Class 9 Science Chapter 9 Very Short Answer Type Questions
Question 1. When do we use the term force of gravity rather than force of gravitation?
Answer:
We use the term force of gravity rather than force of gravitation for the force of attraction between two bodies in which one body has an infinitely large mass.
Question 2. Name the scientist who determined the value of the universal gravitational constant.
Answer:
Henry Cavendish determined the value of the universal gravitational constant.
Question 3. Which force brings tides into the ocean?
Answer:
The gravitational force of the moon brings tides into the ocean.
Question 4. Is it possible to shield a body from gravitational effects?
Answer:
No, as gravitational interaction does not depend on the nature of the intervening medium.
Question 5. Which force keeps the moon in a uniform circular motion around the Earth?
Answer:
The gravitational force between the moon and the Earth keeps the moon in a uniform circular motion around the Earth.
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Question 6. Suppose gravity of the earth suddenly becomes zero, then which direction will the moon begin to move, if no other celestial body affects it?
Answer:
The moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of the moon is due to the centripetal force provided by the gravitational force of the earth.
Question 7. A stone and the earth attract each other with an equal and opposite force. Why, then we see only the stone falling towards the earth but not the earth rising towards the stone?
Answer:
As the mass of the earth is very large, the acceleration produced in the earth is too small, hence we see only the stone falling towards the earth but not the earth rising towards the stone.
Question 8. Two objects kept in a room do not move towards each other as per the universal law of gravitation. Why?
Answer:
The size of the bodies is very small. Therefore the force of attraction between them is very small. So, both objects do not move towards each other.
Question 9. The earth is acted upon by the gravitation of the sun even, then it does not fall into the sun. Why?
Answer:
The earth does not fall into the sun because the earth remains in its circular orbit due to the gravitational force acting on it.
Question 10. Write the direction of acceleration due to gravity.
Answer:
The direction of acceleration due to gravity is downwards i.e. towards the centre of the earth.
Question 11. Two objects of masses m1 and m2 are dropped in a vacuum from a height above the surface of the earth (m1 is greater than m2). Which one will reach the ground first and why?
Answer:
Both will reach the ground at the same time because the acceleration due to gravity is independent of the masses of freely falling bodies.
Question 12. When a body is thrown upwards, its velocity becomes zero at the highest point. What will be its acceleration at this point?
Answer:
The acceleration at this point is equal to the value of the downward direction.
Question 13. Does the velocity of a body during free fall remain constant? Why/Why not?
Answer:
No, the velocity of a body increases at every point of its motion during free fall as acceleration due to gravity acts on it.
Question 14. At which place on the earth, the acceleration due to gravity is zero?
Answer:
At the centre of the earth, the acceleration due to gravity is zero.
Question 15. Name the place on the earth’s surface, where the weight of a body is maximum and minimum.
Answer:
Weight is maximum at the poles and minimum at the equator.
Question 16. Anu buys 300 g of gold at the poles. What will be the weight of gold at the equator?
Answer:
The value of acceleration due to gravity (g) is less at equators than at poles, so the weight of 300 g gold will be less at the equator
Question 17. The astronauts in space feel weightless. Why?
Answer:
They do not exert any force/weight on their spaceship in the absence of gravity in space.
Question 18. What is the final velocity when a body is dropped from a height?
Answer:
The final velocity is zero when a body is dropped from a height.
Question 19. Which of the two will double the pressure? Doubling the area or making the area half.
Answer:
Making the area half \(\left[ p=\frac{F}{A}\right]\)
Question 20. A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways i.e. when length and breadth form the base, when breadth and width form the base, and when width and length form the base. In which condition, the pressure exerted by the box will be maximum?
Answer:
Pressure =\(\frac{\text { Thrust }}{\text { Area }}\)
The pressure exerted by the box will be maximum when the area is small.
The area will be minimal when breadth and width form the base.
Question 21. When a floating body is pressed down a little, which force will increase at that time?
Answer:
Upthrust.
Question 22. A hydrometer is made heavy near the bottom.
Answer:
A hydrometer is made heavy near its bottom so that it can float with the stem in the vertical position.
Question 23. A bucket of water weighs less inside the well water. Why?
Answer :
Due to the upthrust exerted by the well water on the bottom of the bucket in an upward direction.
Question 24. If the density of a body is 800 kg m-3. Will it sink or float when dipped in a bucket of water? (Take, density of water = 1000 kg-3)
Answer:
Since its density is less than that of water, hence it will float.
Question 25. If 100 cc of water is heated to three different temperatures i.e. 4°C, 20°C and 45°C. At what temperature, its density will be maximum?
Answer:
4° C; because density increases with a decrease in temperature.
