is defined as the minimum angle of inclination of a plane with the horizontal, such that a body placed on the plane just begins to slide down.

AB is an inclined plane such that a body placed on it just begins to slide down.

⇒ \(\angle \mathrm{BAC}=\theta\)= angle of repose

⇒  In equilibrium, f =\(m g \sin \theta\) and R =\(m g \cos \theta\)

⇒ \(\frac{f}{R}  =\frac{m g \sin \theta}{m g \cos \theta}=\tan \theta\)

∴ \(\mu=\tan \theta\)

Question 54. A heavy uniform chain lies on the horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum friction of the length of the chain that can hang over one edge of the table is:

1. 20%
2. 25%
3. 35%
4. 15%

Answer: 1. 20%

The force of friction should balance the weight of the chain hanging. If M is the mass of the whole chain of length L and x is the length of the chain hanging to balance, then

⇒ \(\mu \frac{M}{L}(L-x) g=\frac{M}{L} x g\)  or

⇒ \(\mu(L-x)=x   or\)

x=\(\frac{\mu L}{\mu+1}=\frac{0.25 L}{1.25}\)

x=\(\frac{L}{5}\)  or

∴ \(\frac{x}{L} =\frac{1}{5}=\frac{1}{5} \times 100\) =20 \%

Question 55. A block of mass 10 kg is in contact with the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationery when the cylinder is vertical and rotating about its axis will be (g = 10 m/s²):

1. \(\frac{10}{2 \pi} \mathrm{rad} / \mathrm{s}\)
2. \(10 \mathrm{rad} / \mathrm{s}\)
3. \(10 \pi \mathrm{rad} / \mathrm{s}\)
4. \(\sqrt{10} \mathrm{rad} / \mathrm{s}\)

Answer: 2. \(10 \mathrm{rad} / \mathrm{s}\)

⇒  To keep the block stationary Frictional force\( \geq \)weight

Here,\(\mu N \geq M g\)

⇒ N=\(\mathrm{M} \omega^2 r \quad\{\text { Hence, } r=1 m, \mu=0.1\}\)

⇒  \(M \omega^2 r  =M g\)

⇒  \(\omega =\sqrt{\frac{g}{\mu r}}=\sqrt{\frac{10}{0.1 \times 1}}\)

=\(10 \mathrm{rad} \mathrm{s}^{-1}\)

Question 56. Two particles A and B are moving in a uniform circular motion in the concentric circles of radii r_A and r_B with speeds vA and v_B respectively. Their period of rotation is the same. The ratio of the angular speed of A to that of B will be :

1. v_A: v_B
2. r_B: r_A
3. 1: 1
4. r_A: r_B

Answer: 3. 1: 1

We know that:

Time period\(T=\frac{2 \pi}{\omega}\)

Where, ω = angular speed

According to the question,

⇒ \(T_A  =T_B\)

⇒ \(\frac{2 \pi}{\omega_{\mathrm{A}}} =\frac{2 \pi}{\omega_{\mathrm{B}}}\)

⇒ \(\omega_{\mathrm{A}}=\omega_{\mathrm{B}}\)

∴ \(\omega_{\mathrm{A}}: \omega_{\mathrm{B}}\) =1: 1

Question 57. In the given figure a = 15 m/s² represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 0.2 m at a given instant of time. The speed of the particle is:

1. 4.5 m/s
2. 5.0 m/s
3. 5.7 m/s
4. 6.2 m/s

Answer: 3. 5.7 m/s

The particle is moving in a circle of radius R. So acceleration will be:

⇒ \(a_c=\frac{\mathrm{v}^2}{\mathrm{R}}\)

from figure ac = a cos 30°

⇒ \(\frac{v^2}{R}\)= a cos 30°

v² = a cos 30° x R = 15 x cos 30° x 2.5

v² = 15×2.5x\(\frac{\sqrt{3}}{2}\)

∴ v = 5.7 m/s

Question 58. A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tires of the car and the road is ps. The maximum safe velocity on this road is:

1. \(\sqrt{\frac{g}{R} \frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}}\)
2. \(\sqrt{\frac{g}{R^2} \frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}}\)
3. \(\sqrt{g R^2 \frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}}\)
4. \(\sqrt{g R \frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}}\)

Answer: 4. \(\sqrt{g R \frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}}\)

For vertical equilibrium on the road, N cosθ = mg +sinθ

mg = N cosθ – fsinθ →  Equation  1

⇒ Centripetal force for safe turning, \(N \sin \theta+f \cos \theta=\frac{m v^2}{R}\)

⇒ From Equations 1 And 2, We get \(\frac{v^2}{R g}=\frac{N \sin \theta+f \cos \theta}{N \cos \theta-f \sin \theta}\)

⇒ \(\frac{v_{\max }^2}{R g}=\frac{N \sin \theta+\mu_s N \cos \theta}{N \cos \theta-\mu_s N \sin \theta}\)

∴ \(\mathrm{v}_{\max }=\sqrt{R g\left(\frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)}\)

Question 59. Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius r/2 and the lighter one in a radius r. The tangential speed of lighter stones is n times that of the value of heavier stones when they experience the same centripetal forces. The value of n is:

1. 4
2. 1
3. 2
4. 3

Answer: 3. 2

⇒ Let v be the tangential speed of the heavier stone. Then, centripetal force experienced by lighter stone is\(\left(F_c\right)_{\text {lighter }}=\frac{m(n v)^2}{r}\)

and that of heavier stone is \(\left(F_c\right)_{\text {heavier }}=\frac{2 m v^2}{r / 2}\)

⇒ \(\text { But, } \left(F_c\right)_{\text {lighter }}=\left(F_c\right)_{\text {heavier }}\)

⇒ \(\quad \frac{m(n v)^2}{r} =\frac{2 m v^2}{(r / 2)}\)

⇒ \(\text { or, } \quad n^2\left(\frac{m v^2}{r}\right) =4\left(\frac{m v^2}{r}\right)\)

∴ \(n^2 =4 \text { or } n=2\)

Question 60. A car is moving in a circular horizontal track with a radius of 10 m with a constant speed of 10 m/s. A bob is suspended from the roof of the car by a light wire of length 1.0 m. The angle made by the wire with the vertical is:

1. π/3
2. π/6
3. π/4
4. 0°

Answer: 3. π/4

⇒Let θ be the angle made by the wire with the vertical.\(\tan \theta=\frac{v^2}{r g}\)

⇒ Here, v= 10 m/s, r = 10 m, g= 10 m/s²

⇒ \(\tan \theta =\frac{(10 m / s)^2}{10 m\left(10 m / s^2\right)}\)=1

∴ \(\theta =\tan ^{-1}(1)=\frac{\pi}{4}\)

Question 61. A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45°, the speed of the car is:

1. 20 m s^-1
2. 30 m s^-1
3. 5 m s^-1
4. 10 m s^-1

Answer: 2. 30 m s^-1

Here, m = 1000 kg, R = 90 m, 0 = 45°

∴ For banking, \(\tan \theta=\frac{v^2}{R g}\) = 30 ms^-1

Question 62. A car of mass m is moving on a level circular track of radius R. If ps represents the static friction between the road and tires of the car, the maximum speed of the car in circular motion is given by:

1. \(\sqrt{\mu_s m R g}\)
2. \(\sqrt{\frac{R g}{\mu_s}}\)
3. \(\sqrt{\frac{m R g}{\mu_s}}\)
4. \(\sqrt{\mu_s R g}\)

Answer: 4. \(\sqrt{\mu_s R g}\)

The force of friction provides the necessary centripetal force

⇒ \(\frac{m v^2}{R} \leq \mu_{\mathrm{s}} \mathrm{N}\)

⇒ \(v^2 \leq \frac{\mu_s R N}{m}\)

⇒ \(v^2 \leq \mu_{\mathrm{s}} \mathrm{Rg}\)

∴ The maximum speed of the car in circular motion is \(\mathrm{v}_{\max }=\sqrt{\mu_s R g}\)

Question 63. A car of mass 1000 kg negotiates a banked curve of radius 9.0 m on a frictionless road. If the banking angle is 45°, the speed of the car is:

1. 20 ms^-1
2. 30 ms^-1
3. 5 ms^-1
4. 10ms^-1

Answer: 2. 30 ms^-1

Given banking angle = 45°

Radius = 90 m and g = 10 m/s²

Now banking angle, tan \(\theta=\frac{v^2}{r g}\)

⇒ \(\tan 45^{\circ} =\frac{v^2}{90 \times 10}\)

⇒ \(v^2 =90 \times 10 \times \tan 45^{\circ}\)

v =\(\sqrt{90 \times 10 \times \tan 45^{\circ}}=30\)

∴ Speed of car v = 30 m/s

Question 64. A car runs at a constant speed on a circular track with a radius of 100 m, taking 62.8 seconds for every circular lap. The average velocity and average speed for each circular lap respectively is:

1. 10 m/s, 0
2. 0, 0
3. 0, 10 m/s
4. 10 m/s, 10 m/s

Answer: 1. 10 m/s, 0

Distance traveled in one revolution = 2 r

⇒ \(\text { Average speed }=\frac{\text { Totaldistance }}{\text { Total time }}\)

⇒ \(\frac{2 \pi r}{t}=\frac{2 \times 3.14 \times 100}{62.8}\)

Displacement in one revolution is zero

∴ \(\text { Average Velocity }=\frac{\text { Net displacement }}{\text { Time }}=\frac{0}{t}\)=0

Question 65. A ball of mass 0.25 kg attached to the end of a string of length 1.96 m is moving in a horizontal circle. The string will break if the tension is more than 25 N. What is the maximum speed with which the ball can be moved?

1. 14 m/s
2. 3 m/s
3. 3.92 m/s
4. 5 m/s

Answer: 1. 14 m/s

For a ball to move in a horizontal circle, the ball should satisfy the condition

Tension is the string = Centripetal force

⇒ \(T_{\max }=\frac{M v_{\max }^2}{R}\)

⇒ \(v_{\text {max }}=\sqrt{\frac{T_{\max } \cdot R}{M}}\)

∴ Making substitution, we obtain,\(v_{\max }=\sqrt{\frac{25 \times 1.96}{0.25}}=\sqrt{196}=14 \mathrm{~m} / \mathrm{s}\)

Question 66. Two racing cars of masses m and 4 m are moving in circles of radii r and 2r respectively. If their speeds are such that each makes a complete circle at the same time, then the ratio of the angular speeds of the first to the second car is:

1. 8 :1
2. 4: 1
3. 2 :1
4. 1: 1

Answer: 4. 1: 1

⇒ We have, \(\omega=\frac{2 \pi}{t}\)

∴ Since both cars take the same time to complete the circle. Therefore the ratio of angular speeds of the cars will be 1: 1

Question 67. A balloon with mass m is descending with an acceleration a (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration?

1. \(\frac{2 m a}{g+a}\)
2. \(\frac{2 m a}{g-a}\)
3. \(\frac{m a}{g+a}\)
4. \(\frac{m a}{g-a}\)

Answer: 1. \(\frac{2 m a}{g+a}\)

Let F be the upthrust of the air. As the balloon is descending with an acceleration a, mg – F = ma  Equation 1

Let the mass be removed from the balloon so that it starts moving up with an acceleration a. Then, F- (m- m₀)g ={m- m₀)a

F- mg + m₀g = ma- m₀a → Equation 2

⇒ Adding equation. 1 and equation. 2, we get

m₀g = 2ma- m₀a;

m₀g + m₀a = 2ma

⇒ m₀(g + a) = 2ma

∴ \(\mathrm{m}_0=\frac{2 m a}{g+a}\)

Question 68. Three blocks with mass m, 2m, and 3m are F connected by strings, as shown in the figure. T After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m? (g is the acceleration due to gravity)

1. 3 mg
2. 6 mg
3. zero
4. 2 mg

Answer: 3. zero

∴ As all blocks are moving with constant speed, therefore, acceleration is zero. So net force on each block is zero.

Question 69. A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on the control panel. The lift starts moving upwards with an acceleration of 1.0 m/s². If g = 10 m s^-2, the tension in the supporting cable is:

1. 8600 N
2. 9680 N
3. 11000 N
4. 1200 N

Answer: 3. 11000 N

Here, the Mass of the person, m = 60 kg

Mass of lift, M= 940 kg,

a = 1 m/s², g = 10 m/s²

Let T be the tension in the supporting cable.

T-(M+m)g = (M+m)a

T = (M+ m) (a + g)

= (940 + 60) (1 + 10)= 11000N

Question 70. The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is:

1. 4 m s^-2upwards
2. 4 m s^-2 downwards
3. 14 ms^-2 upwards
4. 30 m s^-2 downwards

Answer: 1. 4 m s^-2 upwards

⇒ F- Mg = MA

8000 = 2000a

a = 4

∴ Acceleration is 4 m s^-2 upwards

Question 71. A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block will be (g is the acceleration due to gravity):

1. mg cosθ
2. mg sinθ
3. mg
4. mg/cosθ

Answer: 4. mg/cosθ

The wedge is given an acceleration to the left.

The block has a pseudo acceleration to the right, pressing against the wedge because of which the block is not moving.

 

⇒ \(m g \sin \theta=m a \cos \theta\)

∴ The total reaction of the wedge on the block is N = mg cosθ + ma sinθ

Question 72. A monkey of mass 20 kg is holding a vertical rope. The rope will not break when a mass of 25 kg is suspended from it but will break if the mass exceeds 25 kg. What is the maximum acceleration with which the monkey can climb up along the rope? (g = 10 m/s²):

1. 5 m/s²
2. 10 m/s²
3. 25m/s²
4. 2.5 m/s²

Answer: 4. 2.5 m/s²

Let T be the tension in the rope when the monkey climbs up with an acceleration a. Then, T- mg = ma

25g- 20g = 20a

∴ \(\mathrm{a}=\frac{5 \times 10}{20}=2.5 \mathrm{~m} / \mathrm{s}^2\)

Question 73. A lift of mass 1000 kg which is moving with the acceleration of 1 m/s² in an upward direction, then the tension developed in the string which is connected to the lift is:

1. 9800N
2. 10,800N
3. 11,000 N
4. 10,000 N

Answer: 4. 10,000 N

For a lift that is moving in an upward direction with acceleration a, the tension T developed in the string connected to the lift is given by T- m (g + a). Here m = 1000 kg, a = 1 m/s², g = 9.8 m/s² T= 1000(9.8+ 1)= 10,800 N

Question 74. A mass M is placed on a very smooth wedge resting on a surface without friction. Once the mass is released, the acceleration to be given to the wedge so that M remains at rest is where

1. a is applied to the left and a = g tanθ
2. a is applied to the right and a = g tanθ
3. a is applied to the left and a = g sinθ
4. a is applied to the left and a = g cosθ

Answer: 1. a is applied to the left and a = g tanθ

The pseudo acceleration for the body a’ = a

If the pseudo force Macosθ = Mgsinθ, then the body will be at rest,

∴ This horizontal acceleration should be applied to the wedge to the left.

Question 75. A monkey is descending from the branch of a tree with constant acceleration. If the breaking strength of the branch is 75% of the weight of the monkey, the minimum acceleration with which the monkey can slide down without breaking the branch is:

1. g
2. 3g/4
3. g/4
4. g/2

Answer: 3. g/4

Let T be the tension in the branch of a tree when the monkey is descending with acceleration Thus, mg – T = ma

also, T = 75% of the weight of the monkey

⇒ \(\mathrm{T} =\left(\frac{75}{100}\right) \mathrm{mg}=\frac{3}{4} m g\)

⇒ \(\mathrm{ma} =\mathrm{mg}-\left(\frac{3}{4}\right) \mathrm{mg}=\frac{1}{4} \mathrm{mg}\)

or \(\mathrm{a}=\frac{g}{4}\)

Mechanical Properties Of Fluids

Question 1. A barometer is constructed using a liquid (density = 760 kg/m³). What would be the height of a liquid column, when a mercury barometer reads 76 cm? (Density of mercury = 13600 kg/m³)

1. 1.36 m
2. 13.6 m
3. 136 m
4. 0.76 m

Answer: 2. 13.6 m

From the given question,

Density of liquid, F₁ = 760 kg/m3

Density of mercury, f_m = 13600 kg/m3

Height of liquid column in mercury barometer hv = 760 cm = 0.76 m

If the height of the liquid in the liquid column is h₁ then

⇒ \(\rho_{\text {liquid }} =\rho_{\text {mercury }}\)

⇒ \(h_l \rho_l g =h_m \rho_m g\)

⇒ \(h_l =\frac{h_m \rho_m}{\rho_l}\)

=\(\frac{0.76 \times 13600}{760}\)

=13.6 m.

Question 2. In a w-tube, as shown in the figure, water, and oil are on the left side and right side of the tube respectively. The heights from the bottom for water and oil columns are 15 cm and 20 cm respectively. The density of the oil is [take \(rho\)water= 1000 kg/m³] :

 

1. 1200 kg/m³
2. 750 kg/m³
3. 1000 kg/m³
4. 1333 kg/m³

Answer: 2. 750 kg/m³

According to Pascal’s law, “Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel.”

Read and Learn More NEET Physics MCQs

In the given situation as shown in the figure given below,

Pressure due to water column of height 15 cm = Pressure due to oil column of height 20 cm

⇒ \(h_w \rho_w g =h_0 \rho_0 g\)

⇒ \(15 \rho_w =20 \rho_0\)

⇒ \(\rho_0 =\frac{15}{20} \rho_w\)

⇒ \(\rho_0 =\frac{15}{20} \times 1000\)

⇒ \(\left(\text { Given } \rho_w=1000 \mathrm{~kg} \mathrm{~m}^{-3}\right)\)

=\(750 \mathrm{kgm}^{-3}\)

Question 3. A U tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile, the water rises by 65 mm from its original level (see diagram). The density of the oil is:

1. 650 kg m^-3
2. 425 kg m^-3
3. 800 kg m^-3
4. 928 kg m^-3

Answer: 4. 928 kg m^-3

Both the ends of the U-tube are open, so the pressure on both the free surfaces is equal,

⇒ \(P_1 =P_2 \)

⇒ \(h_{\text {oil }} \cdot S_{\text {oil }} \cdot g =h_{\text {water }} \cdot S_{\text {water }} \cdot g \)

⇒ \(\mathrm{~S}_{\text {oil }} =\frac{h_{\text {water }} \cdot \mathrm{S}_{\text {water }}}{\mathrm{h}_{\text {oil }}}\)

∴ \(S_{\text {oil }} =\frac{(65+65)+1000}{(65+65+10)}=928 \mathrm{kgm}^{-3}\)

Question 4. Two non-mixing liquids of densities p and np(n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length l, and density d is put in this container. The cylinder floats with its axis vertical and length \(\rho \mathrm{L}\), (\((\rho<1)\) ) in the denser liquid. The density d is equal to:

1. \(\{2+(n+1) \rho\} \rho\)
2. \(\{2+(n-1) \rho\} \rho\)
3. \(\{1+(n-1) \tilde{\rho}\} \rho\)
4. \(\{1+(n+1) \rho\} \rho\)

Answer: 3. \(\{1+(n-1) \tilde{\rho}\} \rho\)

The given figure explains the situation of the question,

Applying Archimedes principle, we have weight of cylinder=\((\text { upthrust })_1+(\text { upthrust })_2\)

⇒ \(L A d g=(P L) A(n \rho) \mathrm{g}+(1-P) L A \rho g \)

d=\((1-P) \rho+(P n) \rho=[1+(n-1) P] \rho\)

Question 5. The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 x 10^-11 Pa^-1 and the density of water is 103 Kg/m³. What fractional compression of water will be obtained at the bottom of the ocean?

