NEET Physics Multiple Choice Questions – Units And Measurements 

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, \(\mathrm{K}=\frac{L}{A \Delta T} \frac{d Q}{d V}=\frac{\text { Metre }}{(\text { Metre })^2 \times \text { Kelvin }} \times \text { watt }\)

= \(\mathrm{W} m^{-1} K^{-1}\)

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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

∴ \(\text { unit of } b=\frac{\text { unit of } x}{\text { unit of } t^2} \mathrm{~km} / \mathrm{s}^2\)