Question 26. The density of glass is 3.5 g cm-3. What does it mean?
Answer:
3.5 g cm-3 density of glass means that the volume of 3.5 glass is 1 cm3.
Question 27. An object is dropped one by one in three liquids having different densities. The object floats with \(\frac{1}{9}, \frac{2}{11}\) and \(\frac{3}{7}\) parts of their volumes outside the liquid surface, in liquids of densities \(d_1, d_2\) and \(d_3\), respectively. Arrange them in increasing order.
Answer:
In a liquid of higher density, more part of the object remains outside the liquid. Since the order of part of their volume outside the liquid is \(\frac{1}{9}<\frac{2}{11}<\frac{3}{7}\)
Thus, the order of densities is \(d_1<d_2<d_3\).
Question 28. Name the instrument which is used to determine the density of the liquid.
Answer:
The instrument which is used to determine the density of liquid is a “hydrometer.”
Question 29. Arrange the following in the increasing order of their relative densities.
Iron, air and water
Answer:
The increasing order of the relative densities is as follows: Air < water < iron
Question 30. A body projected horizontally moves with some horizontal velocity although it is under the action of the force of gravity, why?
Answer:
The force of gravity has no effect on the horizontal velocity because it acts in a vertically downward direction. So the body moves with uniform horizontal velocity.
Question 31. A stone dropped from a tree takes 2 seconds to reach the ground. Find its velocity on striking the ground.
Answer:
Given, u =0, t =2 s, a = g =9.8 m/s², v = ?
From the first equation of motion, v =u + at
= 0 + 9.8x 2 = 19.6 m/s
Question 32. Find pressure, when a thrust of 20 N is applied on a surface area of 10 cm².
Answer:
Area, A=10 \(\mathrm{~cm}^2=10 \times 10^{-4} \mathrm{~m}^2[1 \mathrm{~m}=100 \mathrm{~cm}]\)
Pressure(p) =\(\frac{\text { Force }(F)}{\text { Area }(A)}=\frac{20}{10 \times 10^{-4}}\)
=2 \(\times 10^4 \mathrm{~Pa}\)
Pressure is 2 \(\times 10^4 \mathrm{~Pa}\).
Question 33. A metal cuboid of mass M kg rests on a table. A surface area of 40 cm² is in contact with the table. The pressure exerted by the cuboid on the table surface is 10000 Pa. Find the value of M is (given that, g=10 \(\mathrm{~ms}^{-2}\) )
Answer:
Given, A=40 \(\mathrm{~cm}^2=0.004 \mathrm{~m}^2, p=10000 \mathrm{~Pa}\)
p=\(\frac{F}{A}=\frac{M g}{A}\) [e F=M g]
M=\(\frac{p \times A}{g}=\frac{10000 \times 0.004}{10}=4 \mathrm{~kg}\)
Question 34. Find the mass of 5 \(\mathrm{~m}^3\) of cement of density 3000 kg \(\mathrm{~m}^{-3}\).
Answer:
Given, volume, V=5 \(\mathrm{~m}^3\)
Density, \(\rho=3000 \mathrm{~kg} \mathrm{~m}^{-3}\), mass, m= ?
As, \(\rho=\frac{m}{V}\) or m=\(\rho\) V
= 3000×5 =15000 kg
Question 35. An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. What will be the weight of the liquid displaced by the object?
Answer:
Weight of liquid displaced, F =10 – 8 = 2 N
Class 9 Science Chapter 9 Gravitation Short Answer Type Questions
Question 1. “All the objects in the universe attract each other.”
- What is the force of attraction called?
- Name any two factors on which this force of attraction depends.
Answer:
- This force of attraction is called as force of gravitation.
- The force of gravitation depends on two factors:
- Directly proportional to the product of their masses.
- Inversely proportional to the square of the distance between them.
Question 2. What is the source of centripetal force that a planet requires to revolve around the sun? On what factors does that force depend?
Answer:
The motion of the planet around the sun is due to the centripetal force. This centripetal force is provided by the gravitational force between the planet and the sun.
This force depends on the mass of the sun and mass of the planet and the distance between the sun and the planet.
Question 3. State the source of centripetal force that a planet requires to revolve around the sun. On what factors does the force depend? Suppose this force suddenly becomes zero, then in which direction will the planet begin to move, if no other celestial body affects it?
Answer:
The source of centripetal force is the gravitation force. It depends upon the following factors:
- Mass of the planet and the sun.
- Distance between the planet and the sun.
If this force suddenly becomes zero, then the planet will begin to move in a straight line in the direction in which it was moving at that instant.
Question 4. (1) Seema buys a few grains of gold at the poles as per the instruction of one of her friends. She hands over the same when she meets her at the equator. Will the friend agree with the weight of gold bought? If not, why?