1. 0.8×10^-2
2. l.0x 10^-2
3. 1.2 x 10^-2
4. 1.4 x 10^-2

Answer: 3. 1.2 x 10^-2

Depth of the ocean, d=2700 \(\mathrm{~m}\)

Compressibility of water, k=\(45.4 \times 10^{-14} \mathrm{pa}^{-1}\)

Density of water, \(\rho=10^3|g| \mathrm{m}^3\)

The pressure at the bottom of the ocean is,

P = \(\rho g d\)

= \(10^3 \times 10 \times 2700 \)

= \(27 \times 10^6 \mathrm{pa}\)

Fractional compression

= \({ Compressibility } \times { pressure }\)

= \(45.4 \times 10^{-11} \times 27 \times 10^6\)

= \(1.2 \times 10^{-2}\)

Question 6. The heart of a man pumps 5 L of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury. If the density of mercury be 13.6 x 10³ kg/m³ and g = 10 m/s², then the power of heart in watt is:

1. 1.70
2. 2.35
3. 3.0
4. 1.50

Answer: 1. 1.70

Pressure =150 \(\mathrm{~mm} \text { of } \mathrm{Hg}\)

Pumping rate = \(\frac{d v}{d t}=\frac{5 \times 10^{-3}}{60} \mathrm{~m}^3 / \mathrm{s}\)

Power of heart= \(P. \frac{d v}{d t}=\rho g h \cdot \frac{d v}{d t}\)

= \((13.6 \times 10^3 \mathrm{~g}(\mathrm{~m}^3)(10)\).

⇒ \((0.15) \times \frac{5 \times 10^{-3}}{60} \)

= \(\frac{13.6 \times 5 \times 0.15}{60}=1.70 \text { Watt }\)

Question 7. The cylinder tube of a spray pump has a radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is v, the speed of the ejection of the liquid through the holes is:

1. \(\frac{v \mathrm{R}^2}{n^2 r^2}\)
2. \(\frac{v \mathrm{R}^2}{n r^2}\)
3. \(\frac{v \mathrm{R}^2}{n^2 r^2}\)
4. \(\frac{v^2 \mathrm{R}}{n r}\)

Answer: 2. \(\frac{v \mathrm{R}^2}{n r^2}\)

Using the equation of continuity, Av = constant

Where, A = area of cylindrical tube

v = velocity of liquid in the tube

Volume in flow rate = Volume outflow rate

⇒ \(\pi R^2 v =n \pi r^2 v^{\prime}\)

∴ \(v^{\prime} =\frac{R^2 E}{n r^2}\)

Question 8. A small hole of area of cross-section 2 mm² is present near the bottom of a filled open tank of height 2 m. Taking g = 10 m/s², the rate of flow of water through the open hole would be nearly:

1. \(8.9 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}\)
2. \(2.23 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}\)
3. \(6.4 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}\)
4. \(12.6 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}\)

Answer: 4. \(12.6 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}\)

Rate of flow of liquid,

⇒ \(\mathrm{Q}=a u=a \sqrt{2 g h}\)

a=\(2 \mathrm{~mm}^2 \)

=\(2 \times 10^{-6} \mathrm{~m}^2 \times 2 \mathrm{~m} / \mathrm{s}\)

=\(2 \times 2 \times 3.14 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}\)

=\(12.56 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}\)

=\(12.6 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}\)

Question 9. A wind with a speed of 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming that the pressure inside the house is atmospheric, the force exerted by the wind on the roof and the direction of the force will be (\(\left(\rho_{\text {air }}=1.2 \mathrm{~kg} / \mathrm{m}^3\right) =\) 1.2 kg/m²):

1. 2.4 x 10⁵ N, upwards
2. 2.4 x 10⁵ N, downwards
3. 4.8 x 10⁵ N, downwards
4. 4.8 x 10⁵  N, upwards

Answer: 1. 2.4 x 10⁵ N, upwards

⇒ \(P-\frac{1}{2} \rho v^2\)= constant

Inside the pressure is \(\mathrm{P}_{\mathrm{arm}}\)

The pressure outside is \(\mathrm{P} \)

v is zero inside,

⇒ \(\mathrm{P}-\mathrm{P}_{\mathrm{arm}}=\frac{1}{2} \rho \mathrm{v}^2=\frac{1}{2} \times 1.2 \times 40^2\)=960

Force is F=\(\left(P-P_{\text {arm }}\right) \times \text { Area }=960 \times 250=2.4 \times 10^5 \mathrm{~N}\)upward direction as pressure goes from higher to lower pressure area.

Question 10. A fluid is in streamlined flow across a horizontal pipe of the variable area of cross-section. For this which of the following statements is correct?

1. The velocity is maximum at the narrowed part of the pipe and pressure is maximum at the widest part of the pipe.
2. Velocity and pressure both are maximum at the narrowest part of the pipe.
3. Velocity and pressure both are maximum at the widest part of the pipe.
4. The velocity is minimum at the narrowest part of the pipe and the pressure is minimum at the widest part of the pipe.

Answer: 1. The velocity is maximum at the narrowed part of the pipe and pressure is maximum at the widest part of the pipe.

The velocity is maximum at the narrow part of the pipe and pressure is maximum at the widest part of the pipe.

Using the equation of continuity, Av = constant.

And According to the Bernoulli’s theorem,\(P+\frac{1}{2} \rho v^2\) = constant

Question 11. A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball (v) as a function of time (t) is:

1. 2
2. 3
3. 4
4. 1

Answer: 1. 2

1. Viscous force is the force of sliding friction between two solid surfaces in a fluid. Viscosity is frequently referred to as fluid friction because of this. Viscous forces, like other frictional forces, impede the relative motion of nearby fluid layers.

2. When we drop a ball in liquid that is highly viscous and when it falls then initially gravitational force acts. As it goes inside liquid then viscous force increases so its speed will be constant with time after sometimes because gravitational force is balanced by drag or viscous force

3. When gravitational force is balanced by drag force then the acceleration of the particle which can be measured by the slope of the v and t graph initially becomes negative but after some time it becomes zero which means velocity becomes constant with time.

The v -t graph can be shown by

∴ It shows motion as the B part of the graph

Question 12. The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is \(\frac{d}{2}\), then the viscous force acting on the ball will be:

1. \(\frac{\mathrm{Mg}}{2}\)
2. \(\mathrm{Mg}\)
3. \(\frac{3}{2} \mathrm{Mg}\)
4. \(2 \mathrm{Mg}\)

Answer: 1. \(\frac{\mathrm{Mg}}{2}\)

Given the mass of the ball = M

Density of ball = d

Density of glycerine = d/2

⇒ \(\mathrm{F}_{\mathrm{V}}+\mathrm{F}_{\mathrm{B}} =\mathrm{Mg}\)

⇒ \(\mathrm{F}_{\mathrm{B}} =\frac{4}{3} \pi \mathrm{R}^3 g d g\)

=\(\mathrm{Vg} d g \)

⇒ \(\mathrm{Mg} =\mathrm{V} d g\)

⇒ \(\mathrm{~F}_{\mathrm{V}} =\mathrm{Mg}-\mathrm{F}_{\mathrm{B}}\)

=\(\mathrm{Vg}\left(d-\frac{d}{2}\right)\)

=\(\frac{v g d}{2} \)

∴ \(\mathrm{~F}_{\mathrm{V}} =\frac{\mathrm{Mg}}{2}\)

Question 13. Two small spherical metal balls, having equal masses, are made from materials of densities \(\rho_1 \text { and } \rho_2\), \(\left(\rho_1=8 \rho_2\right)\) and have radii of 1 mm and 2 mm, respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals t) and whose density is 0.1 \(\rho_2\). The ratio of their terminal velocities would be:

1. \(\frac{79}{72}\)
2. \(\frac{19}{36}\)
3. \(\frac{39}{72}\)
4. \(\frac{79}{36}\)

Answer: 4. \(\frac{79}{36}\)

The terminal velocity of the ball in a viscous fluid is, \(v_T=\frac{2}{9 \eta} r^2 g(\sigma-\rho)\)

Where, \(\eta= coeff\). of viscosity of the medium

r = radius of ball

⇒ \(\sigma\)= density of ball

⇒ \(\rho \)= density of viscous medium

From the formula, \(\mathrm{v}_{\mathrm{T}} \propto r^2(\sigma-\rho)\)

From this formula

⇒ \(\frac{\text { Terminal velocity of } \mathrm{I}^{\mathrm{st}} \text { ball }}{\text { Terminal velocity } \mathrm{II}^{\text {nd }} \text { ball }}=\frac{\mathrm{v}_1}{\mathrm{v}_2}\)

=\(\frac{1^2}{2^2} \frac{\left(8 \rho_2-0.1 \rho_2\right)}{\left(\rho_2-0.1 \rho_2\right)}\)

∴ \(\frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{1}{4} \times \frac{7.9}{0.9}=\frac{79}{36}\)

Question 14. A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to:

1. \(r^5\)
2. \(r^2\)
3. \(r^3\)
4. \(r^4\)

Answer: 1. \(r^5\)

If, F= Viscous force

⇒ \(v_T \)= terminal velocity

F =\(6 \pi r \eta v_T\)

⇒ \(\frac{d \mathrm{Q}}{d t} =6 \pi \eta r v T \times v_T \)

=\(6 \pi \eta r v_T^2\)  →  Equation 1

We also know that terminal velocity,

⇒ \(v_T=\frac{2}{9} \frac{r^2(\rho-\sigma)}{\eta} g\)

⇒ \( v_T \propto r^2\) →  Equation 2

From (1) and (2),

⇒ \(\frac{d \mathrm{Q}}{d t} \propto r .\left(r^2\right)^2 \)

∴ \(\frac{d \mathrm{Q}}{d t} \propto r^5 \)

Question 15. If a soap bubble expands, the pressure inside the bubble:

1. increases
2. remains the same
3. is equal to the atmospheric pressure
4. decreases

Answer: 4. decreases

As we know the pressure of the soap bubble is written as; \(\mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{o}}=\frac{4 T}{r}\)  →  Equation 1

When the soap bubble expands, the radius of the soap bubble will increase, therefore from equation (1), we see that the change in pressure is inversely proportional to the radius of the soap.

Soap bubble expands, the radius will increase and the inside pressure will decrease

Question 16. A liquid does not wet the solid surface if the angle of contact is:

1. equal to 45°
2. equal to 60°
3. greater than 90°
4. zero

Answer: 3. greater than 90°

A liquid does not wet the solid surface if the angle of contact is obtuse i.e., greater than 90°. In this case, cohesive forces will be greater than adhesive forces and so, the liquid does not wet the surfaces of the solid.

Question 17. A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of water in the capillary tube is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is:

1. 5.0 g
2. 10.0 g
3. 20.0 g
4. 2.5 g

Answer: 4. 2.5 g

The height of water rise in a capillary tube is h =\(\frac{2 \mathrm{~T} \cos \theta}{r \rho g}\)

⇒ \(h \propto \frac{1}{r} \)

⇒ \(\frac{h_1}{h_2} =\frac{r_2}{r_1}=\frac{2 r}{r}\)=2

m =\({ A.h. } \rho\)

⇒ \(\frac{m_2}{m_1} =\frac{A h_2 \rho}{A h_1 \rho}=\frac{h_2}{h_1}=\frac{1}{2}\)

∴ \(m_2 =\frac{m_1}{2}=\frac{5}{2}\)=2.50

Question 18. A soap bubble, having a radius of 1 mm, is blown from a detergent solution having a surface tension of 25 x 10^-2 N/m. The pressure inside the bubble equals at a point Z0 below the free surface of water in a container.
Taking, g = 10 m/s², density of water = 103 kg/m³, The value of Z0 is :

1. 10 cm
2. 1 cm
3. 0.5 cm
4. 100 cm

Answer: 2. 1 cm

Excess Pressure =\(\frac{4 T}{R}\),

Gauge Pressure =\(\rho g Z_o\)

⇒ \(P_0+\frac{4 \mathrm{~T}}{\mathrm{R}}=P_0+\rho g \mathrm{Z}_o\)

⇒ \(Z_0=\frac{4 T}{R \rho g}\)

⇒ \(Z_0=\frac{4 \times 2.5 \times 10^{-2}}{10^{-3} \times 1000 \times 10} \mathrm{~m}\)

∴ \(Z_0=1 \mathrm{~cm}\)

Question 19. A rectangular film of liquid extended from (4 cm x 2 cm) to (5 cm x 4 cm). If the work done is 3 x 10^-4 J, the value of the surface tension of the liquid is:

1. 0.250 Nm^-1
2. 0.125 Nm^-1
3. 0.2 Nm^-1
4. 8.0 Nm^-1

Answer: 2. 0.125 Nm^-1

We know that, Surface energy = Work done \(\times\) Surface Tension

Surface energy, \( \Delta \mathrm{E}=(5 \times 4-4 \times 2) \times 2 \ \) film has two surface.

=\(24 \mathrm{~cm}^2=24 \times 10^{-4} \mathrm{~m}^2\)

Work done =T. \(\Delta E\)

⇒ \(3 \times 10^{-4}=T \times 24 \times 10^{-4}\)

T=\(\frac{3 \times 10^{-4}}{24 \times 10^{-4}}=0.125 \mathrm{~N} / \mathrm{m}\)

Question 20. Three liquids of densities, \(\rho_1, \rho_2 \text { and } \rho_3\) (with \(\rho_1>\rho_2>\left.\rho_3\right)\), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact \(\theta_1, \theta_2 \text { and } \theta_3\) obey:

1. \(\frac{\pi}{2}>\theta_1>\theta_2>\theta_3 \geq 0\)
2. \(0 \leq \theta_1<\theta_2<\theta_3<\frac{\pi}{2}\)
3. \(\frac{\pi}{2}<\theta_1<\theta_2<\theta_3<\pi\)
4. \(\pi>\theta_1>\theta_2>\theta_3>\frac{\pi}{2}\)

Answer: 2. \(0 \leq \theta_1<\theta_2<\theta_3<\frac{\pi}{2}\)

From surface tension, the height of capillary,

h=\(\frac{2 \mathrm{~T} \cos \theta}{r \rho g}\)

From above equation, \(\frac{\cos \theta_1}{\rho_1}=\frac{\cos \theta_2}{\rho_2}=\frac{\cos \theta_3}{\rho_3}\)

Thus, \(\cos \theta \propto \rho \)

⇒ \(\rho_1>\rho_2>\rho_3 \)

⇒ \(\cos \theta_1>\cos \theta_2>\cos \theta_3\)

∴ \(0 \leq \theta_1 \leq \theta_2 \leq \theta_3<\frac{\pi}{2}\)

Question 21. Water rises to a height of ‘h’ in a capillary tube. If the length of the capillary tube above the surface of the water is made less than ‘h’ then:

1. water rises to the tip of the capillary tube and then starts overflowing like a fountain
2. Water rises to the top of the capillary tube and stays there without overflowing.
3. water rises to a point a little below the top and stays there.
4. water does not rise at all.

Answer: 1. water rises to the tip of the capillary tube and then starts overflowing like a fountain

Water rises to the tip of the capillary tube and then starts overflowing like a fountain.

Question 22. A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid then:

1. Energy =\(4 V T\left(\frac{1}{r}-\frac{1}{\mathrm{R}}\right)\) is released
2. Energy =\(3 V T\left(\frac{1}{r}+\frac{1}{\mathrm{R}}\right)\) is absorbed
3. Energy =\(3 V T\left(\frac{1}{r}-\frac{1}{\mathrm{R}}\right)\) is released
4. Energy is neither released nor absorbed

Answer: 3. Energy =\(3 V T\left(\frac{1}{r}-\frac{1}{\mathrm{R}}\right)\) is released

According to the question,

⇒ \(A_f=4 \pi R^2=\frac{3}{34} \times \pi \times \frac{R^2}{\frac{R}{V}}=\frac{3 V}{R}\)

⇒ \(A_i=n \times 4 \pi r^2=\frac{V}{\frac{4}{3} \pi r^3} \times 4 \pi r^2=\frac{3 V}{S}\)

∴ Hence, energy released, \(\left(A_i-A_f\right) T=3 V T\left(\frac{1}{r}-\frac{1}{R}\right)\)

Question 23. The wettability of a surface by a liquid depends primarily on:

1. viscosity
2. surface tension
3. density
4. angle of contact between the surface and the liquid

Answer: 4. angle of contact between the surface and the liquid

The wettability of a surface by a liquid depends primarily on the angle of contact between the surface and the liquid

Motion In A Plane Question And Answers

Question 1. Which of the following is not a vector quantity?

1. Speed
2. Velocity
3. Torque
4. Displacement

Answer: 1. Speed

Speed is a scalar quantity. It gives no idea about the direction of motion of the object. Velocity is a vector quantity, as it has both magnitude and direction. Displacement is a vector as it possesses both magnitude and direction. When an object goes on the path ABC (in figure), then the displacement of the object is AC. The arrowhead at C shows that the object is displaced from A to C.

Torque is the turning effect of force which is a vector quantity.

Question 2. Vectors \(\vec{A}, \vec{B}\) ,and \(\vec{C}\) are such \(\vec{A} \cdot \vec{B}=0 \quad \vec{A} \cdot \vec{C}=0\). Then the vector parallel to A is:

1. \(\vec{A}×\vec{B}\)
2. \(\vec{B}+\vec{C}\)
3. \(\vec{B}×\vec{C}\)
4. \(\vec{B}\) and \(\vec{C}\)

Answer: 3. \(\vec{B}×\vec{C}\)

From Vector Product,\(\vec{A} \times(\vec{B} \times \vec{C})=(\vec{A} \cdot \vec{C}) \vec{B}-(\vec{A} \cdot \vec{B}) \vec{C}\)

Given that:

⇒  \(\vec{A} \cdot \vec{B}=0, \vec{A} \cdot \vec{C}\)=0,

then\(\vec{A} \times(\vec{B} \times \vec{C})\)=0

∴Thus vector A is paralle to \(\vec{B} \times \vec{C}\).