(2) If the moon attracts the Earth, then why does the Earth not move towards the moon?
Answer:
- No, her friend will not agree with the weight of the gold bought because the weight at the poles is greater than the weight at the equator.
- We know that the gravitational force is always attractive, still, the moon does not fall on the Earth because the gravitational force between the Earth and the moon works as the necessary centripetal force for the moon to make it revolve around the Earth.
Question 5. Give two reasons for the variation of g at the equator and at the poles.
Answer:
The variation of g at the equator and at the poles are
- due to the difference in the radius and
- due to the rotation of the earth.
Question 6. Two solid objects of masses 1 kg and 2 kg are dropped from a helicopter at the same time. Which one will reach the ground earlier? Justify your answer with a suitable reason.
Answer:
Both will reach the ground at the same time as we know that an object experiences acceleration during free fall. This acceleration experienced by an object is independent of its mass because g=\(\frac{G M}{R^2}\)
As they are dropped at the same time, they will reach the ground at the same time.
Question 7. Give reasons.
- The moon does not have an atmosphere.
- If you jump on the moon, you will rise much higher than, if you jump on the earth.
Answer:
Moon does not have strong gravity to hold atmospheric gases.
Acceleration due to gravity g is much less on the moon’s surface than that of the earth’s surface. v² -u²
Hence, h=\(\frac{v^2-u^2}{2 g}\) is larger.
Question 8. Give one example of each where the same force acting on
- a smaller area exerts a larger pressure.
- a larger area exerts a smaller pressure.
Answer:
- Needles have sharp tips having smaller areas to exert a larger pressure.
- School bags have broad bases and wide straps having larger areas to exert less pressure.
Question 9. Explain, why a camel walks easily on a sandy surface than a man in spite of the fact that a camel is much heavier than a man.
Answer:
The feet of the camel are larger and so cover a larger area, which results in low pressure on the sand bed as compared to human beings and thus, enables them to walk easily on sand without sinking.
Question 10. Give a brief description of why the bottom of dams is broad.
Answer:
As we know, pressure is the force acting per unit area. Dams have large water storage. If the bottom of the dams is not made broad, the large hydraulic pressure may cause dams to sink into the Earth’s basin. So, an increase in the base area decreases the pressure exerted by large water storage.
Question 11. Why pressure exerted by the solid and fluid are different?
Answer:
- Solids exert pressure because of their weight, i.e. pull of mass by the centre of the Earth with an acceleration of g = 9.8 ms-2.
- Similarly, fluids, i.e. gases and liquids both exert equal pressure in all directions over the inner walls of the container in which it is kept.
Question 12. Name two forces which act on a body immersed in a liquid. Give the directions in which they act.
Answer:
The two forces are
- Weight of the body acting downwards.
- The buoyant force acts upwards.
Question 13. State the condition under which an object floats on the surface of a liquid. What is the volume of the liquid displaced by the object?
Answer:
- When the upthrust on the body in a liquid is greater than the weight of the body, then the body floats on the surface of the liquid.
- The volume of the body = volume of liquid displaced.
Question 14. (1) Name the SI units of thrust and pressure.
(2) In which situation, do we exert more pressure on the ground when we stand on one foot or on both feet? Justify your answer.
Answer:
- The SI unit of thrust is Newton (N). The SI unit of pressure is Nm-2or pascal (Pa).
- We exert more pressure on the ground when we stand on one foot than both feet, as the area of one foot is half that of two feet as p \(\propto \frac{1}{A}\).
Question 15. (1) Explain, why a completely immersed bottle in water bounces back on the surface.
(2) Why does a bucket of water weigh less inside the well water?
Answer:
- Since it is known that a body can sink in water only when its weight is greater than the upthrust acted on it by the water. But in this case, the upthrust act on the bottle is greater than its weight, that’s why, it bounces back on the water surface.
- A bucket of water weighs less inside the well water, it is because when the bucket immersed in the water fully, upthrust “acts on it by water which reduces its actual weight.
Question 16. When a fresh egg is put into a beaker filled with water, it sinks in water. But when a lot of salt is dissolved in the water, the egg begins to rise and then floats. Why?
Answer:
On dissolving a lot of salt in water, the density of salt solution becomes higher than that of pure water. Due to its much higher density, the salt solution exerts much more upward buoyant force on the egg, making the egg rise and then float.
Question 17. Two different bodies are completely immersed in water and undergo the same loss in weight. Is it necessary that their weights in the air should also be the same? Justify your answer.
Answer:
No, their weights in air don’t need to also be the same. This is because the two bodies have undergone the same loss in weight on completely immersing in water due to their equal volumes, not due to their equal weights. So, they may have different weights in the air.
Question 18. (1) Why does a bucket of water feel heavier when taken out of water?
(2) Lead has greater density than iron and both are denser than water. Is the buoyant force on a lead object greater than, less than or equal to the buoyant force on an iron object of the same volume?
Answer:
- A bucket of water feels heavier when taken out of water because when immersed in water, an upward force, i.e. buoyant force acts on it which is equal to the weight of water displaced by the bucket.
- The buoyant force on a lead object is lesser than the buoyant force on an iron object because lead has greater density, so it displaces a lesser amount of water consequently lesser amount of buoyant force acts on it.
Question 19. Verify Archimedes’ principle of buoyancy with an activity. For the activity, you are provided with a piece of stone, a rubber string and a container filled with water.
Answer:
- First of all, tie up the stone with the rubber string and hold it against a scale fixed on a wall. Put a mark on the elongated rubber string when the stone is tied.
- Repeat this experiment but this time suspend the stone in a beaker, filled with water. Now, compare the markings.
- Explanation: As the buoyant force is acting on the stone in an upward direction, this, gravitational pull decreases and in turn, the stretch of the rubber is lost.
Question 20. State Archimedes’ principle. Write its two applications.
Answer:
Archimedes’ principle states that “when a body is immersed fully or partially in a liquid, it experiences an upward force that is equal to the weight of the liquid displaced by it.”
Applications
- It is used in designing ships and submarines.
- It is used in making lactometers, which are used to determine the purity of milk.
Question 21. When a metallic block is immersed below the surface of a liquid, state and define the upward force acting on it.
Answer:
When a metallic block is immersed below the surface of the liquid, it experiences an upward force known as buoyant force.
Buoyant force acting on the block =Weight of liquid displaced by object = mg=(V \(\rho\)) g
where, m – mass of the metallic block
⇒ \(\rho\) = density of metallic block V = volume
g = acceleration due to gravity.
Question 22. If a body is compressed to half its previous volume, what will be the effect on its density and why?
Answer:
- Since, density \((\rho)=\frac{\text { mass }(m)}{\text { volume }(V)}\)
- Therefore, if the volume of a body is compressed to half of its previous volume, then the density will be doubled.
Question 23. A block of ice is floating in a bucket of water full up to the brim. Some portion of the ice is visible above the water level. As it melts completely do you expect water to spill out? Give a reason for your answer.
Answer:
We know that the volume of ice is greater than that of water. So in this case when the ice melts, it occupies a volume less than that occupied by the ice dipped in it, so the water level will not spill out.
Question 24. Identical packets are dropped from two aeroplanes, one above the equator and the other above the North Pole, both at height h. Assuming all conditions are identical, will those packets take the same time to reach the surface of the earth? Justify your answer.
Answer:
- No, those packets do not take the same time to reach the surface of the earth.
- As the value of g is less at the equator than poles, so packets dropped at the poles reach the surface of the earth first.
Question 25. How will the weight of a body of mass 100 g change, if it is taken from the equator to the poles? Give reasons.
Answer:
Since, the acceleration due to gravity increases from the equator to the poles, so its weight increases because the radius of the earth is less at the poles than at the equator.
Question 26. When you immerse an empty plastic bottle in a bucket of water, it comes above the surface of the water. Why does this happen? How can it remain immersed in water and why?
Answer:
- When we immerse an empty plastic bottle in a bucket of water, the upward force (upthrust or buoyant force) exerted by water on the bottle is greater than its own weight, therefore it comes above the surface of the water.
- To keep the bottle completely immersed, an external force which is equal to the difference between the upward force and the weight of the bottle, must be applied to the bottle in a downward direction. This is because the upthrust on the bottle due to water must be balanced.
Question 27. What happens to the magnitude of the force of gravitation between two objects, if (1) distance between the objects is tripled? (2) mass of both objects is doubled?
Answer:
As we know,F=\(\frac{G m_1 m_2}{r^2}\) [symbols have their usual meanings]
(1) \(r^{\prime}=3 r \Rightarrow F^{\prime}=\frac{G m_1 m_2}{9 r^2}=\frac{F}{9}\) [force decreases by 9 times]
(2) \(m_1^{\prime}=2 m_1 and m_2^{\prime}=2 m_2\)
⇒ \(F^{\prime}=\frac{4 G m_1 m_2}{r^2}=4 F\) [force increases by 4 times]
To maintain the same force one of the mass is to be increased by 36 times.
Question 29. A body weighs 25 kg on the surface of the earth. If the mass of the earth is \(6 \times 10^{24}\) kg then the radius of the earth is 6.4 \(\times 10^{6}\) and the gravitational constant 6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\).
Calculate
- the mutual force of attraction between the body and the earth.