Question 3. If \(\vec{A}=\vec{A}+\vec{C}\) and their magnitude in ratio 5:4:3 respectively, then angle between vector \(\vec{A} \text { and } \vec{C}\) is:

1. 53°
2. 90°
3. 60°
4. 18°

Answer: 1. 53°

From question,\(\vec{A}=\vec{B}+\vec{C}\) means \(\vec{A}-\vec{C} =\vec{B}\)

Read and Learn More NEET Physics MCQs

⇒  Now, \((\vec{A}-\vec{C})(\vec{A}-\vec{C}) =\vec{B} \cdot \vec{B}\)

\(\overrightarrow{A^2}+\overrightarrow{C^2}-2 \vec{A} \cdot \vec{C} =B^2\)(Self product at both sides)

⇒  \(\overrightarrow{A^2}+\overrightarrow{C^2}-2 \vec{A} \vec{C} \cos \theta=\overrightarrow{B^2}\)

Where θ is the angle between A and C.

⇒  \(\cos\theta=\frac{\overrightarrow{A^2}+\overrightarrow{C^2}-\overrightarrow{B^2}}{2 A C}\)

=\(\frac{(5)^2+(3)^2-(4)^2}{2 \times 5 \times 3}=\frac{18}{30}=\frac{3}{5}\)

⇒  \(\cos \theta=\frac{3}{5}\)

∴ \(\theta=\cos ^{-1}\left[\frac{3}{5}\right]=53^{\circ}\)

Question 4.\(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) are two vectors and 0 is the angle between them, if \(|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|=\sqrt{3}(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}})\)), the value of θ is:

1. 45°
2. 30°
3. 90°
4. 60°

Answer: 4. 60°

⇒ \(|\vec{A}+\vec{B}|=\sqrt{3}(\vec{A} \cdot \vec{B})\)

⇒  \(\vec{A} \vec{B} \sin \theta=\sqrt{3} \vec{A} \vec{B} \cos \theta\)

⇒ \(\tan \theta=\sqrt{3}\)

∴ \(\theta=60^{\circ}\)

Question 5. The vectors \(\vec{A} \text { and } \vec{B}\) are such that \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\). The angle between the two vectors is:

1. 45°
2. 90°
3. 60°
4. 75°

Answer: 2. 90°

According to the question, \(|\vec{A}+\vec{B}| =|\vec{A}-\vec{B}|\)

⇒ \(\quad|\vec{A}+\vec{B}|^2 =|\vec{A}-\vec{B}|^2\)

⇒  \(\overrightarrow{A^2}+\overrightarrow{B^2}+2 \vec{A} \vec{B}=\overrightarrow{A^2}+\overrightarrow{B^2}-2 \vec{A} \vec{B}\)

⇒  \(4 \vec{A} \cdot \vec{B}=0\)

∴ \(\vec{A} \cdot \vec{B}\) =0

⇒ \(\quad \vec{A} \vec{B} \cos \theta\) =0

⇒ \( \quad \theta =90^{\circ}\)

∴ \([As  \vec{A}=\vec{B} \neq 0 ]\)

Question 6. If a vector \(2 \hat{i}+3 \hat{j}+8 \hat{k}\) is perpendicular to the vector \(4 \hat{i}-4 \hat{j}+\alpha \hat{k}\) then the value of a is:

1. \(\frac{1}{2}\)
2. A
3. 1
4. -1

Answer: 2. A

⇒  \(\vec{a}=2 \hat{i}+3 \hat{i}+8 \hat{k}\)

⇒ \(\vec{b}=4 \hat{i}-4 \hat{j}+\alpha \hat{k}\)

⇒ Since \(\vec{a} \text { and } b \text { are perpendicular, }\)

∴ So,\(\vec{a} \cdot \vec{b}\)=0,

⇒  \((2 \hat{i}+3 \hat{j}+8 \hat{k}) \cdot(4 i+4 j+2 k) =\frac{1}{2}\)

⇒ \(8+12+8 \alpha\)=0

∴ \(\alpha =-\frac{1}{2}\)

Question 7. If the angle between the vector\(\vec{A} \text { and }(\vec{B} \times \vec{A}) \cdot \vec{A}\) 0, the value of the product \((\vec{B} \times \vec{A}) \cdot \vec{A}\) is equal to:

1. BA² sin θ
2. BA² cos θ
3. BA² sin θ cos θ
4. zero

Answer: 4. zero

⇒ \((\vec{B} \times \vec{B}) \vec{A}=(B A \sin \theta)(\vec{n} \cdot \vec{A})\)=0

∴ Since n is perpendicular to both \(\vec{A} \text { and } \vec{B}\)

Question 8. If \(|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B} \text { then the value of }|\vec{A} \times \vec{B}| \text { is: }\)

1. \(\left(A^2+B^2+A B\right)^{1 / 2}\)
2. \(\left(A^2+B^2+\frac{A B}{\sqrt{3}}\right)^{\frac{1}{2}}\)
3. A+B
4. \(\left(A^2+B^2+\sqrt{3} A B\right)^{\frac{1}{2}}\)

Answer: 1. \(\left(A^2+B^2+A B\right)^{1 / 2}\)

According To Question, \(\vec{A} \times \vec{B} =\sqrt{3} \vec{A} \cdot \vec{B}\)

⇒ \(\vec{A} \vec{B} \sin \theta =\sqrt{3} \vec{A} \vec{B} \cos \theta\)

⇒ \(\quad \tan \theta=\sqrt{3}\)

⇒  \(\quad \theta =60^{\circ}\)

∴ \(\quad|\vec{A}+\vec{B}| =\sqrt{|\vec{A}|^2+|\vec{B}|^2+2|\vec{A}||\vec{B}| \cos \theta}\)

Question 9. What is the linear velocity, if angular velocity vector ω = \(3 \hat{i}-4 \hat{j}+\hat{k}\) and position vector r = \(5 \hat{i}-6 \hat{j}+6 \hat{k}\)

1. \((A)6 \hat{i}+2 \hat{j}-3 \hat{k}\)
2. \(-18 \hat{i}-13 \hat{j}+2 \hat{k}\)
3. \(18 \hat{i}+13 \hat{j}-2 \hat{k}\)
4. \(6\hat{i}-2 \hat{j}+8 \hat{k}\)

Answer: 2. \(-18 \hat{i}-13 \hat{j}+2 \hat{k}\)

The relation between linear velocity v, angular velocity to, and position vector r is given by,

∴ This confirms from the above that velocity remains constant.

Question 10. The resultant of A x 0 will be equal to:

1. zero
2. A
3. zero vector
4. unit vector

Answer: 3. zero vector

According to the properties of vector product, the cross product of any vector with zero is a null vector or zero vector.

Question 11. If the magnitude of the sum of two vectors is equal to the magnitude of the difference between the two vectors, the angle between these vectors is:

1. 90°
2. 45°
3. 180°
4. 0°

Answer: 1. 90°

From the question, if \(\vec{A} \text { and } \vec{B}\) are two vectors, then

⇒ \(|\vec{A}+\vec{B}|=|\vec{A}+\vec{B}|\)

A² + B² + 2AB cos θ = A²+ B²– 2AB cos θ

⇒  4AB cos θ = 0

cos θ = 0

∴θ = 90°

Question 12. The vector sum of two forces is perpendicular to their vector differences. In this case, the force:

1. are equal to each other
2. are equal to each other in magnitude
3. are not equal to each other in magnitude
4. cannot be predicted

Answer: 2. are equal to each other in magnitude

⇒ \((\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})\cdot(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})=0\)

⇒ \(\mathrm{~A}^2-\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}-\mathrm{B}^2 =0\)

∴ \(\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{B}} \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}})\)

Question 13. The magnitudes of vectors A, B, and C are 3, 4, and 5 units respectively. If A + B = C, the angle between A and B is:

1. \(\frac{\pi}{2}\)
2. \(\cos ^{-1}(0.6)\)
3. \(\tan ^{-1}\left(\frac{7}{5}\right)\)
4. \(\frac{\pi}{4}\)

Answer: 1. \(\frac{\pi}{2}\)

From Figure, \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{C}}\)

⇒  Also,\(|\vec{A}|=3,|\vec{B}|=4,|\vec{C}|\)=5

⇒  As \(\quad \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{C}}\)

So,\(\quad 5^2=3^2+4^2+2.4 .3 \cos \theta\)

⇒  \(\cos \theta\)=0

⇒ \(\quad \theta=\frac{\pi}{2}\)

∴ \(\vec{A} \text { is perpendicular to } \vec{B}\)

Question 14. If vectors A=\(\cos \omega t \hat{i}+\sin \omega t \hat{j} and B=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}\) are functions of time, then the value of t at which they are orthogonal to each other:

1. \(t=\frac{\pi}{4 \omega}\)
2. \(\frac{\pi}{2 \omega}\)
3. t=\(\frac{\pi}{\omega}\)
4. t=0

Answer: 3. t=\(\frac{\pi}{\omega}\)

⇒  Given,\(\vec{A}=\cos \omega t \hat{i}+\sin \omega t \hat{j}\) and \(\vec{B}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}\)

Since, they are orthogonal to each other, \(\vec{A} \cdot \vec{B}=0\)

⇒  \((\cos \omega t \hat{i}+\sin \omega t \hat{j}) \cdot\left(\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}\right)\)=0

⇒  \(\cos \omega t \cdot \cos \frac{\omega t}{2}+\sin \omega t \cdot \sin \frac{\omega t}{2}\)=0

⇒ \(\{\hat{i} \cdot \hat{i}=1, \hat{j} \cdot \hat{j}=1\}\)

⇒  \(\cos \left(\omega t-\frac{\omega t}{2}\right)\)=0

⇒  \( \cos \mathrm{A} \cos \mathrm{B}+\sin A \cdot \sin B=\cos (A-B)\}\)

∴ ⇒\(\cos \frac{\omega t}{2}\)=0

⇒  \(\quad \frac{\omega t}{2}=\frac{\pi}{2}\)

∴ \(\quad t=\frac{\pi}{\omega}\)

Question 15. The between the vectors \(\mathrm{A}=3 \hat{i}+4 \hat{j}+5 \hat{k} \text { and } \mathrm{B}=3 \hat{i}+4 \hat{j}-5 \hat{k}\) will be:

1. 0°
2. 45°
3. 90°
4. 180°

Answer: 3. 90°

The angle between two vectors is given as from dot product A.B = |A| |B| cosθ

cosθ = \(\frac{\mathrm{A} \cdot \mathrm{B}}{\mathrm{AB}}\)

∴ Here,\(\mathrm{A}=3 \hat{i}+4 \hat{j}+5 \hat{k}\)

⇒  \(\mathrm{~B}=3 \hat{i}+4 \hat{j}-5 \hat{k}\)

⇒ \(\mathrm{~A}=\sqrt{(3)^2+(4)^2+(5)^2}=\sqrt{50}\)

⇒ \(\mathrm{~B}=\sqrt{(3)^2+(4)^2+(-5)^2}=\sqrt{50}\)

And,\(\mathrm{~A} \cdot \mathrm{B}=(3 \hat{i}+4 \hat{j}+5 \hat{k}) \cdot(3 \hat{i}+4 \hat{j}-5 \hat{k})\)=9+16-25=0

⇒ \(\cos \theta=\frac{0}{\sqrt{50} \cdot \sqrt{50}}\)=0

∴\(\theta=90^{\circ}\)

Question 16. The magnitudes of vectors A, B, and C are 3, 4, and 5 units respectively. If A + B = C, the angle between A and B is:

1. \(\frac{\pi}{2}\)
2. \(\cos ^{-1}(0.6)\)
3. \(\tan ^{-1}\left(\frac{7}{5}\right)\)
4. \(\frac{\pi}{4}\)

Answer: 1. \(\frac{\pi}{2}\)

In the Figure shown, A+B=C

⇒  Also,\(\quad|\mathrm{A}|=3,|\mathrm{~B}|=4,|\mathrm{C}|=5\)

⇒  As\(\quad \mathrm{A}+\mathrm{B}=\mathrm{C}\)

⇒  So,\(\quad 5^2=3^2+4^2+2 \cdot 4 \cdot 3 \cos \theta\)

⇒ \(\cos \theta=0=90^{\circ}\)

∴A=is perpendicular to B.

Question 17. Six vectors a through / have the magnitudes and direction indicated in the figure. Which of the following statements is true?

1. \(\vec{b}+\vec{b}=\vec{f}\)
2. \(\vec{d}+\vec{c}=\vec{f}\)
3. \(\vec{d}+\vec{e}=\vec{f}\)
4. \(\vec{b}+\vec{e}=\vec{f}\)

Answer: 3. \(\vec{d}+\vec{e}=\vec{f}\)

According to the parallelogram law of vector addition, we can say that \(\vec{d}+\vec{e}=\vec{f} \text {correct}\)

Question 18. A car starts from rest and accelerates at 5m/s ². At t=4s,  a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s?

1. 20 m/s, 5 m/s²
2. 20 m/s, 0
3. 20 √2 m/s, 0
4. 20 √2 m/s, 10 m/s²

Answer: 4. 20√2 m/s, 10 m/s²

Given that initial velocity u=0

⇒  Acceleration a=5 m/s²

At t=4 ball is dropped.

⇒  velocity of the car at t = 4 sec,

v=u+at

⇒  v=0+5(4):20m/s

At t = 6 sec, acceleration is due to gravity.

⇒  a = g = 10mls

vx=20 m/s (due to car)

⇒  vy=u+at =0+g(2)(downward)

=20 m/s(downward)

v = sq.rt (20² + 20²) = 20 sq.rt²

Question 19. A particle moves in the x-y plane according to the rule x = a sin ω and y = a cos ωt. The particle follows:

1. an elliptical
2. a circular path
3. a parabolic path
4. a straight line path inclined equally to the x and y-axis

Answer: 2. a circular path

It is given that,x=\(a \sin \omega t\)

⇒  \(\frac{x}{a} =\sin \omega t\)    → equation..1

y =\(b \cos \omega t\)

⇒  \(\frac{x}{b} =\cos \omega t\)   →  equation.. 2

From equation 1,2

⇒  \(\frac{x^2}{a^2}+\frac{y^2}{b^2} =1\)

∴ \(x^2+y^2 =a^2\)

Question 20. A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is (g = 10m/s²):

1. 340m
2. 320m
3. 300m
4. 360m

Answer: 3. 300m

Given that u=20 m/s, v=80 m/s and h = ? using third equation of motion

⇒  \(v^2 =u^2+2 g h\)

h=\(\frac{v^2-u^2}{2 g}\)

=\(\frac{(80)^2-(20)^2}{2 \times 10}\)=300m

Question 21. When an object is shot from the bottom of a long smooth inclined plane kept at an angle of 60° horizontal it can travel a distance x1 along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x2 distance. Then x1 : x2 will be

1. √2:1
2. 1:√3
3. 1:2√3
4. 1:√2

Answer: 2. 1:√3

⇒ Stopping distance, x1=\(\frac{u^2}{2 g \sin 60^{\circ}}\)

⇒  Stopping distance, x2=\(\frac{u^2}{2 g \sin 30^{\circ}}\)

⇒  \(\frac{x_1}{x_2}=\frac{\sin 30^{\circ}}{\sin 60^{\circ}}\)

=\(\frac{1 \times 2}{2 \times \sqrt{3}}=\frac{1}{\sqrt{3}}\)

=\(1:\sqrt{3}\)

Question 22. A particle moving with velocity v is acted by three forces shown by the vector triangle PQR. The velocity of the particle will

1. decreases
2. remain constant
3. change according to the smallest force QR
4. increase

Answer: 1. decreases

Net Force=0

⇒  \(\vec{a} =0\)

⇒  F =\(m vec{a}\)

a=\(\frac{F}{m}\)

∴ \(\vec{a} =0\)

Question 23. An object flying in air with velocity \((20 \hat{i}+25 \hat{j}-12 \hat{k})\) suddenly breaks into two pieces whose masses are in the ratio 1:5. The smaller mass flies off with a velocity \((100 \hat{i}+25 \hat{j}-8 \hat{k})\). The velocity of the larger piece will be :

1. \(4 \hat{i}+23 \hat{j}-8 \hat{k}\)
2. \(-100 \hat{i}-35 \hat{j}-8 \hat{k}\)
3. \(20 \hat{i}+15 \hat{j}-80 \hat{k}\)
4. \(-20 \hat{i}-15 \hat{j}-80 \hat{k}\)

Answer: 1. \(4 \hat{i}+23 \hat{j}-8 \hat{k}\)

Let m be the mass of the object that is flying with velocity \((20 \hat{i}+25 \hat{j}-12 \hat{k})\) is in the air. According to the question, the mass breaks into two pieces of ratio 1: 5, which means the small piece is \(\frac{7 m}{6}\)and another piece is \(\frac{5 m}{6}.\)

Using the law of conservation of momentum,

Initial momentum = Final momentum

⇒  \(\not h(20 \hat{i}+25 \hat{j}-12 \hat{k})=\frac{\not h}{6}(100 \hat{i}+35 \hat{j}-8 \hat{k})+\frac{5}{6} \not h v\)

⇒  \((120 \hat{i}+150 \hat{j}-72 \hat{k}) =(100 \hat{i}+35 \hat{j}+8 \hat{k})+5 v\)

v =\(\frac{1}{5}(20 \hat{i}+115 \hat{j}-80 \hat{k})\)

∴v=\(4 \hat{i}+23 \hat{j}-16 \hat{k}\)

Question 24. A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to:

1. \(\frac{7}{5} \mathrm{D}\)
2. D
3. \(\frac{3}{2} D\)
4. \(\frac{5}{4} D\)

Answer: 4. \(\frac{5}{4} D\)

Since the track is frictionless.

⇒  So, total mechanical energy remains constant.

Total Mechanical Energy₁ Total Mechanical Energy₂

⇒ 0+mgh =\(\frac{1}{2} m v_{\mathrm{L}}^2+m g \Delta\)

⇒  h =\(\frac{v_{\mathrm{L}}{ }^2}{2 g}\)

For completing the verticle circle,

⇒  \(v_L \geq \sqrt{g \mathrm{R}}=\sqrt{\frac{\Delta g}{2}}\)

mg =\(\frac{1}{2} m \frac{\Delta}{2} g+m g \Delta\)

∴ h =\(\frac{5}{4} \Delta\)

Question 25. One end of the string of length 1 is connected to a particle of mass m and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v, the net force on the particle (directed towards the center) will be (T represents the tension in the string):

1. T
2. \(T+\frac{m v^2}{l}\)
3. \(T-\frac{m v^2}{l}\)
4. Zero

Answer: 1. T

According to the question, string l connected to a particle is shown below.