- the acceleration produced in the body.
Answer:
Given, \(M_e=6 \times 10^{24} \mathrm{~kg}\),m=25 kg
⇒ \(R_e=6.4 \times 10^6 \mathrm{~m}\) and G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
(1) Mutual force, F=G \(\frac{M_e}{R_e^2} m \)
= \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 25}{\left(6.4 \times 10^6\right)^2}=244 \mathrm{~N}\)
(2) Acceleration produced in the body,
a=\(\frac{F}{m}=\frac{244}{25}=9.8 \mathrm{~m} / \mathrm{s}^2\)
Question 30. Two bodies of masses 3 kg and 12 kg are placed at a distance of 12 m. A third body of mass 0.5 kg is to be placed at such a point that the force acting on this body is zero. Find the position of that point, for Given, mx =3 kg and m² =12 kg
Answer:
Given, \(m_1\)=3 kg and m_2=12 kg
Let the mass, \(m_3\)=0.5 kg be placed at a distance of x from m_1 as shown
Then, the force acting on \(m_3\) due to \(m_1\) is equal and opposite to the force acting on \(m_3\) due to \(m_2\).
⇒ \(F_{31}=F_{32}\)
⇒ \(\frac{G m_1 m_3}{x^2}=\frac{G m_3 m_2}{(12-x)^2} \Rightarrow \frac{3}{x^2}=\frac{12}{(12-x)^2}\)
⇒ \(\left(\frac{12-x}{x}\right)^2=\frac{12}{3}\) =4
⇒ \(\frac{12-x}{x}\) =2
12-x=2 x \(\Rightarrow\) 12=3 x \(\Rightarrow x=4 \mathrm{~m}\)
The required point is at a distance of 4 m from the mass of 3 kg.
Question 31. Calculate the acceleration due to gravity on the surface of satellite having mass 7.4 x \(10^22\) kg and radius 1.74 x \(10^6\) cm. (Take, G = 6.7 x \(10^{-11} \mathrm{~N}-\mathrm{m} / \mathrm{kg}^2\))
Answer:
As we know, acceleration due to gravity, g=\(\frac{G M}{R^2}\).
For the satellite, R =1.7 .4 \(\times 10^6 \mathrm{~cm}=\frac{1.74 \times 10^6}{100}=1.74 \times 10^4 \mathrm{~m}\)
M =7.4 \(\times 10^{22} \mathrm{~kg}\)
g =\(\frac{6.67 \times 10^{-11} \times 7.4 \times 10^{22}}{1.74 \times 10^4 \times 1.74 \times 10^4}\)
=\(\frac{6.67 \times 7.4}{1.74 \times 1.74} \times 10^3 \mathrm{~g}=16.30 \times 10^3 \mathrm{~m} / \mathrm{s}^2\)
Question 32. What height above the surface of the earth, the value of g becomes 64% of its value at the surface of the earth? Take, the radius of the earth = 6400 km. r? Let g – acceleration due to gravity at the earth’s surface.
Answer:
g, = acceleration due to gravity at height
⇒ \(g_h\) = acceleration due to gravity at height
h =\(\frac{64}{100} \times g=0.64 g\)
g =\(\frac{G M}{R_e^2}\)
Similarly, \(g_h=\frac{G M}{\left(R_e+h\right)^2}\)
From Eqs. (1) and (2), we get
⇒ \(g_h=\frac{g R_e{ }^2}{\left(R_e+h\right)^2}\)
0.64 g=\(\frac{g R_e{ }^2}{\left(R_e+b\right)^2}\)
0.64\(\left(R_e+b\right)^2=R_e^2\)
0.8\(\left(R_e+h\right)=R_e\)
0.8 h=\(R_e-0.8\)
⇒ \(R_e=0.2 R_e\)
h=\(\frac{2 \times 6400}{8}=1600 \mathrm{~km} \left[ R_e=6400 \mathrm{~km}\right]\)
Thus, the height above the surface of the earth is 1600 km.
Question 33. Estimate the gravitational force between two protons (1.6x \(10^{-27}\) kg) separated by a distance of I A.
Answer:
Since, gravitational force, F=\(\frac{G M_p M_p}{r^2}\)
Given, mass of proton, \(M_p=1.6 \times 10^{-27} \mathrm{~kg}\)
Distance, r=1 \(Å=10^{-10} \mathrm{~m}\)
F =\(\frac{6.67 \times 10^{-11} \times 1.6 \times 10^{-27} \times 1.6 \times 10^{-27}}{\left(10^{-10}\right)^2}\)
=17.1 \(\times 10^{-45} \mathrm{~N}\)
Question 34. A ball is dropped from a height half of the earth’s radius. Find the value of g at this point.
Solution:
According to the question, the distance of the ball from the earth’s centre,
⇒ \((R+h)=R+\frac{R}{2}=\frac{3 R}{2} [height, h=R / 2]\)
From the formula of acceleration due to gravity,
g =\(\frac{G M}{(R+h)^2}=\frac{G M}{(3 R / 2)^2}=\frac{4 G M}{9 R^2}\)
= \(\frac{4}{9} \times 10=4.44 \mathrm{~m} / \mathrm{s}^2 \left[ \frac{G M}{R^2}=10 \mathrm{~m} / \mathrm{s}^2\right]\)
Thus, the value of g is 4.44 \(\mathrm{~m} / \mathrm{s}^2\).
Question 35. A body is dropped from a height of 100 m. What is its height above the ground after 2 seconds of its fall? (Take, g=10m/s² )
Solution:
Given, initial velocity, u = 0 Time taken, t – 2 s
Acceleration due to gravity, a = g
From the second equation of motion,
⇒ \(s=u t+\frac{1}{2} g t^2=0+\frac{1}{2} \times 10 \times(2)^2=20 \mathrm{~m}\)
The height of the body above the ground after 2 s of its fall, h =100 – 20 =80 m
Question 36. A ball is thrown up with a velocity of 19.6 m/s.
- How long will it take to reach the maximum height?
- How high will it go?
Answer:
Given, initial velocity, u = 19.6 m/s
(1) Final velocity, v = 0 at maximum height.
So, t=\(\frac{u}{g}=\frac{19.6}{9.8}=2 \mathrm{~s}\) [ v=u-g t and v=0]
From the third equation of motion,
⇒ \(v^2=u^2-2 g h\) [negative sign is taken due to upward motion]
⇒ \(u^2=2 g h\) {[ v=0]}
h=\(\frac{u^2}{2 g}=\frac{(19.6)^2}{2 \times 9.8}=19.6 \mathrm{~m}\)
Question 37. Your mass on the earth is 50 kg. Planet M has two times the force of gravity of that of the Earth. What will be your mass and weight on planet M? (Take, g = 9.8 m/s²)
Answer:
As the mass remains constant, so on the planet M your mass will be 50 kg.
Given, that acceleration due to gravity on planet M is two times that of the Earth.
So, weight on planet M will be 50 x 2 x g = 50x2x9.8 =980
Question 38. Suppose that the radius of the earth becomes twice its original radius without any change in its mass. Then, what will happen to your weight?
Answer:
The weight of a body is the force with which a body is attracted towards the earth, w=\(\frac{G M m}{R^2}\)
If the radius of the earth becomes twice its original radius, then w=G \(\frac{M m}{(2 R)^2}=\frac{G M m}{4 R^2}=\frac{w}{4}\)
Thus, the weight will be reduced to one-fourth of the original
Question 39. 39 Find the weight of an 80 kg man on the surface of the moon. What should be his mass on the earth and on the moon?
(Take, \(g_e=9.8 \mathrm{~m} / \mathrm{s}^2, g_m=1.63 \mathrm{~m} / \mathrm{s}^2\) )
Answer:
Given, mass on the earth = mass on the moon =80 kg
Man’s weight on the earth,
⇒ \(w_e=9.8 \times\) 80=784 N
Man’s weight on the moon,
⇒ \(w_m=1.63 \times\) 80=130.4 N
Thus, the weight of the moon is 130.4 N.
Question 40. 40 The weight of any person on the moon is about l/6th of that on the earth. He can lift a mass of 15 kg on the earth. What will be the maximum mass which can be lifted by the same force applied by the person on the moon?
Answer:
Maximum weight which can be lifted = mg
=15 \(\mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^2=147 \mathrm{~N}\)
Mass which can be lifted on the moon
=147 \(\mathrm{~N} \times \frac{1}{g_{\text {earth }} / 6}\)
[at moon, acceleration due to gravity is 1/6 th to that of the earth]
= \(\frac{147 \times 6}{9.8}=\frac{147}{1.63}=90 \mathrm{~kg}\)
Thus, the maximum mass which can be lifted is 90 kg.
Question 41. Calculate the average density of the earth in terms of g, G and R.
Answer:
According to the formula, density =\(\frac{\text { mass }}{\text { volume }}\).