⇒  Since the particle is in a uniform circular motion, the net force is equal to the centripetal force.

Centripetal force = Net force

∴ \(\frac{m v^2}{l}=T\)

Question 26. Two cars P and Q start from a point at the same time in a straight line and their positions are represented by X_p(t) = at + bt² and X_qt =ft – t². At what time do the cars have the same velocity?

1. \(\frac{a-t}{1+b}\)
2. \(\frac{a+t}{2(b-1)}\)
3. \(\frac{a+f}{2(1+b)}\)
4. \(\frac{f-a}{2(1+b)}\)

Answer: 4. \(\frac{f-a}{2(1+b)}\)

Position of car P, Xp(t) = at + bt²

velocity of car P,\(v_P=\frac{d x_p(t)}{d t}=a+2 b t\)

⇒ Position of car Q,\(x_{\mathrm{Q}}(t)=t-t^2\)

velocity of car Q,\(v_Q=f-2 t\)

Since, \( V_P=V_Q\)

a+2bt=t-2 t

∴ t=\(\frac{f-a}{2(b+1)}\)

Question 27. Two particles A and B move with constant velocities v₁ and v₂. At the initial moment, their position vectors are r₁ and r₂ respectively. The condition for particles A and B for their collision is:

1. \(\frac{r_1+r_2}{\left|r_1-r_2\right|}=\frac{v_2+v_1}{\left|v_2-v_1\right|}\)
2. \(r_1 \cdot v_1=r_2 \cdot v_2\)
3. \(r_1 \times v_1=r_2 \times v_2\)
4. \(r_1-r_2=v_1-v_2\)

Answer: 1. \(\frac{r_1+r_2}{\left|r_1-r_2\right|}=\frac{v_2+v_1}{\left|v_2-v_1\right|}\)

Displacement OF Particle is \(R=r_1+v_1 t=r_2+v_2 t\)

⇒  \(\text { i.e. } r_1-r_2=\left(v_2-v_1\right) t\)

= \(\frac{r_1-r_2}{\left|r_1-r_2\right|}=\frac{\left|v_2-v_1\right|}{\left|v_2-v_1\right|} t\)

∴ \(\frac{r_1-r_2}{\left|r_1-r_2\right|}=\frac{\left(v_2-v_1\right)}{\left|v_2-v_1\right|}\)

Question 28. A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2 s and (13 m, 14 m) at time t = 5 s. Average velocity vector (Vav) from / = 0 to t = 5 s is:

1. \(\frac{1}{5}(13 \hat{i}+13 \hat{j})\)
2. \(\frac{7}{3}(\hat{i}+\hat{j})\)
3. \(2(\hat{i}+\hat{j})\)
4. \(\frac{11}{5}(\hat{i}+\hat{j})\)

Answer: 4. \(\frac{11}{5}(\hat{i}+\hat{j})\)

Here, Net displacement,\(\overrightarrow{\mathrm{S}}_1=(13-2) \hat{i}+(4-3) \hat{j}=11 \hat{i}+11 \hat{j}\) and Time taken (5-0) = 5 sec

⇒  Average Velocity ,V_av=\(\frac{\text { Net displacement }}{\text { Taken time }}\)

= \(\frac{11 \hat{i}+11 \hat{j}}{5}\)

= \(\frac{11}{5}(\hat{i}+\hat{j})\)

Question 29. The velocity of a projectile at the initial point A is \((2 \hat{i}+3 \hat{j})\) m/s. Its velocity (in m/s) at point B will be:

1. \(-2 \hat{i}-3 \hat{j}\)
2. \(-2 \hat{i}+3 \hat{j}\)
3. \(2 \hat{i}-3 \hat{j}\)
4. \(2 \hat{i}+3 \hat{j}\)

Answer: 3. \(2 \hat{i}-3 \hat{j}\)

From the graph, the x-component remains unchanged and the component is reversed. So, the velocity at point B is (21 -3 j) ms^-1.

∴ [Here x-compound does not change and change y-component]

Question 30. A particle has initial velocity \((3 \hat{i}+4 \hat{j})\) and has . acceleration \((0.4 \hat{i}+0.3 \hat{j})\). Its speed after 10 s is:

1. 7 unit
2. 1√2 unit
3. 8.5 unit
4. 10 unit

Answer: 2. 1√2 unit

We Know That the First Equation OF Motion

v =u+a t

⇒  \(\vec{v} =(3 \hat{i}+4 \hat{j})+(0.4 \hat{i}+0.3 \hat{i}) 10\)

=\(3 \hat{i}+4 \hat{j}+4 \hat{i}+3 \hat{j}\)

⇒ \(\vec{v} =7 \hat{i}+7 \hat{j}\)

⇒  \(|v|=\sqrt{7^2+7^2}=\sqrt{49+49}\)

= \(7 \sqrt{2} \mathrm{~m} / \mathrm{s}\)

Question 31. The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is:

1. 30 ms^-2 downwards
2. 4 ms^-2 upwards
3. 4 ms^-2 downwards
4. 14 ms^-2 upwards

Answer: 2. 4 ms^-2 upwards

⇒ When the lift is accelerating upward, Apparent weight > actual weight the equation becomes, R – mg = ma

28000 – 20000 = 2000 a

∴ \(a=\frac{8000}{2000}=4 \mathrm{~ms}^{-2} \text { upwards }\)

Question 32. An object of mass 3 kg is at rest. Now a force \(\vec{F}=6 t^2 \hat{i}+4 t \hat{j}\) is applied on the object, then the velocity of the object at t = 3 s is:

1. \(18 \hat{i}+3 \hat{j}\)
2. \(18 \hat{i}+6 \hat{j}\)
3. \(3 \hat{i}+6 \hat{j}\)
4. \(18 \hat{i}+4 \hat{j}\)

Answer: 4. \(18 \hat{i}+4 \hat{j}\)

Let the velocity along x and y-axes be v_x and v_y respectively\(v_x=\frac{d x}{d t} \text { and } v_{\mathrm{y}}=\frac{d y}{d t}\)

⇒  From figure,\(\tan \alpha=\frac{y}{x}\)

⇒  \(y=x \tan \alpha\)

Differentiating Equation 1 We Get

⇒ \(\frac{d y}{d t}=\frac{d x}{d t} \tan \alpha\)

⇒ \(v_y=v_x \tan \alpha\)

∴ \(v_x=10 \tan 60^{\circ} =10 \sqrt{3}=17.3 \mathrm{~m} / \mathrm{s}\)

Question 33. Two particles A and B are connected by a rigid rod AB. The rod slides along perpendicular rails as shown here. The velocity of A to the right is 10 m/s. What is the velocity of B when angle a = 60°?

1. 9.8m/s
2. 10m/s
3. 5.8m/s
4. 17.3m/s

Answer: 2. 10m/s

From Question, \(\vec{a}=\frac{\vec{F}}{m}=2 t^2 \hat{i}+\frac{4}{3} t \hat{j}\)

⇒ \(d \vec{V}=\left(2 t^2 \hat{i}+\frac{4}{3} t \hat{j}\right) d t\)

⇒  Integrate On Both,we get\(\vec{V} =2\left[\frac{t^3}{3}\right] \hat{i}+\frac{4}{3}\left[\frac{t^2}{2}\right] \hat{j}\)

∴ \(\text { At } t=3 \sec , \vec{V} =\frac{2}{3}(3)^3 i+\frac{4}{6}(3)^2 j =18 \hat{i}+6 \hat{j}\)

Question 34. The position vector of a particle is r=\((\mathrm{a} \cos \omega t) \hat{i}+ (\mathrm{a} \sin \omega t) \hat{j}\) The velocity of the particle is

1. directed toward the origin
2. directed away from the origin
3. parallel to the position vector
4. perpendicular to the position vector

Answer: 4. perpendicular to the position vector

velocity is rate of change of position vector, i.e.,\(\mathrm{v}=\frac{d \boldsymbol{r}}{d t}\) where r is the position vector.\(\mathbf{v}=\frac{d}{d t}[(\mathrm{a} \cos \omega t) \hat{i}+(\mathrm{a} \sin \omega t) \hat{j}]\)

=\((-\mathrm{a} \omega \sin \omega t) \hat{i}+(\mathrm{a} \omega \cos \omega t)\)

=\(\omega[(-\mathrm{a} \sin \omega t) \hat{i}+(\mathrm{a} \cos \omega t) \hat{j}]\)

Slope Of Position Vector\(=\frac{a \sin \omega t}{a \cos \omega t}=\tan \omega t\)

And Slope OF Velocity Vector=\(\frac{-a \cos \omega t}{a \sin \omega t}= -\frac{1}{\tan \omega t}\)

∴ Velocity Is Perpendicular To The Displacement.

Question 35. A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time t₁ if the elevator is at rest and in time t₂ if the elevator is moving uniformly. Then:

1. t₁ < t₂ or t₁ > t₂ depending on whether the lift is going up or down
2. t₁ < t₂
3. t₁ > t₂
4. t₁ = t₂

Answer: 4. t₁ = t₂

⇒  In both cases, elevation is an internal frame of reference. So, effective gravity remains the same in both frames.

∴ Hence, the time of fall remains the same in both cases. (Since initial velocity is the same in both frames.) Hence, t₁ = t₂

Question 36. A block of mass m is placed on a smooth inclined wedge ABC of inclination 0 as shown in the figure. The wedge is given an acceleration towards the right. The relation between a and θ for the block to remain stationary on the wedge is:

1. a=\(g \cos \theta\)
2. a=\(\frac{g}{\sin \theta}\)
3. a=\(\frac{g}{{cosec} \theta}\)
4. a=\(g \tan \theta\)

Answer: 4. a=\(g \tan \theta\)

In a non-inertial frame,

⇒  At sin θ = ma  →  Equation-1

N cos θ = mg  →  Equation -2

⇒ From equation. (1) and (2), we get \(\tan \theta =\frac{a}{g}\)

∴ \(a=g \tan \theta\)

Question 37. The x and y coordinates of the particle at any time are x = 5t² – 2t² and y = 10t respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 s is:

1. 0
2. 5 m/s²
3. – 4 m/s²
4. – 2m/s²

Answer: 4. – 2m/s²

Given,x=\((t+5)^{-1}\)

⇒  Velocity, v=\(\frac{d x}{d t}=(t+5)^{-2}\)

⇒  acceleration,a=\(\frac{d v}{d t}\)

⇒ \(a \propto(t+5)^{-3}\)

∴ \(\quad a^2 \propto(t+5)^{-3+2}\)

Question 38. The position vector of a particle R as a function of time is given by:
\(R=4 \sin (2 \pi t) \hat{i}+4 \cos (2 \pi t) \hat{j}\) where R in meter, t is in seconds, and \(\hat{i} \text { and } \hat{j}\)denote unit vectors along x and y-directions, respectively. Which one of the following statements is wrong for the motion of a particle?

1. Acceleration is along – R.
2. The magnitude of the acceleration vector is \(\frac{v^2}{R}\), where v is the velocity of the particle.
3. The magnitude of the velocity of the particle is 8 m/s.
4. The path of the particle is a circle of radius 4 m.

Answer: 3. Magnitude of the velocity of the particle is 8 m/s.

Here,\(\vec{R}=4 \sin (2 \pi t) \hat{i}+4 \cos (2 \pi t) \hat{j}\)

⇒The velocity of the particle is,\(\vec{v}=\frac{d \vec{R}}{d t}=\frac{d}{d t}[4 \sin (2 \pi t) \hat{i}+4 \cos (2 \pi t) \hat{j}]=8 \pi \cos 2 \pi t \hat{i}-8 \pi \sin (2 \pi t) \hat{j}\)

Its magnitude is\(|\vec{v}|=\sqrt{(8 \pi \cos (2 \pi t))^2+(-8 \pi \sin (2 \pi t))^2}\)

=\(\sqrt{64 \pi^2 \cos ^2(2 \pi t)+64 \pi^2 \sin ^2(2 \pi t)}\)

=\(\sqrt{64 \pi^2\left[\cos ^2(2 \pi t)+\sin ^2(2 \pi t)\right]}\)

=\(\sqrt{64 \pi^2}=8 \pi \mathrm{m} / \mathrm{s}\)

∴\(\quad\left(\text { as } \sin ^2 \theta+\cos ^2 \theta=1\right)\)

Question 39. A particle has initial velocity \((2 \hat{i}+3 \hat{j})\) and acceleration\((0.3 \hat{i}+02 \hat{j})\). The magnitude of velocity after 10 s will be:

1. 9√2 units
2. 5√2
3. 5 units
4. 9 units

Answer: 1. 9√2 units

Given that, velocity, \(\vec{u}=2 \hat{i}+3 \hat{j}\)

Acceleration, \(\vec{a}=(0.3 \hat{i}+0.2 \hat{j})\)

By the first equation of motion, v =u + at

⇒ \(v=(2 \hat{i}+3 \hat{j})+(0.3 \hat{i}+0.2 \hat{j}) \times 10 \\=5 \hat{i}+5 \hat{j}\)

∴ v =\(5 \sqrt{2} \text { units }\)

Question 40. A particle moves a distance x in time / according to equation x = (t + 5)^–¹. The acceleration of the particle is proportional to:

1. (velocity )^3/2
2. (density)²
3. (distance)^-2
4. (velocity)^2/3

Answer: 3. (distance)^-2

Given,x =5t-2t²

⇒ Velocity \(v_x =\frac{d x}{d t}=\frac{d}{d t}\left(5 t-2 t^2\right)=5-4 t\)

Acceleration,\(a_x =\frac{d}{d t}(v)=\frac{d}{d t}(5-4 t)=-4 \mathrm{~m} / \mathrm{s}^2\)

Again, y=10t

⇒ Velocity \(v_y=\frac{d}{d t}(10 t)=10\)

Acceleration,\(a_y=\frac{d}{d t}\left(v_y\right)=\frac{d}{d t}(10)=0\)

∴ Net acceleration of particle,\(⇒{a_{n e t}}=a_x \hat{i}+a_y \hat{j}=-4 \hat{i}+0=-4 \hat{i} \mathrm{~ms}^{-2}=4 \mathrm{~m} / \mathrm{s}^2\)

Question 41. A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 m-1 while passing through a distance of 135 m in t second. The value of t is:

1. 10
2. 1.8
3. 12
4. 9

Answer: 4. 9

For Constant Acceleration, \(v^2=u^2+2 as\)

⇒ \((2)^2 =(10)^2+2 \times 9 \times 135\)

⇒a =\(\frac{300}{270} \mathrm{~ms}^{-2}\)

⇒ As, v =\(u+a t\)

⇒ \(0+a \times t\)

⇒ 10 =\(\frac{300}{270} \times t\)

∴ t =\(9 \mathrm{sec}\)

Question 42. A ball is projected with a velocity, of 10 ms-1, at an angle of 60° with the vertical direction. Its speed and the highest point of its trajectory will be:

1. 5√3 ms^-1
2. 5 ms¹
3. 10 ms^-1
4. zero

Answer: 1. 5 √3 ms^-1

Given: velocity of ball =10 m/s,

Angle with vertical direction = 60°

⇒ From equation (1) we can say that the speed of the ball at the highest point = v cosθ where θ = 90 – 60 = 30°

⇒ \(\mathrm{v}=10 \times \cos (30)=10 \times((\sqrt{3}) / 2)\)

∴ \(\mathrm{v}=5 \sqrt{3} \mathrm{~m} / \mathrm{s}\)

Question 43. A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle ‘θ’ to the horizontal, the maximum height attained by it equals 4R. The angle of projection, 0, is then given by:

1. \(\theta=\cos ^{-1}\left(\frac{g \mathrm{~T}^2}{\pi^2 \mathrm{R}}\right)^{\frac{1}{2}}\)
2. \(\theta=\cos ^{-1}\left(\frac{\pi^2 \mathrm{R}}{g \mathrm{~T}^2}\right)^{\frac{1}{2}}\)
3. \(\theta=\sin ^{-1}\left(\frac{\pi^2 \mathrm{R}}{g \mathrm{~T}^2}\right)^{\frac{1}{2}}\)
4. \(\theta=\sin ^{-1}\left(\frac{2 g \mathrm{~T}^2}{\pi^2 \mathrm{R}}\right)^{\frac{1}{2}}\)

Answer: 4. \(\quad \theta=\sin ^{-1}\left(\frac{2 g \mathrm{~T}^2}{\pi^2 \mathrm{R}}\right)^{\frac{1}{2}}\)

Given, Radius = R

⇒ Time = T for one revolution Angle of projection = 0

Hmax=4R , θ=?

⇒ \(\mathrm{v}=\frac{2 \pi \mathrm{R}}{\mathrm{T}}\) from circular motion.

⇒ \({H}_{\max }=\frac{\mathrm{v}_y^2}{2 g}=\frac{\mathrm{v}^2 \sin ^2 \theta}{2 g}=4 \mathrm{R}\)

⇒ \(\frac{4 \pi^2 \mathrm{R}^2}{\mathrm{~T}^2} \times \frac{\sin ^2 \theta}{2 g}=4 \mathrm{R}\)

⇒ \(\sin ^2 \theta=\frac{8 g \mathrm{~T}^2}{4 \pi^2 \mathrm{R}}=\frac{2 g \mathrm{~T}^2}{\pi^2 \mathrm{R}}\)

∴ \(\theta=\sin ^{-1}\left(\frac{2 g \mathrm{~T}^2}{\pi^2 \mathrm{R}}\right)^{1 / 2}\)

Question 44. The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is:

1. 60°
2. 15°
3. 30°
4. 45°

Answer: 1. 60°

The speed of a projectile at its maximum height\(v^{\prime}=v_0 \cos \theta\)

⇒ \(\frac{v_0}{2}=v_0 \cos \theta\)

⇒ \(\cos \theta=\frac{1}{2}\)

⇒ \(\theta=60^{\circ}\)

Question 45. A projectile is fired from the surface of the earth with a velocity of 5 ms^-1 and angle δ with the horizontal. Another projectile fired from another planet with a velocity of 3 ms^-1 at the same angle follows a trajectory that is identical to the trajectory of the projectile fired from the Earth. The value of the acceleration due to gravity on the planet is (in ms^-2) is (given, g = 9.8 ms^-2):

1. 3.5
2. 5.9
3. 16.3
4. 110.8

Answer: 1. 3.5

The equation of trajectory is y =\(x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta},\) where 0 is the angle of projection and u is the velocity with which projectile is projected.

⇒ For equal trajectories and for same angles of projection,\(\frac{g}{u^2}=\text { constant }\)

As per Question,\(\frac{9.8}{(5)^2}=\frac{g^{\prime}}{(3)^2}\)

where θ is the acceleration due to gravity on the planet,\(g^{\prime}=\frac{9.8 \times 9}{25}=3.5 \mathrm{~m} / \mathrm{s}^2\)

Question 46. The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is:

1. \(\theta=\tan ^{-1}\left(\frac{1}{4}\right)\)
2. θ = tan^-1 (4)
3. θ = tan^-1 (2)
4. θ = 45°

Answer: 2. θ = tan’1 (4)

We know that,Range,\(R=\frac{u^2(2 \sin \theta \cos \theta)}{g}\)

⇒ H=\(\frac{u^2 \sin ^2 \theta}{2 g}\)

According To the Question, R=H

⇒ \(\frac{u^2(2 \sin \theta \cos \theta)}{g}=\frac{u^2 \sin ^2 \theta}{2 g}\)

⇒ \(2 \cos \theta=\frac{\sin 0}{2}\)

∴ \(\tan \theta =4 \theta =\tan ^{-1}(4)\)

Question 47. A missile is fired for maximum range with an initial velocity of 20 m/s, if g = 10 m/s², the range of the missile is:

1. 50 m
2. 60 m
3. 20 m
4. 40 m

Answer: 4. 40 m

∴ As we know Maximum Range, \(R_{\max }=\frac{u^2}{g}=\frac{(20)^2}{10}=40 \mathrm{~m}\)

Question 48. A projectile is fired at an angle of 45° with the horizontal. The elevation angle of the projectile at its highest point as seen from the point of projection is:

1. \(60^{\circ}\)
2. \(\tan ^{-1}\left(\frac{1}{2}\right)\)
3. \(\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
4. \(45^{\circ}\)

Answer: 2. \(\tan ^{-1}\left(\frac{1}{2}\right)\)

Height of Projectile,

⇒ H=\(\frac{u^2 \sin ^2 \theta}{2 g}\)

=\(\frac{u^2 \sin ^2 45^{\circ}}{2 g}=\frac{u^2}{4 g}\) Equation 1

⇒ And Range Of Projectile ,R=\(\frac{u^2 \sin 2 \theta}{g}\)

=\(\frac{u^2 \sin 90^{\circ}}{g}=\frac{u^2}{g}\)    →   Equation 2

⇒ \(\frac{R}{2}=\frac{u^2}{2 g}\)  →  Equation 2

⇒ \(\quad \tan \alpha=\frac{H}{R / 2}\)

⇒ considering eq. (1) and eq. (2)

⇒ \(\tan \alpha=\frac{1}{2}\)

∴ \(\alpha=\tan ^{-1}\left(\frac{1}{2}\right)\)

Question 49. A particle of mass is projected with velocity v making an angle of 45° with the horizontal. When the particle lends on the level ground the magnitude of the change in its momentum will be:

1. 2 mv
2. \(\frac{m v}{\sqrt{2}}\)
3. \(m v \sqrt{2}\)
4. zero

Answer: 3. \(m v \sqrt{2}\)

The situation is shown in the figure,

Change in momentum \(\vec{\Delta} P =\vec{P}_f-\vec{P}_i =m\left(\overrightarrow{\hat{v}}_f-\vec{v}_i\right)\)

= \(m\left[\left(v \cos 45^{\circ} i-v \sin 45^{\circ} j\right)\right.\left.-\left(v \cos 45^{\circ} i+v \sin45^{\circ} j\right)\right]\)

= \(m\left[\left(\frac{v}{\sqrt{2}} i-\frac{v}{\sqrt{2}} j\right)-\left(\frac{v}{\sqrt{2}} \mathrm{i}+\frac{v}{\sqrt{2}} j\right)\right]=-\sqrt{2} m v j\)

∴ \(\quad|\overrightarrow{\Delta P}|=\sqrt{2} \mathrm{mv}\)

Question 50. For angles of projection of the projectile at angles (45° + θ) and (45° – θ), the horizontal range described by the projectile is in the ratio of:

1. 2: 1
2. 1:1
3. 2:3
4. 1:2

Answer: 2. 1:1

⇒ Horizontal range,\(R=\frac{u^2 \sin ^2 \theta}{g}\)

When the angle of projection is (45° – 0) then

⇒ \(R_1=\frac{u^2 \sin 2\left(45^{\circ}-\theta\right)}{g}\)

= \(\frac{u^2 \sin \left(90^{\circ}-\theta\right)}{g}\)

= \(\frac{u^2 \cos 2 \theta}{g}\)

∴ Now,\(\frac{R_1}{R_2}=\frac{u^2 \cos \frac{2 \theta}{8}}{u^2 \cos \frac{2 \theta}{g}}=1: 1\)

Question 51. The maximum range of a gun on horizontal terrain is 16 km. If g = 10 ms^-2, then the muzzle velocity of a shell must be:

1. 160 ms^-1
2. 200 V2 ms^-1
3. 400 ms^-1
4. 800 ms^-1

Answer: 3. 400 ms^-1

Range of projectile is given by,\(R=\frac{u^2 \sin 2 \theta}{g}\)

⇒  For maximum range, the angle should be 45°.\(R_{\max }=\frac{u^2 \sin 2 \times 45}{g}\)

=\(\frac{u^2 \sin 2 \theta}{g}\)

⇒  \(R_{\text {max }}=\frac{u^2}{g}\)

⇒ \(R_{\max }=\frac{u^2}{g}=16 \mathrm{~km}=16000 \mathrm{~m}\) or

∴ u =\(\sqrt{16000 \times g}=\sqrt{16000 \times 10}=400 \mathrm{~ms}^{-1}\)

Question 52. An electric fan has blades of length 30 cm measured from the axis of rotation. If the fan is rotating at 120 rev/ min, the acceleration of a point on the tip of the blade is:

1. 1600 ms^-2
2. 47.4 ms^-2
3. 23.7 ms^-2
4. 50.55 ms^-2

Answer: 2. 47.4 ms^-2

We have, centripetal acceleration of rotating body \(a_c=\frac{v^2}{r}=\frac{v^2 \omega^2}{r}=r \omega^2\)

Since,ω= 2πv,

∴ where v is the frequency of rotation. a_c = r (2πv)² = r x 4π²V² ⇒  r =30cm = 30x 10^-2m = 0.30m

⇒ \(v=120 \mathrm{rev} / \mathrm{m}=\frac{120}{60} \mathrm{rev} / \mathrm{s}=2 \mathrm{rev} / \mathrm{s}\)

∴ a_c =(0.30x4x3.14×3.14x2x2) = 47.4 ms^-2

Question 53. A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:

1. the wire is horizontal.
2. the mass is at the lowest point.
3. inclined at an angle of 60° from vertical.
4. the mass is at the highest point

Answer: 2. the mass is at the lowest point.

From question, \(T-m g=\frac{m u^2}{l}\)

⇒ \(T=m g+\frac{m u^2}{l}\)

∴ The tension is maximum at the lowest position of the mass, so the chance of breaking is maximum.

Question 54. A particle moves so that its position vector is given by r = cos cot x + sin cot y, where co is a constant. Which of the following statements is true?

1. Velocity and acceleration both are parallel for r.
2. Velocity is perpendicular to r and acceleration is directed toward the origin
3. Velocity is perpendicular to r and acceleration is directed away from the origin
4. Velocity and acceleration both are perpendicular to

Answer: 4. Velocity and acceleration both are perpendicular to r.

According to the question,

⇒ \(\mu m g \geq \frac{m v^2}{R}\)

⇒ \(v \geq \sqrt{\mu R g}\)

∴ \(v_{\max }=\sqrt{\mu R g}\)

Question 55. A disc and a sphere of the same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?

1. Sphere
2. Both reach at the same time
3. Depends on their masses
4. Discs

Answer: 2. Both reach at the same time

⇒ \(\vec{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}\)

⇒ \(\vec{v}=\frac{d r}{d t}=\omega \sin \omega t \hat{x}+\omega \cos \omega t \hat{y}\)

⇒ \(\vec{a}=\frac{d v}{d t}=\omega^2 \cos \omega t \hat{x}+\omega \sin \omega t \hat{y}\)

=\(-\omega^2 \vec{r}\)

⇒ \(\vec{r} \cdot \vec{v}=0 \)

Hence,\(\quad \vec{r} \perp \vec{v}\)

∴ \(\vec{a}\) is directed towards the origin.

Question 56. A car of mass m is moving on a level circular track of radius R. If π_s represents the static friction between the road and tires of the car, the maximum speed of the car in circular motion is given by:

1. \(\sqrt{\mu_s m R g}\)
2. (\(\sqrt{\frac{\frac{R g}{\mu_s}}{\mu_s G g}}\)
3. \(\sqrt{m R g / \mu_s}\)
4. \(\sqrt{\mu_s N g}\)

Answer: 1.

We have the acceleration of an object rolling down an inclined plane is, a=\(\frac{g \sin \theta}{1+\frac{K^2}{R^2}}\)

⇒ For Disc, \(\frac{K^2}{R^2}=\frac{1}{2}=0.5\)

For Sphere, \(\frac{K^2}{R^2}=\frac{2}{5}=0.4\)

⇒  a (sphere) > a (disc)

∴ Hence, the sphere will reach at bottom first.

Motion In A Straight Line

Question 1. Two particles move with initial speeds of 4 ms^-1 and 2 ms^-1 and their accelerations are 1 ms^-2 and 2 ms^-2 respectively. If both reach at final position at the same time, then the length of the path travelled is:

1. 12 m
2. 24 m
3. 36 m
4. 48 m

Answer: 2. 24 m

According to the question, two particles reach at final position in the same time distance covered S and time t will remain the same.

⇒ From equation of motion,\(S=u t+\frac{1}{2} a t^2\)

For ^st particle S=\(4(t)+\frac{1}{2}(1) t^2 =4 t+\frac{t^2}{2}\)

⇒ For 2^nd particle S=\(2(t)+\frac{1}{2}(2) t^2 =2 t+t^2\)

Equating equation .1 and equation .2 we get,\(4 t+\frac{t^2}{2}=2 t+t^2\) t=4s

∴ Putting the value of t in equation .1 we have\(S=4(4)+\frac{16}{2}=24 \mathrm{~m}\)

Question 2. A particle starting from the origin (0,0) moves in a straight line in the (x,y)plane. Its coordinates at a later time are (√3,3) The path of the particle makes with the x-axis an angle of:

1. 45°
2. 60°
3. 0°
4. 30°

Answer: 2. 60°

Here, \(\tan \theta=\frac{3}{\sqrt{3}}=\sqrt{3}\)

∴ θ=60

Question 3. The displacement-time graph of moving particles is shown below.  The instantaneous velocity of the particle is negative at the point:

1. D
2. F
3. C
4. E

Answer: 4. E

∴ Instantaneous velocity is the slope of the displacement time graph. At point E, the slope is negative so the instantaneous velocity of the particle is negative. At points C and 4 the slope is positive and at D, the slope is zero.

Read and Learn More NEET Physics MCQs

Question 4. A body starts from rest, what is the ratio of the distance travelled by the body during the 4^th and 3^rd s?

1. \(\frac{7}{5}\)
2. \(\frac{5}{7}\)
3. \(\frac{7}{3}\)
4. \(\frac{3}{7}\)

Answer: 1. \(\frac{7}{5}\)

Distance travelled by the body in the second is given by \(s_n=u+\frac{a}{2}(2 n-1)\)

Here, u=0

⇒ For 4th second,\(s_4=\frac{a}{2}(2 \times 4-1)\)

For3’d second,\(\frac{a}{2}(2 \times 3-1)\)

∴ Hence,\(\frac{s_4}{s_3}=\frac{(2 \times 41)}{(2 \times 3 \quad 1)}=\frac{7}{5}\)

Question 5. What will be the ratio of the distance moved by a freely falling body from rest in the 4^th and 5^th second of the journey?

1. 4: 5
2. 7: 9
3. 16: 25
4. 1: 1

Answer: 2. 7: 9

As distance travelled in nth sec is given by, \(s_n=u+\frac{1}{2} a(2 n-1)\)

Here, w= 0, acceleration due to gravity a = 9.8 m/s²

⇒ For 4^th second,\(s_4=\frac{1}{2} \times 9.8(2 \times 4-1)\)

And for 5^th second,\(s_5=\frac{1}{2} \times 9.8(2 \times 5-1)\)

∴ \(\frac{s_4}{s_5}=\frac{7}{9}\)

Question 6. A person travelling in a straight line moves with a constant velocity of v₁ for a certain distance V and with a constant velocity of v₂ for the next equal distance. The average velocity v is given by the relation :

1. \(\frac{1}{v}=\frac{1}{v_1}+\frac{1}{v_2}\)
2. \(\frac{2}{v}=\frac{1}{v_1}+\frac{1}{v_2}\)
3. \(\frac{v}{2}=\frac{v_1+v_2}{2}\)
4. \(v=\sqrt{v_1 v_2}\)

Answer: 2. \(\frac{2}{v}=\frac{1}{v_1}+\frac{1}{v_2}\)

⇒ \(t_1 =\frac{x}{v_1} \text { and } t_2=\frac{x}{v_2}\)

v =\(\frac{x+x}{t_1+t_2}=\frac{2 x}{\frac{x}{v_1}+\frac{x}{v_2}}=\frac{2 v_1 v_2}{v_1+v_2}\)

∴ \(\frac{2}{v}=\frac{1}{v_1}+\frac{1}{v_2}\)

Question 7. A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E. Due to the force qE, its velocity increases from 0 to 6 m/s in one second. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds respectively will be:

1. 1 m/s, 3.5 m/s
2. 1 m/s, 3 m/s
3. 2 m/s, 4 m/s
4. 1.5 m/s, 3 m/s

Answer: 2. 1 m/s, 3 m/s

0s < t < Is: velocity increases from 0 m/s to 6 m/s

⇒ ls<t<2s: velocity decreases from 6 m/s to 0 m/s but car continues to move forward

⇒ 2s < t < 3s: Since field strength is the same.

That is the same acceleration.

Therefore, the car’s velocity increases from 0 m/s to -6m/s

Displacement in three seconds = Area under graph = Area of AOPO’ + Area of APO’Q – Area of QRS

⇒ \(\frac{1}{2} \times 1 \times 6+\frac{1}{2} \times 1 \times 6-\frac{1}{2} \times 6 \times 1=3 \mathrm{~m}\)

Average velocity =\(\frac{3}{3}\)= 1 ms-1

Total travelled distance = 9 m

∴Average speed = \(\frac{9}{3}\) = 3 ms-1

Question 8. Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time ty. On other days, if she remains stationary on the moving escalator, then the escalator takes up time f₂. The time taken by her to walk up on the moving escalator will be:

1. \(\frac{t_1+t_2}{2}\)
2. \(\frac{t_1+t_2}{t_2-t_1}\)
3. \(\frac{t_1 t_2}{t_1+t_2}\)
4. \(t_1-t_2\)

Answer: 3. \(\frac{t_1 t_2}{t_1+t_2}\)

Velocity of girl with respect to elevator= \(\frac{d}{t_1}=\mathrm{v}_1\)

Velocity of elevator with respect to ground =\(\frac{d}{t_2}=\mathrm{v}_2\)

The velocity of the girl concerning ground = vg=v₁+v₂

⇒ \(\frac{d}{t}=\frac{d}{t_1}+\frac{d}{t_2}\)

⇒ \(\frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_2}\)

∴ \(\mathrm{t}=\frac{t_1 t_2}{t_1+t_2}\)

Question 9. A particle covers half of its total distance with speed v₁ and the rest half the distance with speed v₂. Its average speed during the complete journey is:

1. \(\frac{v_1 v_2}{v_1+v_2}\)
2. \(\frac{2 v_1 v_2}{v_1+v_2}\)
3. \(\frac{2 v_1 v_2}{v_1+v_2}\)
4. \(\frac{v_1+v_2}{2}\)

Answer: 2. \(\frac{2 v_1 v_2}{v_1+v_2}\)

We know the formula, s=vt from v=\(\frac{s}{t}\)

∴The Average speed of particle is,\(v_{a v}=\frac{s+s}{\frac{s}{v_2}+\frac{s}{v_2}}=\frac{2 v_1 v_2}{v_1+v_2}\)

Question 10. A ball is dropped from a high-rise platform at t = 0 starting from rest. After 6 s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (take g = 10 ms^-2)

1. 74 ms^-1
2. 55 ms^-1
3. 40 ms^-1
4. 60 ms^-1

Answer: 1. 74 ms”

According to the question, For the first ball, u = 0

⇒ \(s_1=\frac{1}{2} g t_1^2=\frac{1}{2} \times g(18)^2\)

For second ball, \(s_2=v t_2+\frac{1}{2} g t^2\)

⇒ \(t_2=18-6=125\)

⇒ \(s_2=v \times 12+\frac{1}{2} g(12)^2\)

Here, s¹-s² \(\frac{1}{2} g(18)^2=12 v+\frac{1}{2} g(12)^2\)

∴On Solving We get, V= 74ms^-1

Question 11. Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speed is v and 2v at any instant then the speed of the centre of mass of the system will be:

1. 2v
2. 0
3. 1.5v
4. v

Answer: 2. 0

The conditional main acceleration of parts is zero means COM=0

This shows that the speed of the centre of mass of the system is always zero.

Question 12. A particle moving along the x-axis has acceleration f at time t, given by \(f=f_0\left(1-\frac{t}{T}\right)\), where f₀ and T are constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle’s velocity (VR) is:

1. \(\frac{1}{2} f_0 T^2\)
2. f₀t₂
3. \(\frac{1}{2} f_0 T\)
4. f₀T

Answer: 2. f₀t₂

⇒ \(\frac{d v}{d t}=f =f_0\left(1-\frac{t}{T}\right)\)

⇒ \(\int_0^v d v =f_0 \int_0^T\left(1-\frac{t}{T}\right) d t\)

v =\(f_0\left(t-\frac{t^2}{2 T}\right)_0^T\)

= \(f_0\left(T-\frac{T^2}{2 T}\right)=\frac{1}{2} f_0 T\)

Question 13. Three different objects of masses mls m2 and m3 are allowed to fall from rest and the same point 0 along three different frictionless paths. The speeds of the three objects on reaching the ground will be in the ratio of:

1. m₁ : m₃ : m₃
2. m₁ : 2m₂ : 3m₃
3. 1: 1: 1
4. \(\frac{1}{m_1}: \frac{1}{m_2}: \frac{1}{m_3}\)

Answer: 3. 1 : 1 : 1

When an object falls freely under gravity, then its speed “depends only on its height of fall and is. independent of the mass of the object. As all objects are falling through the same height, therefore their speeds on reaching the ground will be in the ratio of 1:1:1.