If the radius of the earth is R, then its volume =\(\frac{4}{3} \pi R^3\). [The shape of the earth is spherical]
Mass of the earth, M=\(\frac{g R^2}{G} [ g=\frac{G M}{R^2}]\)
Density =\(\frac{g R^2}{G} / \frac{4}{3} \pi R^3=\frac{3 g}{4 \pi R G}\)
Thus, the average density of the earth is \(\frac{3 g}{4 \pi R G}\)
Question 42. Which will exert more pressure, 100 kg mass on 10 m2 or 50 kg mass on 4 m2? Give reason. (Take, g = 10 ms-2)
Answer:
For 100 kg of mass,
Pressure, \(p_1=\frac{\text { Force }}{\text { Area }}=\frac{m_1 g}{A_1}=\frac{100 \times 10}{10}=100 \mathrm{~Pa}\)
For 50 kg of mass,
Pressure, \(p_2=\frac{m_2 g}{A_2}=\frac{50 \times 10}{4}=125 \mathrm{~Pa}\)
⇒ \(p_2>p_1\)
Hence, a 50 kg mass will exert more pressure.
Question 43. A block of wood of mass 5 kg and dimensions 40 cm x 20 cm x 10 cm is placed on a tabletop. Find the pressure exerted, if the block lies on the tabletop with sides of dimension
- 40 cm x 20 cm
- 40 cm x 10 cm (Take, g = 10 \(\mathrm{~ms}^{-2}\))
Answer:
Given, mass, m =5 kg, acceleration due to gravity, g =10 \(\mathrm{~ms}^{-2}\)
g=10 \(\mathrm{~ms}^{-2}\)
Weight =m g=5 \(\times 10=50 \mathrm{~N}\)
(1) Pressure =\(\frac{\text { Weight }}{\text { Area }}=\frac{50}{40 \times 20 \times 10^{-4}} \mathrm{~Pa}\)
=6.25 \(\times 10^2 \mathrm{~Pa}\)
(2) Pressure =\(\frac{\text { Weight }}{\text { Area }}=\frac{50}{40 \times 10 \times 10^{-4}} \mathrm{~Pa}\)
=1.25 \(\times 10^3 \mathrm{~Pa}\)
Question 44. If two forces in the ratio 5:9 act on two areas in the ratio 10:3, find the ratio of pressure exerted.
Answer:
Given, ratio of areas =10: 3$, i.e. \(A_1 / A_2\)=10 / 3
Ratio of forces =5: 9, i.e. \(F_1 / F_2\)=5 / 9
Ratio of pressures =\(\frac{p_1}{p_2}=\frac{F_1}{A_1} \times \frac{A_2}{F_2}\)
= \(\frac{F_1}{F_2} \times \frac{A_2}{A_1}=\frac{5}{9} \times \frac{3}{10}=\frac{1}{6}\)
Therefore, the ratio of the pressures is 1:6.
Question 45. A ball filled with air has a volume of 500 cm3. Calculate the minimum force applied by a child to put it completely inside the water. (Take, g = 10 \(\mathrm{~ms}^{-2}\))
Answer:
Given that,
Volume, V=500 cm3 =500 x 10-6 m3,
g = 10 ms-2, F =?
The force required to put the ball inside the water = Buoyant force
= Weight of water displaced – mg … (1)
Now, we know that,
Mass of water = Density of water x Volume m =pV
On substituting this value in Eq. (1), we get Force = p Vg
= (1000 kg m-3) X (500 X 10-6 m3) X (10 ms-2)
= 1000 x 500 x 10~6 x 10 N =5 N
The minimum force applied by a child to put the ball completely inside the water is 5 N.
Question 46. A ball weighs 80 g in air, 60 g in water and 50 g in liquid. If the density of water is 1 g cm-3, find the density of this liquid.
Answer:
As given, when immersed in water, the ball displaces 80 -60 = 20 g of water.
So, its volume = volume of water displaced
=\(\frac{20 \mathrm{~g}}{1 \mathrm{~g} \mathrm{~cm}^{-3}}=20 \mathrm{~cm}^3\)
Density of ball =\(\frac{80 \mathrm{~g}}{20 \mathrm{~cm}^3}=4 \mathrm{~g} \mathrm{~cm}^{-3}\)
So, the ball displaces (80-50) \(\mathrm{g}=30 \mathrm{~g}\) of liquid with density =\(\frac{30 \mathrm{~g}}{20 \mathrm{~cm}^3}=1.5 \mathrm{~g} \mathrm{~cm}^{-3}\).
Question 47. Prove that, if a body is thrown vertically upwards, then the time of ascent is equal to the time of descent.
Answer:
For the upward motion,
v=u-g \(t_1\), 0=u-g \(t_1\), \(t_1=\frac{u}{g}\)
and the downward motion,
v=u+g \(t_2\), v=0+g \(t_2\)
The body falls back to the earth at the same speed as it was thrown vertically upwards.
v=u, u=0+g \(t_2 \Rightarrow t_2=\frac{u}{g}\)
From Eqs. (1) and (2), we get \(t_1=t_2 \Rightarrow\) Time of ascent = Time of descent
Question 48. A ball is dropped from the edge of a roof. It takes 0.1s to cross a window of height 2.0 m. Find the height of the roof above the top of the window.