Question 14. A car accelerates from rest at a constant rate for some time, after which it decelerates at a constant rate of P and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is:

1. \(\left(\frac{\alpha^2+\beta^2}{\alpha \beta}\right)t\)
2. \(\left(\frac{\alpha^2-\beta^2}{\alpha \beta}\right)t\)
3. \(\left(\frac{\alpha-\beta}{\alpha \beta}\right)t\)
4. (\(\left(\frac{\alpha \beta t}{\alpha+\beta}\right)\)

Answer: 4. (\(\left(\frac{\alpha \beta t}{\alpha+\beta}\right)\)

This situation is plotted on a (v-t) graph. In the (v-t) graph, OA represents the accelerated part and AB represents the decelerated part.

⇒Let t₁ and t₂ be the times for parts OA and AB respectively. At point A velocity is maximum and let it be v max. But,t = t₁ +t₂

=\(v_{\max }\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=v_{\max }\left(\frac{\alpha+\beta}{\alpha \beta}\right)\)

Question 15. A car covers the first half of the distance between two places at 40 km/h and the other half at 60 km/h. The average speed of the car is:

1. 40 km/hr
2. 48 km/hr
3. 50 km/hr
4. 60 km/hr

Answer: 2. 48 km/hr

Let the distance between two places be d and t₁ is the time taken by the car to travel the first-half length, and t₂ is the time taken by the car to travel the second-half length. Time taken by car to travel first half

length,\(t_1=\frac{\left(\frac{d}{2}\right)}{40}=\frac{d}{80}\)

⇒ Time taken by car to travel second-half length,\(t_2=\frac{\left(\frac{d}{2}\right)}{60}=\frac{d}{120}\)

Total Time ,t=t₁+t₂ = \(\frac{d}{80}+\frac{d}{120}\)

⇒ \(d\left(\frac{1}{80}+\frac{1}{120}\right)=\frac{d}{48}\)

∴Average speed, =\(\frac{d}{t_1+t_2}=\frac{d}{\left(\frac{d}{48}\right)}\) = 48 km/h

Question 16. A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with a velocity of 30 km/h and 40 km/h respectively. The velocity of the car midway between P and Q is:

1. 33.3 km/h
2. 20 V₂ km/h
3. 35 V₂ km/h
4. 0.35 km/h

Answer: 3. 35 V₂ km/h

Let x be the total distance between points P and Q and v be the velocity of the car while passing a certain middle point of PQ.

If a is the acceleration of the car then,

For part PQ, 40² – 30² = 2ax

⇒For part RQ, \(40^2-v^2=\frac{2 a x}{2}\)

We have,\(40^2-v^2=2\left(\frac{350}{\mathrm{x}}\right) \frac{x}{2}\)

∴v²= 1250 \(v=35 \sqrt{2} \mathrm{~km} / \mathrm{h}\)

Question 17. A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude of P₀. The instantaneous velocity of the car is proportional to:

1. \(t^2 P_0\)
2. \(t^{\frac{1}{2}}\)
3. \(t^{-\frac{1}{2}}\)
4. \(\frac{t}{\sqrt{m}}\)

Answer: 2. \(t^{\frac{1}{2}}\)

We know that power, P₀= F. v.

⇒ \(P_0=m v \cdot \frac{d v}{d t}\)

Interesting we get; \(\int m v d v=\int \mathrm{P}_0 d t\)

⇒\(\frac{m v^2}{2}=P_0 t\)

∴\(v \propto \sqrt{t}\)

Question 18. A bus is moving at a speed of 10 ms^-1 on a straight road, Ascooterist wants to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should be sooterist chase the bus?

1. 20 ms^-1
2. 40 ms^-1
3. 25 ms^-1
4. 10ms^-1

Answer: 1. 20 ms^-1

S=Scooter

B=bus

Let v = relative velocity of S w.r.t. B

v = v_s – v_b

⇒ v_b = V + V_{g → 1}

Relative Velocity = \(\frac{\text { Displacement }}{\text { time }}\)

⇒\(\frac{1000}{100}=10 \mathrm{~ms}^{-1}\)

∴ Putting this value in the equation. 1, Vs= 10 + 10 = 20 ms

Question 19. A particle shows the distance-time curve given in the figure. The maximum instantaneous velocity of the particle is around the point:

1. 2
2. 3
3. 4
4. 1

Answer: 3. 4

In the distance-time graph, the speed at the instant is expressed by the slope at that instant. The slope is maximum at C.

Question 20. A car moves from X to Y with a uniform speed vu and returns to Y with a uniform speed v. The average speed for this round trip is:

1. \(\sqrt{v_u v_d}\)
2. \(\frac{v_d v_u}{v_d+v_v}\)
3. \(\frac{v_u+v_d}{2}\)
4. \(\frac{2 v_d v_u}{v_d+v_u}\)

Answer: 4. \(\frac{2 v_d v_u}{v_d+v_u}\)

Average speed=\(\frac{\text { Total distance }}{\text { Total time }}\)

=\(\frac{2 d}{\frac{d}{v_u}+\frac{d}{v_d}}=\frac{2 v_d v_u}{v_d+v_u}\)

Question 21. Which of the following curves does not represent motion in one dimension? The instantaneous velocity of the particle is negative at the point:

Answer: 4.

In option 3, particles have two velocities at a particular instant of time, which is impossible.

Question 22. A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = β^-2n where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by:

1. – 2nβ² x^-2n-1
2. – 2nβ² x^-4n-1‘
3. – 2nβ² x^-2n+1
4. – 2nβ²e^-4n+1

Answer: 2. – 2nβ² x^-4n-1‘

The velocity (v) as a function position of a particle is given by,

v = bx^-2n

Acceleration, \(a=\frac{d v}{d t}=\frac{d x}{d t} \cdot \frac{d v}{d x}\)

⇒ \(a=\frac{d v}{d x} \cdot v=v \cdot \frac{d v}{d x}\)

=\(\beta x^{-2 n} \cdot\left(-2 n \beta x^{-2 n-1}\right)\)

∴ a=\(-2 n \beta^2 x^{-4 n-1}\)

Question 23. A balloon with mass m depends on an acceleration a (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration?

1. \(\frac{2 m a}{g+a}\)
2. \(\frac{2 m a}{g-a}\)
3. \(\frac{m a}{g+a}\)
4. \(\frac{m a}{g-a}\)

Answer: 1. \(\frac{2 m a}{g+a}\)

According to the question when it descends with acceleration a,

mg – N=ma → 1

N = Buoyant force

In the second case let a new mass of between in m,

removed mass = m – m’ ⇒N-mg = ma →  2

Solving eq. 1 and 2,

mg-N=ma

N- m’g = m’a

mg- m’g = m’a + m’a

= ma + m’a

mg-ma -m'(g + a)

m(g-a) = m'(g + a)

\(m^{\prime}=\frac{m(g-a)}{g+a}\) → 3

Now removed mass is m0- m’ = Am

From eq. 3 we get \(m-m^{\prime} =m-\frac{m(g-a)}{g+a}\)

=\(m\left[1-\frac{g-a}{g+a}\right]\)

=\(m\left[\frac{g+a-(g-a)}{g+a}\right]\)

=\(m\left[\frac{2 a}{g+a}\right]\)

∴ \(\Delta m =\frac{2 m a}{g+a}\)

Question 24. A body is moving with a velocity of 30 m/s towards the east. After 10 s its velocity becomes 40 m/s towards the north. The average acceleration of the body is:

1. 7 m/s²
2. \(\sqrt{7} \mathrm{~m} / \mathrm{s}^2\)
3. 5 m/s²
4. 1 m/s²

Answer: 3. 5 m/s²

Average Acceleration

⇒\(\vec{a}_{\mathrm{avg}}=\frac{\overrightarrow{\mathrm{v}}_f-\overrightarrow{\mathrm{v}}_i}{t}=\frac{40 \hat{j}-30 \hat{i}}{10}\)

∴ \(\vec{a}_{\text {avg }}=\sqrt{\frac{40^2+30^2}{10}}=5 \mathrm{~m} / \mathrm{sec}^2\)

Question 25. The position of a particle varies with time t, as x = at²-bt³. The acceleration of the particle will be zero at time t equals:

1. zero
2. \(\frac{a}{3 b}\)
3. \(\frac{2 a}{3 b}\)
4. \(\frac{a}{b}\)

Answer: 2. \(\frac{a}{3 b}\)

Acceleration, \(a=\frac{d v}{d t}=\frac{d^2 x}{d t^2}\)

Velocity, \(v=\frac{d x}{d t}\)

The given equation is, x= at² – bt³

Velocity = \(v=\frac{d x}{d t}=2 a t-3 b t^2\)

Acceleration \(a=\frac{d v}{d t}=2 a-6 b t\)

but a=0,

∴ 2a- 6bt = 0

Question 26. The ratio of the distances travelled by a freely falling body in the 1^st, 2^nd, 3^rd and 4^th second:

1. 1: 4 : 9: 16
2. 1 : 3: 5: 7
3. 1 : 1: 1: 1
4. 1: 2 : 3: 4

Answer: 2. 1 : 3 : 5: 7

⇒ \(\mathrm{S}_{\mathrm{nth}}= u+\frac{a}{2}(2 n-1)\)

=\(0+\frac{a}{2}(2 n-1)\)

\(\mathrm{S}_{\mathrm{nth}} \alpha(2 n-1)\)⇒\(\quad \mathrm{S}_{1 \mathrm{st}}: \mathrm{S}_{2 \mathrm{nd}}: \mathrm{S}_{3 \mathrm{rd}}: \mathrm{S}_{4 \text { th }} \)

=\({[2(1)-1]:[2(2)-1]:[2(3)-1]:[2(4)-1] }\)

= 1: 3: 5: 7

Question 27. The displacement-time graphs of two moving particles make angles of 30° and 45° with the x-axis as shown in the figure. The ratio of their respective velocity is:

1. 1:1
2. 1:2
3. \(1: \sqrt{3}\)
4. \(\sqrt{3}: 1\)

Answer: 3. \(1: \sqrt{3}\)

Let displacement be s and time be t and y-intercept is zero. S = ut + 0 ⇒ S = ut,

where u is velocity as per laws of motion Also u is the slope of the graph s vs t.

u = tan θ

u₁ = tan 30°

⇒ \(\mathrm{u}_1=\frac{1}{\sqrt{3}}\) ⇒\(\mathrm{u}_2=\tan 45^{\circ}\)

⇒ \(\mathrm{u}_2=1\)

∴u₁: u₂=\(\frac{1}{\sqrt{3}}: 1\)

∴u₁: u₂ =1:\(\sqrt{3}\)

Question 28. A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let S_n be the distance travelled by the block in the interval t = n – 1 to t = n. Then the ratio \(\frac{\mathrm{S}_n}{\mathrm{~S}_{n+1}}\) is:

1. \(\frac{2 n-1}{2 n}\)
2. \(\frac{2 n-1}{2 n+1}\)
3. \(\frac{2 n+1}{2 n-1}\)
4. \(\frac{2 n}{2 n-1}\)

Answer: 2. \(\frac{2 n-1}{2 n+1}\)

Given,Distance =S_n

Time t=n-1 to t=n-1

⇒\(\frac{\mathrm{S}_n}{\mathrm{~S}_{n+1}}=?\)

∴a=g sinθ

Initial time\(\mathrm{S}=u t+\frac{1}{2} a t^2\)

⇒\(\mathrm{~S}=u t+\frac{1}{2} g \sin \theta t^2\)

Since u=0

⇒\({S}=\frac{1}{2} a t^2\)

Case 1 time interval t=n-1 to t=n

⇒\(\mathrm{S}_n =\frac{1}{2} a n^2-\frac{1}{2} a(n-1)^2\)

=\(\frac{1}{2} a\left(n^2-n^2-1+2 n\right)\)

⇒\(\mathrm{S}_n =\frac{1}{2} a(2 n-1)\)

\(\mathrm{S}_{n+1} =\frac{1}{2} a(n+1)^2-\frac{1}{2} a n^2\) =\(\frac{1}{2} a\left(n^2+1+2 n-n^2\right)\)

=\(\frac{1}{2} a(2 n+1)\)

∴\(\frac{\mathrm{S}_n}{\mathrm{~S}_{n+1}} =\frac{2 n-1}{2 n+1}\)

Question 29. A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is p = 0.5. The distance that the box will move relative to the belt before coming to rest on it taking g = 10 ms^-2, is:

1. 1.2 m
2. 0.6 m
3. zero
4. 0.4 m

Answer: 4. 0.4 m

Here, F = μmg Retardation of the block on the belt,\(a=\frac{F}{m}=\frac{\iota m g}{m}=\mu \mathrm{g}\)

⇒From third equation of motion, v² = w² + 2as ⇒ 0 = (2)² – 2(μg)s

∴\(s=\frac{4}{2 \times 0.5 \times 10}=0.4 \mathrm{~m}\)

Question 30. Ship A is moving westwards with a speed of 10 km h-1 and ship B 100 km South of A, is moving Northwards with a speed of 10 km hr1. The time after which the distance between them becomes shortest is:

1. O h
2. 5 h
3. \(5 \sqrt{2} h\)
4. \(10 \sqrt{2} h\)

Answer: 2. 5 h

From the diagram, the distance S between the ships is \(S=\sqrt{(100-x)^2+x^2}\)[From A To B]

⇒\(for_{\min }, \quad \frac{d s}{d x}=0\)

[- 2 (100-x) + 2x] = 0 ⇒ 4x – 200 = 0 ⇒ x = 50

⇒\(\quad \frac{d s}{d x}=\frac{1}{2\left[(100-x)^2+x^2\right]^{-1 / 2}}\)

∴So after both the ships have covered 50 m. The time taken is \(t=\frac{x}{10}=\frac{50}{10}=5 h\)

Question 31. A stone falls freely under gravity. It covers distances h₁, h₂ and h₃ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁,h₂ and h₃ is:

1. \(h_1=2 h_2=3 h_2\)
2. \(h_1=\frac{h_2}{3}=\frac{h_3}{5}\)
3. \(h_2=3 h_1 and h_3=3 h_2\)
4. \(4 i=h_2=h_2\)

Answer: 2. \(h_1=\frac{h_2}{3}=\frac{h_3}{5}\)

Here, \(h_1=\frac{1}{2} g t^2=\frac{1}{2} g(5)^2=125\)

⇒\(h_2=\frac{1}{2} g t^2=\frac{1}{2} g(10)^2=375\)

⇒\(h_3=\frac{1}{2} g t^2=\frac{1}{2} g(15)^2=625\)

∴\(h_1=\frac{h_2}{3}=\frac{h_3}{5}\)

Question 32. A particle starts its motion from rest under the action of a constant force. If the distance covered in the first 10 s is s₁ and that covered in the first 20 s is s₂, then:

1. s₂ = 2 S₁
2. s₂ = 3S₁
3. s₂ = 4 S₁
4. s₂ = S₁

Answer: 3. s₂ = 4 S₁

We know the second equation of motion,\(S=u t+\frac{1}{2} a t^2\)

⇒When, \(u=0: S=\frac{1}{2} a t^2\)

⇒ \(S_1=\frac{1}{2} a(10)^2\)

⇒\(S_2=\frac{1}{2} a(20)^2\)

\(\frac{S_1}{S_2}=\frac{(10)^2}{(20)^2}=\frac{1}{4}\) ⇒ s₂=4s₁

Question 33. The distance travelled by a particle starting from rest and moving with an acceleration\(\frac{4}{3}\)ms~2, in the third second is:

1. 6 m
2. 4m
3. \(\frac{10}{3} \mathrm{~m}\)
4. \(\frac{19}{3} \mathrm{~m}\)

Answer: 3. \(\frac{10}{3} \mathrm{~m}\)

Distance covered in nth second,

⇒\(\mathrm{S}_n=u+\frac{1}{2} a(2 n-1)\)

=\(0+\frac{1}{2} \times \frac{4}{3}(2 \times 3-1)=\frac{1}{2} \times \frac{4}{3} \times 5\)

=\(\frac{10}{3} \mathrm{~m}\)

Question 34. A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by x = 40 + 12t-P. How long would the particle travel before coming to rest?

1. 16 m
2. 24 m
3. 40 m
4. 56 m

Answer: 1. 16 m

The given equation is X = 40 + 127- 13

On differentiating, we get \(u=\frac{d x}{d t}=12-3 t^2\)

When the particle comes to rest means, v = 0

12-3t₂=0 ⇒ t=2

The Distance travelled before coming to rest \(\int_0^5 d s=\int_0^2 v d t\)

⇒\({[s]_0^5 }=\int_0^2\left(12-3 t^2\right) d t\)

=\(\left[12 t-\frac{3 t^3}{3}\right]_0^2\)

=12 x 2 -(8-0) = 16 m

Question 35. The displacement x of a particle varies with time t as x = ae^-at + b^βt, where a, b, a and (3 are positive constants. The velocity of the particle will:

1. be independent of β
2. drop to zero when α =β
3. go on decreasing with time
4. go on increasing with time

Answer: 4. go on increasing with time

From question, x = ae^-αt + be^βt

⇒ \(v=\frac{d x}{d t}=-a \alpha e^{-\propto t}+b \beta_e^{\mathrm{B}+}\)

⇒ v=-a α e^-t + b βξ^βt

∴Velocity will decrease for a certain value of t.

Question 36. If a ball is thrown vertically with speed u, the distance covered during the last t seconds of its ascent is:

1. ut
2. \(\frac{1}{2} g t^2\)
3. \(u t-\frac{1}{2} g t^2\)
4. \((u+g t) t\)

Answer: 2. \(\frac{1}{2} g t^2\)

Let the time of flight be T, Then \(T=\frac{u}{g}\)

Again, Let h be the distance covered during the last second of its ascent.

⇒\(=u-g\left(\frac{u}{g}-t\right)=g t\)

∴Velocity at point B, v_B =u – g (T -t)

⇒h=\(h=v_B t-\frac{1}{2} g t^2\)

Question 37. A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so, more than two balls are in the sky at any time?

1. more than 19.6 m/s
2. at least 9.8 m/s
3. any speed less than 19.6 m/s
4. only with a speed of 19.6 m/s

Answer: 1. more than 19.6 m/s

From question Interval of a ball thrown = 2 sec

If we want a minimum of three balls to remain in the air then the time of flight of the first ball must be greater than 4 sec. ⇒T > 4sec

⇒\(\frac{2 u}{g}>4 \mathrm{sec}\)

∴4 > 19.6 m/s

Question 38. A car moving with a speed of 40 km/h can be stopped after 2 m by applying brakes. If the same car is moving at a speed of 80 km/h, what is the minimum stopping distance?

1. 8m
2. 2m
3. 4m
4. 6m

Answer: 1. 8m

According to the conservation of energy, the kinetic energy of the car = work done in stopping the car

⇒\(\frac{1}{2} m v^2=F s\)

where F is the retarding force and s is the stopping distance. For the same retarding force s α v²

⇒\(\frac{s_2}{s_1}=\left(\frac{v_2}{v_1}\right)=\left(\frac{80}{40}\right)^2=4\)

∴s₂=4s₁=4×2

Question 39. If a ball is thrown vertically upwards with a velocity of40 m/s, then velocity of the ball after 2s will be(g = 10 m/s²):

1. 15 m/s
2. 20 m/s
3. 25 m/s
4. 28 m/s

Answer: 2. 20 m/s

Here, the initial velocity of the ball, u = 40 m/s²

Acceleration of ball, a=-gm/s² = – 10 m/s²

Time = 2s

⇒From the first equation of motion,

∴v = u + at = 40 – 10 x 2⇒ v = 20 m/s

Question 40. The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing to the east. If he is standing on the south bank and wishes to cross the river along the shortest path the angle at which he should make his strokes w.r.t. north is given by:

1. 0°
2. 60° west
3. 45° west
4. 30° west

Answer: 4. 30° west

Here, vSR = 20 m/s

VRG = 10 m/s

⇒\(\sin \theta=\left|\frac{\vec{v}_{R G}}{\overrightarrow{v_{S R}}}\right|=\frac{10}{20}=\frac{1}{2}\)

∴θ= 30° west.

Question 41. The two bullets are fixed horizontally and simultaneously towards each other from the rooftops of two buildings 100 m apart and of the same height of 200 m with the same velocity of 25 m/s. When and where will the two bullets collide (g = 10 m/s²)?

1. After 2s at a height of 180 m
2. After 2s at a height of 20 m
3. After 4s at a height of 120 m
4. They will not collide

Answer: 1. After 2s at a height of 180 m

Let bullet collide at time t,

x₁ + X₂ = 100 m 25/+ 25/= 100

t = 2s

\(y=\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times 2^2\) = 20 m ⇒h = 200-20= 180 m

∴Hence, bullets will collide after 2s at a height of 180 m above the ground.

Question 42. An object of mass 500 g, initially at rest acted upon by a variable force whose X components vary with X in the manner shown. The velocities of the object a point X = 8 m and X = 12 m, would be the respective values of (nearly):

1. 18 m/s and 24.4 m/s
2. 23 m/s and 24.4 m/s
3. 23 m/s and 20.6 m/s
4. 18 m/s and 20.6 m/s

Answer: 3. 23 m/s and 20.6 m/s

From the work-energy theorem,

K = work = area under F-X graph from

X = 0 to X=8m

⇒ \(\frac{1}{2} m v^2\)100 + 30

⇒\(\frac{1}{2} \times \frac{1}{2} \times v^2 =130\)

⇒\(v^2 =520\)

⇒\(v =\sqrt{520}=23 m/s\)

from X = 0 to X =12m ⇒\(\frac{1}{2} m v^2 =100+30-47.5+2\)

⇒v =\(\sqrt{410}\)

∴v= 20.6m/s

Question 43. Two boys are standing at the ends A and B of a ground where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity v₁. The boy at A starts running simultaneously with velocity v and catches the other in time t, where t is equal to:

1. \(\frac{a}{\sqrt{v^2+v_1^2}}\)
2. \(\frac{a}{v+v_1}\)
3. \(\frac{a}{v-v_1}\)
4. \(\sqrt{\frac{a^2}{v^2-v_1^2}}\)

Answer: 4. \(\sqrt{\frac{a^2}{v^2-v_1^2}}\)

∴t=\(\frac{a}{v^{\prime}}=\frac{a}{\sqrt{v^2-v_1^2}}\)

Units And Measurements

Question 1. Plane angle and solid angle have:

1. Dimensions but no units
2. No units and no dimensions
3. Both units and dimensions
4. Units but no dimensions

Answer: 4. Units but no dimensions

Radians and Steradians are measured in either pi or degrees. Because (7t) and degrees are pure numbers, planar and solid angles have no dimensions. Plane and solid angles both have units but no dimensions.

Question 2. The unit of thermal conductivity is:

1. Jm^-1 K^-1
2. WmK^-1
3. Wm^-1K^-1
4. JmK^-1

Answer: 3. Wm^-1K^-1

Thermal conductivity, \(\mathrm{K}=\frac{L}{A \Delta T} \frac{d Q}{d V}=\frac{\text { Metre }}{(\text { Metre })^2 \times \text { Kelvin }} \times \text { watt }\)

= \(\mathrm{W} m^{-1} K^{-1}\)

Read and Learn More NEET Physics MCQs

Question 3. The unit of Stefan’s constant is:

1. Wm² K⁴
2. Wm² K ^-4
3. Wm^-2 K^-1
4. Wm^-2 K^-4

Answer: 4. Wm^-2 K^-4

The value of the Stefan-Boltzmann constant is approximately 5.67 x 10^-8 watts per meter squared per kelvin to the fourth of absolute temperature. (W.m^-2 K^-4 ).

Question 4. The angle of T (minute of arc) in radian is nearly equal to:

1. 2.91 x 10^-4rad
2. 4.85 x KT^-4 rad
3. 4.80 x 10^-8rad
4. 175 x 10^-2 rad

Answer: 1. 2.91 x 10^-4rad

We have, \(360^{\circ}=2 \pi \mathrm{rad}\)

⇒ \(1^{\circ}=\left(\frac{\pi}{180}\right) \mathrm{rad}=1.745 \times 10^{-2} \mathrm{rad}\)

⇒ \(1^{\circ}=60^{\prime}=1.745 \times 10^{-2} \mathrm{rad}\)

⇒ \(1^{\prime}=\frac{1.745}{60} \times 10^{-2}\)

⇒ \(1^{\prime}=2.908 \times 10^{-4} \mathrm{rad}=2.91 \times 10^{-4} \mathrm{rad}\)

Question 5. The length of the wire between the two ends of the sonometer is 100 cm. What should be the positions of two bridges below the wire, so that the three segments of the wire have their fundamental frequencies in the ratio 1:3:5?

1. \(\frac{1500}{23} \mathrm{~cm}, \frac{500}{23} \mathrm{~cm}\)
2. \(\frac{1500}{23} \mathrm{~cm}, \frac{300}{23} \mathrm{~cm}\)
3. \(\frac{300}{23} \mathrm{~cm}, \frac{1500}{23} \mathrm{~cm}\)
4. \(\frac{1500}{23} \mathrm{~cm}, \frac{2000}{23} \mathrm{~cm}\)

Answer: 2. \(\frac{1500}{23} \mathrm{~cm}, \frac{300}{23} \mathrm{~cm}\)

According to the length of the wire. L = 100 cm and ratio of fundamental frequencies = 1:3:5

⇒  \(L_1: L_2: L_3=\frac{1}{1}: \frac{1}{3}: \frac{1}{5}\)

= 15: 5: 3

Let x= common factor, then \(15 \mathrm{x}+5 \mathrm{x}+3 x=100\)

23 x =100

x = \(\frac{100}{23}\)

⇒  \(L_1 =15 \times \frac{100}{23}\)

= \(\frac{1500}{23} \mathrm{~cm}\)

⇒  \(L_2 =5 \times \frac{100}{23}\)

= \(\frac{500}{23} \mathrm{~cm}\)

⇒  \(L_3 =3 \times \frac{100}{23}\)

= \(\frac{300}{23} \mathrm{~cm}\)

This shows that the bridge should be placed from A at \(\frac{1500}{23} \mathrm{~cm} \text { and } \frac{300}{23}\) cm respectively.

Question 6. A screw gauge gives the following readings when used to measure the diameter of a wire, main scale reading: 0 mm, Circular scale reading: 52 divisions. Given that 1 mm on the main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is:

1. 0.52 cm
2. 0.026 cm
3. 0. 26 cm
4. 0.052 cm

Answer: 4. 0.052 cm

Given, Meter Scale Reading = 0

Circular Scale Reading = 52

Least Count = mm = \(\frac{1}{100}\) mm

d = MSR + CSR x Least Count

= 0 + 52 x 0.01

= 0.52 mm ord = 0.052 cm

Question 7. Time intervals measured by a clock gave the following readings: 1.25 s, 1.24, 1.27 s, 1.21 and 1.28 s. What is the percentage error of the observation?

1. 2%
2. 4%
3. 16%
4. 1.6%

Answer: 4. 1.6%

Meantime Interval, \(\bar{a}=\frac{1.25+1.24+1.27+1.21+1.28}{5}\)

⇒ \(\Delta \bar{a}=\frac{6.25}{5}=1.255\)

=\(\frac{\left|\Delta a_1\right|+\left|\Delta a_2\right|+\left|\Delta a_3\right|+\left|\Delta a_4\right|+\left|\Delta a_5\right|}{5}\)

⇒  \(|1.25-1.25|+|1.25-1.94|+\mid 1.25\)

=\(\frac{-1.27|+| 1.25-1.21|+| 1.25-1.25 \mid}{5}\)

=\(\frac{0+0.01+0.02+0.01+0.03}{5}=\frac{0.1}{5}\)

=\(0.02 \mathrm{sec}\)

Now, percentage error=\(\frac{\Delta \bar{a}}{a} \times 100\)

=\(\frac{0.02}{1.25} \times 100=1.6 \%\)

Question 8. A screw gauge has the least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is:

1. 0.25 mm
2. 0.5 mm
3. 1.0 mm
4. 0.01 mm

Answer: 2. 0.5 mm

Given, Least count =0.01 mm

No. of division on circular scale = 50 Formula :

The pitch of the screw gauge = least count x no. of divisions on a circular scale = 0.01 x 50 = 0.5 mm

Question 9. The main scale of a Vernier caliper has n divisions/cm. n divisions of the vernier scale coincide with (n – 1) divisions of the main scale. The least count of the vernier calipers is: In an experiment, the percentage of error that occurred in the measurement of physical quantities A, B, C, and D are 1%, 2%, 3%, and 4% respectively. Then, the maximum percentage of error in the measurement X, where,

1. \(\frac{1}{(n+1)(n-1)} \mathrm{cm}\)
2. \(\frac{1}{\mathrm{n}} \mathrm{cm}\)
3. \(\frac{1}{\mathrm{n}^2} \mathrm{~cm}\)
4. \(\frac{1}{\mathrm{n}(\mathrm{n}+1)} \mathrm{cm}\)

Answer: 3. \(\frac{1}{\mathrm{n}^2} \mathrm{~cm}\)

n main scale divisions = 1cm

1 main scale divisions =\(\frac{1}{n} \mathrm{~cm}\)

⇒  \(1 \mathrm{MSD}=\frac{1}{n} \mathrm{~cm}\)

⇒  \(n \mathrm{VSD}=(n-1) \mathrm{MSD}\)

⇒  \(1 \mathrm{VSD}=\frac{n-1}{n} \mathrm{MSD}\)

⇒ L.C.=\(1 \text { MSD }-1 \text { V.S.D }\)

= \(\frac{1}{n}-\left(\frac{n-1}{n}\right) \mathrm{MSD}\)

L.C.= \(\left(\frac{1}{n}-\frac{n-1}{n^2}\right) \mathrm{cm}\)

L.C.=\(\frac{n-n+1}{n^2}=\frac{1}{n^2} \mathrm{~cm}\)

Question 10. In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C, and D are 1%, 2%, 3%, and 4% respectively. Then, the maximum percentage of error in the measurement X, where,\(X=\frac{A^2 B^{1 / 2}}{C^{1 / 2} D^3}\) Will be:

1. 16%
2. -10%
3. 10%
4. \(\left(\frac{3}{13}\right) \%\)

Answer: 1. 16%

According to the question, \(X=\frac{A^2 B^{1 / 2}}{C^{1 / 3} D^3}\)

Using a percentage error of X, \(\frac{\Delta \mathrm{X}}{\mathrm{X}}\) x 100 = \(2\left(\frac{\Delta \mathrm{A}}{\mathrm{A}}\right) \times 100+\frac{1}{2}\left(\frac{\Delta \mathrm{B}}{B}\right) \times 100\) + \(\left(\frac{\Delta \mathrm{C}}{\mathrm{C}}\right) \times 100+3\left(\frac{\Delta \mathrm{D}}{\mathrm{D}} \times 100\right)\)

Putting the given value, \(\frac{\Delta \mathrm{X}}{\mathrm{X}} \times 100=2(1 \%)+\frac{1}{2}(2 \%)+3\)

=2%+1%+1%+12%

= 16%

The maximum error in X is 16%

Question 11. A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If the screw gauge has a zero error of – 0.004 cm, the correct diameter of the ball is:

1. 0.053 cm
2. 0.525 cm
3. 0.521 cm
4. 0.529 cm

Answer: 4. 0.529 cm

The diameter of the ball = Meter Scale Reading + Circular Scale Reading (Least Count) – zero error

= 0.5+ 25 x 0.001-(-0.009)

= 0.5 + 0.025 + 0.009 = 0.529 cm

Question 12. In an experiment, four quantities a, b, c, and d are measured with percentage errors 1%, 2%, 3%, and 4% respectively. Quantity P is calculated as follows, \(P=\frac{a^3 b^2}{c d} \%\), error in P is:

1. 14%
2. 10%
3. 7%
4. 4%

Answer: 1. 14%

Given, \(P=\frac{a^3 b^2}{c d}\)

Error in P can be written as \(\frac{\Delta \mathrm{P}}{\mathrm{P}} \times 100=\left(3 \frac{\Delta a}{a}+2 \frac{\Delta b}{b}+\frac{\Delta c}{c}+\frac{\Delta d}{d}\right) \times 100\)

=\(3 \times \frac{\Delta a}{a} \times 100+2 \frac{\Delta b}{b} \times 100+\frac{\Delta c}{c}\)

⇒  \(\times 100+\frac{\Delta d}{d} \times 100\)

Putting the given values of percentage errors in the above equation, we get,

= 3 x 1 + 2 X 2 + 3 + 4 = 14%

Question 13. A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time e1 and e2 respectively, the percentage error in the estimation of g is:

1. e₂ – e₁
2. e₁ + 2e₂
3. e₁ + e₂
4. e₁ – 2e₂

Answer: 2. e₁ + 2e₂

We know that \(h =\frac{1}{2} g t^2\)

g = \(\frac{h}{t^2}\)

log g =log h-2 log t

⇒  \(\left(\frac{\Delta g}{g} \times 100\right)_{\max }=\left(\frac{\Delta h}{h} \times 100\right)+2\left(\frac{\Delta t}{t} \times 10\right)\)

= \(e_1+2 e_2\)

Question 14. If the error in the measurement of the radius of a sphere is 2%, then the error in the determination of the volume of the sphere will be:

1. 4%
2. 6%
3. 8%
4. 2%

Answer: 2. 6%

We know that,

Volume of sphere =V=\(\frac{4}{3} \pi R^3\)

Taking logs on both sides \(\log V=\log \frac{4}{3} \pi+3 \log R\)

From error formula, \(\frac{4 V}{V}=0+\frac{3 \Delta V}{R}\)

=3 times 2 %=6 %

∴ \([Given,\frac{\Delta R}{R}=2 \%]\)

Question 15. The density of a cube is measured by measuring its mass and the length of its sides. If the maximum error in the measurement of mass and length are 4% and 3% respectively, the maximum error in the measurement of density will be:

1. 7%
2. 9%
3. 12%
4. 13%

Answer: 4. 13%

As density \(\rho=\frac{m}{v}=\frac{\mathrm{m}}{l^3}\)

⇒ \(\frac{\Delta \rho}{\rho} \times 100= \pm\left(\frac{\Delta m}{m}+3 \frac{\Delta l}{l}\right) \times 100 \%\)

=\(\pm(4+3 \times 3)= \pm 13 \%\)

Question 16. A certain body weighs 22.42 g and has a measured volume of 4.7 cc. The possible errors in the measurement of mass and volume are 0.01 g and 0.1 cc. Then, the maximum error in the density will be:

1. 22%
2. 2%
3. 0.2%
4. 0.02%

Answer: 2. 2%

⇒ \(\text { Density }=\frac{\text { Mass }}{\text { Volume }}\)

⇒ \(\rho=\frac{m}{V}\)

∴ \(\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{\Delta V}{V}\)

Here, \(\Delta m=0.01, m=22.42\)

⇒  \(\Delta V=0.1, V=4.7\)

∴ \(\quad \frac{\Delta \rho}{\rho}=\left(\frac{0.01}{22.42}+\frac{0.1}{4.7}\right) \times 100=2 \%\)

Question 17. The area of a rectangular field (in m²) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digits is:

1. 1382
2. 1382.5
3. 14 x 10²
4. 138 x 10¹

Answer: 3. 14 x 10²

Given that,

length = 55.3 m

breadth = 25 m

To find Area =?

As we know, Area = length x breadth

= 55.3 x 25 = 1382.5 m²

= 13.8 x 10² m² = 14 x 10² m²

Hence, the Resultant should have 2 significant digits.

Question 18. Taking into account the significant figures, what is the value of 9.99 m – 0.0099 m?

1. 9.98 m
2. 9.980 m
3. 2.25 x 10^-15m
4. 2.25 x 10¹⁵m

Answer: 1. 9.98 m

Value = 9.99 m – 0.0099 m = 9.9801 m

Taking significant figures, both the values have two significant figures after the decimal point so the current answer is 9.98 m.

Question 19. The dimensions [MLT^-2 A^-2] belong to the:

1. self-inductance
2. magnetic permeability
3. electric permittivity
4. magnetic flux

Answer: 1. self-inductance

Dimensions [MLT^-2A^-2]

S.I. Unit of mass – kg (M)

S.I. Unit of length – m (L)

S.I. Unit of Time – Seconds (T)

S.I. Unit of current – Ampere (A)

[MLT^-2A^-2] in terms of S.I. Unit is kg ms^-2A^-2 which represents magnetic permeability.

Question 20. Which of the following is a dimensional constant?

1. Refractive index
2. Poisson’s ratio
3. Relative density
4. Gravitational constant

Answer: 4. Gravitational constant

A quantity that has dimensions and also has a constant value is called a dimensional constant. Here, the gravitational constant (G) is dimensional.

Question 21. Match List-1 with List-2

Choose the correct answer from the options given below:

1. A-2,  B-4,  C-1,  D-3
2. A-2,  B-4,  C-3,  D-1
3. A-4,  B-2,  C-1,  D-3
4. A-2,  B-1,  C-4,  D-3

Answer: 1. A-2, B-4, C-1, D-3

Explanation: A-2, B-4, C-1, D-3

Question 22. If E and G respectively denote energy and gravitational constant, then E\G has the dimensions of:

1. [M²] [L^-1] [T⁰]
2. [M] [L^-1] [T^-1]
3. [M] [L⁰] [T⁰]
4. [M²][L^-2][T^-1]

Answer: 1. [M²] [L^-1] [T⁰]

Given, Energy = E , Gravitational const = G

We know that [E] = F.S

⇒ \(\mathrm{kg} \mathrm{ms}^{-2} \mathrm{~m}\)

⇒ \(\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-2}\)

⇒ \(\mathrm{ML}^2 \mathrm{~T}^{-2}\)

⇒ \(\mathrm{G}=\frac{\mathrm{Fr}^2}{m^2}\)

= \(\frac{\mathrm{Kg} m s^{-2} m^2}{\mathrm{~kg}^2}\)

= \(\frac{m^3 \mathrm{~s}^{-2}}{\mathrm{~kg}}\)

= \(\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\)

⇒ \(\frac{[\mathrm{E}]}{[\mathrm{G}]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-1}}\)

=\(\mathrm{M}^{1+1} \mathrm{~L}^{2-3} \mathrm{~T}^{-2+2}\)

=\(\mathrm{M}^2 \mathrm{~L}^{-1} \mathrm{~T}^0\)

Question 23. If force [F], acceleration [A], and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy:

1. [F] [A] [T]
2. [F] [A] [T²]
3. [F] [A] [T^-1]
4. [F] [A^-1] [T]

Answer: 2. [F] [A] [T²]

We know that,

Given, \(\mathrm{E} =\mathrm{F}^\alpha \mathrm{A}^\beta \mathrm{T}^\gamma\)

⇒ \(\mathrm{E}=\mathrm{FS}=\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-2}\)

= \(\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)

⇒ \({[\mathrm{F}] }=\mathrm{kg} \mathrm{m} \mathrm{s}^{-2}\)

=\(\left[\mathrm{MLT}^{-2}\right]\)

⇒ \({[\mathrm{A}] } =\mathrm{ms}^{-2}=\mathrm{M}^0 \mathrm{LT}^{-2}\)

⇒ \({[\mathrm{~T}] } =\mathrm{s}=\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\)

⇒ \(\mathrm{ML}^2 \mathrm{~T}^{-2}=\left(\mathrm{MLT}^{-2}\right)^\alpha\)

⇒ \(\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}=\mathrm{M}^\alpha \mathrm{L}^{\alpha+\beta} \mathrm{T}^{-2 \alpha-2 \beta+\gamma}\)

Comparing a= 1

a + p = 2

-2a – 2p + y = -2

P = 2- a

=2-1=1

-2(l)-2(l) + y = -2

y = 2

E = F²A²T²

Question 24. The dimensional formula of stress is:

1. [ML²T^-2]
2. [ML⁰T^-2]
3. [ML^-1T^-2]
4. [MLT^-2]

Answer: 3. [ML^-1T^-2]

Stress = \(\frac{[\text { Force }]}{[\text { Area }]}\) =\(\frac{\left[\mathrm{MLT}^{-2}\right]}{\mathrm{L}^2}\) = \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)

Question 25. If energy (E), velocity (v), and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be:

1. [EV^-2T^-1]
2. [EV^-1 T^-2]
3. [Ev^-2,T^-2]
4. [E^-2 V^-1 T^-3]

Answer: 3. [Ev^-2, T^-2]

Dimension of energy, E = [ML²T^-2]

Dimension of velocity, v = [LT^-1]

Dimension of time, t = [T]

From the question,

Surface tension, S = E^av^bT^c …1

[S] = [MT^-2]

a, b, c are constants.

Equating dimensions on both sides, [ML⁰T^-2] = [ML²T^-2]^a [LT^-1]^b [T]^c ⇒ [ML⁰T^-2] = [M^aL^2a=b T^-2a-b+c]

On comparing both sides, a = 1, 2a + b = 0 and -2a – b + c = -2

On solving these equations, we get, a= 1,b = – 2, and c =- 2 Putting the values in eq 1 we get,

[S] = [Ev^-2T^-2]

Question 26. The dimensions of (μ₀ε₀)^1/2 are:

1. [L^1/2-T^1/2]
2. [L^-1 T]
3. [LT^-1]
4. [L^1/2T^1/2]

Answer: 3. [LT^-1]

We have \(C=\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)

So, the dimension of (μ₀ε₀)^1/2 is equal to the dimension of C. Therefore, the dimension of C = [LT^-1].

Question 27. The dimension of \(\frac{1}{2} \varepsilon_0 E^2,\) where ε₀ is the permittivity of free space and E is the electric field, is:

1. [ML²T²]
2. [ML^-2 T^-2]
3. [ML²T^-2]
4. [MLT²]

Answer: 2. [ML^-2 T^-2]

Dimension of μ₀= [M^-1L^-3T⁴A²]

Dimension of E = [MLT^-3A^-1]

Dimension of \(\frac{1}{2}\) μ₀ E² = [M^-1L^-3T⁴A²] [MLT^-3A^-1]²

= [ML^-1T^-3]

Question 28. If p represents momentum and d represents position, then the dimensions of Planck’s constant (h) in terms of p and d are:

1. [pd]
2. [pd^-1]
3. [p^-1d]
4. [p^-1d^-1]

Answer: 2. [pd^-1]

Let h=[p^ad^b]

[ML²T^-1] = [MLT^-1]^a[L]^b

[ML²T^-1] = [M^aL^a+bT^-a]

Comparing powers on both sides, we get a = 1, a + b = 2,

b= 1

Putting these values in an equation we get

The dimension of, h = [pd]

Question 29. Which two of the following five physical parameters have the same dimensions?

1. Energy density
2. Refractive index
3. Dielectric constant
4. Young’s modulus
5. Magnetic field

Choose the correct answer:

1. 2 and 4
2. 3 and 5
3. 1 and 4
4. 1 and 5

Answer: 3. 1 and 4

Energy density and Young’s modules have the same dimension and are equal to [ML-1T-2]. The dielectric constant and refractive index are dimensionless.

Question 30. Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T, and of current I, would be:

1. [ML²T^-2]
2. [ML²T^-1I^-1]
3. [ML²T^-3I^-2]
4. [ML²T^-3I^-1]

Answer: 3. [ML²T^-3I^-2]

We know that, \(R =\frac{V}{I}=\frac{\left[M L^2 T^{-2}\right]}{\left[I^2 T\right]}\)

=\(\left[M L^2 T^{-3} I^{-2}\right]\)

Question 31. The ratio of the dimension of Planck’s constant and that of the moment of inertia is the dimension of:

1. time
2. frequency
3. angular momentum
4. velocity

Answer: 2. frequency

⇒ \(\frac{h}{I}=\frac{\mathrm{E} \times \lambda}{\mathrm{C} \times \mathrm{I}}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right][\mathrm{L}]}{\left[\mathrm{LT}^{-1}\right]\left[\mathrm{ML}^2\right]}\)

∴ \(\frac{h}{I}=\left[\mathrm{T}^{-1}\right] \text {, }\)

(which is the dimension of frequency)

Question 32. The dimensions of the universal gravitational constant are:

1. [ML^-1L³T^-2]
2. [ML²T^-1]
3. [ML^-2L³T^-2]
4. [ML^-2L²T^-1]

Answer: 1. [ML^-1L³T^-2]

We have,\(F=\frac{\mathrm{GMm}}{R^2}\)

⇒ \(\mathrm{G}=\frac{F R^2}{M m}\)

= \(\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{\left[\mathrm{M}^2\right]}\)

= \(\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]\)

Question 33. Which of the following will have the dimensions of times?

1. \(\mathrm{LC}\)
2. \(\frac{R}{L}\)
3. \(\frac{L}{R}\)
4. \(\frac{C}{L}\)

Answer: 3. \(\frac{L}{R}\)

⇒ \(\frac{L}{R}\) is the time constant of the R-L circuit so, the dimensions of \(\frac{L}{R}\) is the same as that of time

⇒ \(\frac{\text { Dimensions of } L}{\text { Dimensions of } R}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}{\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]}=[\mathrm{T}]\)

Question 34. The dimensional formula for the permeability of free space is:

1. [MLT^-2A^-2]
2. [ML^-1T^-2A^-2]
3. [ML^-1T^-2A²]
4. [MLT^-2A^-2]

Answer: 1. [MLT^-2A^-2]

From Biot-Savart law,

d B =\(\frac{\mu_0}{4 \pi} \frac{\text { ld } \sin \theta}{r^2}\)

ld l = current element

r \(=\text { displacement vector }\)

⇒ \(\mu_0 =\frac{4 \pi r^2(\mathrm{~dB})}{\text { ldl } \sin \theta}=\frac{\left[\mathrm{L}^2\right]\left[\mathrm{MT}^{-2} \mathrm{~A}^{-1}\right]}{[\mathrm{A}][\mathrm{L}]}\)

= \(\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right]\)

Question 35. The dimensional formula of pressure is:

1. [MLT^-2]
2. [ML^-1T²]
3. [ML^-1T^-2]
4. [MLT^-2]

Answer: 3. [ML^-1T^-2]

⇒ \(\text { Pressure }=\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{F}}{\mathrm{A}} =\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}\)

∴ \(\left[\mathrm{ML}^{-1}\mathrm{~T}^{-2}\right]\)

Question 36. If C and R denote capacitance and resistance respectively, then the dimensional formula of CR is:

1. [M⁰L⁰T]
2. [M⁰L⁰T⁰]
3. [M⁰L⁰T^-1]
4. Not expressible in terms of [MLT]

Answer: 3. [M⁰L⁰T^-1]

⇒ \(\frac{q}{V}=\frac{q}{W}=\frac{q^2}{W}=\frac{(i t)^2}{F \cdot x}=\frac{\left[\mathrm{AT}^2\right.}{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}\)

= \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right] \text { and } R=\frac{V}{i}=\frac{W}{q i}\)

=\(\frac{F \cdot x}{i^2 t}=\frac{7}{5}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\)

The dimensional formula of CR,

= [M^-1L^-1T⁴A²] [ML²T^-3A^-2]

=[M⁰L⁰T^-1]

Question 37. A physical quantity of the dimensions of length that can be formed out of c, G and \(\frac{e^2}{4 \pi \varepsilon_0}\) is [c speed of light, G is the universal constant of gravitation and e is charge]:

1. \(\frac{1}{C^2}\left[G \frac{e^2}{4 \pi \varepsilon_0}\right]^{1 / 2}\)
2. \(C^2\left[G \frac{e^2}{4 \pi \varepsilon_0}\right]^{1 / 2}\)
3. \(\frac{1}{C^2}\left[\frac{e^2}{G 4 \pi \varepsilon_0}\right]^{1 / 2}\)
4. \(\frac{1}{C} G \frac{e^2}{4 \pi \varepsilon_0}\)

Answer: 1. \(\frac{1}{C^2}\left[G \frac{e^2}{4 \pi \varepsilon_0}\right]^{1 / 2}\)

⇒ \(L=[\mathrm{C}]^{\mathrm{x}}[\mathrm{G}]^{\mathrm{y}}\left[\frac{e^2}{4 \pi \varepsilon_0}\right]\)

⇒ \(\quad \frac{e^2}{4 \pi \varepsilon_0}=\mathrm{ML}^3 \mathrm{~T}^{-2}\)

⇒ \(\mathrm{~L}=\left[\mathrm{LT}^{-1}\right]^{\mathrm{x}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{\mathrm{y}}\left[\mathrm{ML}^3 \mathrm{~T}^{-2}\right]^{\mathrm{z}}\)

∴ \({[\mathrm{L}]=\left[\mathrm{L}^{\mathrm{x}+3 \mathrm{y}+3 \mathrm{z}} \mathrm{M}^{-\mathrm{y}+\mathrm{z}} \mathrm{T}^{-\mathrm{x}-2 \mathrm{y}-2 \mathrm{z}}\right]}\)

Comparing both sides,

-y + z = 0

⇒  y = z → 1

x + 3y + 3z = 1→ 2

– x -4z = 0 → 3

From 1,2 and 3 \(z=y=\frac{1}{2}, x=-2\)

∴ \(\mathrm{~L}=\frac{1}{\mathrm{C}^2}\left[\mathrm{G} \cdot \frac{e^2}{4 \pi \varepsilon_0}\right]^{\frac{1}{2}}\)

Question 38. Planck’s constant (h), the speed of light in a vacuum, and Newton’s gravitational constant (G) are the fundamental constants. Which of the following combinations of these quantities has the dimension of length?

1. \(\frac{\sqrt{h \mathrm{G}}}{\mathrm{C}^{3 / 2}}\)
2. \(\frac{\sqrt{h \mathrm{G}}}{\mathrm{C}^{5 / 2}}\)
3. \(frac{\sqrt{h \mathrm{C}}}{\mathrm{G}}\)
4. \(\frac{\sqrt{G C}}{h^{3 / 2}}\)

Answer: 1. \(\frac{\sqrt{h \mathrm{G}}}{\mathrm{C}^{3 / 2}}\)

According to question, L = [h]^a[c]^b[G]^c

Putting the dimensions on both sides, we get,

[M⁰LT⁰] = [ML²T^-1]^a [LT^-1]^b [M^-1L³T^-2]^c

[M⁰LT⁰] = [M^a+c] [L^2a+b+3c] [T^-a-b+2c]

Comparing powers on both sides, we get,

a – c = 0

2a + b + 3c = 1

– a – b – 2c = 0

Solving the above equations, \(a=c=\frac{1}{2} \text { and } b=-\frac{3}{2}\)

Putting equation 1 we get, \(L=h^{1 / 2} \mathrm{C}^{-3 / 2} \mathrm{G}^{1 / 2}=\frac{\sqrt{h \mathrm{G}}}{\mathrm{C}^{3 / 2}}\)

∴ \(L =\frac{\sqrt{h \mathrm{G}}}{\mathrm{C}^{3 / 2}}\)

Question 39. If Force (F), velocity (v), and time (T) are taken as fundamental units, then the dimensions of mass are:

1. [FvT^-1]
2. [FvT^-2]
3. [Fv^-1T^-1]
4. [Fv^-1T]

Answer: 4. [Fv^-1T]

Dimension of force, F=[MLT^-2]

Dimension of velocity, v = [LT^-1]

and the dimension of time, t = [T]

According to the question,

M = F^a V^b T^c   →  (1)

Where a, b, and c are constant putting dimensions.

[M] = [MLT^-2]^a [LT^-1]^b[T]^c

[M] = [M^a L^a+b T^-2a-b+c]

Comparing powers on both sides, we get:

a= 1, a +b = 0, – 2a – b + c = 0

Solving the above equation,

we get a= 1, b = – 1 and c= 1

Putting these values in eq.(1), we get

M = [FV^-1T]

Question 40. The density of material in the CGS system of units is 4 g/cm3. In a system of units in which the unit of length is 10 cm and the unit of mass is 100 g, the value of the density of the material will be:

1. 0.4
2. 40
3. 400
4. 0.04

Answer: 2. 40

From using the formula, p = constant

⇒ \(u_2 =u_1\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right]\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]\)

= \(4\left[\frac{1}{100}\right]\left[\frac{1}{10}\right]^{-3}=40\)

Question 41. If the dimensions of a physical quantity are given M°L*TC, then the physical quantity will be:

1. pressure, if a=1,b = -1,c = -2
2. velocity, if a = 1, b = 0, c = -1
3. acceleration, if a=1,b=1,c = -2
4. force, if a = 0, b = – 1, c= – 2

Answer: 1. pressure, if a=1,b = -1,c = -2

This question will be solved with the help of given options:

X = M^aL^bT^c

1. If a = 1, b = – 1, c = -2

x = [ML^-1T^-2] -> Pressure

2. If a = 1, b = 0, c = -1

X = [MT^-1] It is not the dimension of velocity.

3. If a = 1, b = 1, c = – 2

X = [MLT^-2], It is not the dimension of acceleration.

4. If a = 0, b = – 1, c = – 2

X = [L^-1T^-2], It is not the dimension of force.

Question 42. In Van der Waals’ gas equation \(\left[p+\frac{a}{V^2}\right][V-b]=R T\) dimension of Van der Waal’s constant a is:

1. [ML⁵T^-2]
2. [ML⁵T²]
3. [ML⁵T²]
4. [M^-1L⁵T^-2]

Answer: 2. [ML^-5T²

Using the principle of homogeneity, \({\left[\frac{a}{V^2}\right] } =P\)

\({[a]=[P]\left[V^2\right] }=\left[M L^{-1} T^{-2}\right]\left[L^6\right]\)

=\(\left[M L^5 T^{-2}\right]\)

Question 43. The velocity v of a particle at time t is given by \(\mathbf{v}=a t+\frac{b}{t+c}\) where a, b, and c are constants. The dimensions of a,b, and c are

1. (L), (LT) and (LT²)
2. (LT²), (L) and (T)
3. (L²), (T), and (LT²)
4. (LT²), (LT) and (L).

Answer: 2. (LT²), (L) and (T)

Using the principle of homogeneity,

Dimension of v = Dimension of at = Dimension of

⇒ \(\frac{b}{t+c}\)

at = [LT^-1]

⇒ \(\|a|=\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{T}]}=\left[\mathrm{LT}^{-2}\right]\)

⇒ And \(\frac{[b]}{[t]}\)

= \(\left[\mathrm{LT}^{-1}\right]\)

b = \(\left[\mathrm{LT}^{-1}\right][\mathrm{T}]=\mathrm{L}\)

And c=T

⇒ c=[T]

Question 44. In a particular system, the units of length, mass, and time are chosen to be 10 cm, 10 g, and 0.1 s respectively. The unit of force in this system will be equivalent to:

1. 0.1 N
2. 1 N
3. 10N
4. 100 N

Answer: 1. 0.1 N

Force F = [MLT^-2]

= (10 g) (10 cm) (0.1 S)^-2

Changing these units into the MKS system

F = (10^-2 kg) (10^-1 m) (10-1 s)^-2

=10^-1 N = 0.1 N

Question 45. If x = at + bt², where x is the distance traveled by the body in kilometers while t is the time in seconds, then the unit of b is:

1. km/s
2. km-s
3. km/s²
4. km-s²

Answer: 3. km/s²

As x =at + bt²

According to the concept of dimensional analysis and the principle of homogeneity.

unit of x = unit of bt²

∴ \(\text { unit of } b=\frac{\text { unit of } x}{\text { unit of } t^2} \mathrm{~km} / \mathrm{s}^2\)