Answer:
Let AB be the window and suppose the roof T is at a height y above A. Also, suppose it takes a time TV for the ball to reach A. The velocity of the ball at A is
⇒ \(v_1=0+g t_1=9.8 t_1\) (1)
Now, consider the motion of the ball from A to B.
Here, the initial velocity is v1 the distance covered is 2 m and the time taken is 0.1 s.
From second equation of motion, s=u t+\(\frac{1}{2} g t^2\)
2.0=\(v_1(0.1)+\frac{1}{2} \times 9.8 \times(0.1)^2 =9.8 t_1(0.1)+0.049\)
⇒ \(t_1 =1.99 \approx 2 \mathrm{~s}\)
The height y is y=\(\frac{1}{2} g t_1^2\)
= \(\frac{1}{2} \times 9.8 \times(2)^2=19.6 \mathrm{~m}\)
The roof is at a height of 19.6 m above the top of the window.
Question 49. On the earth, a stone is thrown from a height in a direction parallel to the earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why?
Answer:
For both the stones, initial velocity, u- 0
Acceleration in a downward direction = g
Now, from the second equation of motion,
h=u t+\(\frac{1}{2} g t^2 \Rightarrow h=0+\frac{1}{2} g t^2\)
h=\(\frac{1}{2} g t^2 \quad \Rightarrow t=\sqrt{\frac{2 h}{g}}\)
Both stones will take the same time to reach the ground because the two stones fall from the same height.
Question 50. A ball is thrown with some speed u m/s. Show that under the free fall, it will fall on the ground at the same speed.
Answer:
When the ball is thrown upwards, then it will reach a certain height h and start falling. At maximum height h, the final velocity will be v = 0.
The maximum height reached by the ball,
⇒ \(v^2-u^2=2 g h\)
⇒ \(0-u^2=-2 g h\) [acceleration =-g]
h=\(\frac{u^2}{2 g}\) [using equation]
In the second case, when the ball starts to fall, then the initial velocity u=0. It will accelerate due to gravity, i.e. a=g and reach ground with speed (say \(v_2\) ).
Using equation, \(v_2^2-u^2\) =2 g h
⇒ \(v_2^2-0 \)=2 g h
⇒ \(v_2^2 =2 g\left(\frac{u^2}{2 g}\right)=u^2\) [{ from Eq. (1) })
⇒ \(v_2\) =u
Thus, the ball reaches the ground at the same speed.
Class 9 Science Chapter 9 Gravitation Long Question And Answers
Question 1. (1) Write the formula to find the magnitude of the gravitational force between the earth and an object on the earth’s surface.
(2) Derive how does the value of gravitational force F between two objects change when
- distance between them is reduced to half and
- mass of an object is increased four times.
Answer:
(1) Formula to find the magnitude of gravitational Force, F=\(\frac{G M m}{R^2}\)
M=mass of the earth
m = mass of the object
R = radius of the earth
and universal gravitational constant, G=\(6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
(2) (1) Let the gravitational force be F when the distance between them is R,
F=\(\frac{G M m}{R^2}\) → Equation 1
Now, when the distance is reduced to half,
⇒ \(F^{\prime}=\frac{G M m}{\left(\frac{R}{2}\right)^2}=\frac{4 G M m}{R^2}\) → Equation 2
On dividing Eq. (1) by Eq. (2), we get
⇒ \(\frac{F}{F^{\prime}}=\frac{G M m}{R^2} \times \frac{R^2}{4 G M m}\)
⇒ \(F^{\prime}\) =4 F
(2) When the mass becomes 4 times,
∴ \(\frac{F}{F^{\prime}}=\frac{G M m}{R^2} \times \frac{R^2}{4 G M m} \Rightarrow F^{\prime}=4 F\)
Question 2. (1) Prove that, if the earth attracts two bodies placed at the same distance from the centre of the earth with equal force, then their masses will be the same.
(2) Mathematically express the acceleration due to gravity in terms of mass of the earth and radius of the earth.
(3) Why is G called a universal constant?
Answer:
Let the two bodies have masses m1 and m2 and they are placed at the same distance R from the centre of the earth. According to the question, if the same force acts on both of them, then
⇒ \(F_1=\frac{G M m_1}{R^2}\)
and\(F_2=\frac{G M m_2}{R^2}\)
As, \(F_1=F_2\)
Hence, \(\frac{G M m_1}{R^2}=\frac{G M m_2}{R^2}\)
So, \(m_1=m_2\), their masses will be the same.
(2) Mathematically, g=\(\frac{G M}{R^2}\).
where, g = acceleration due to gravity
G = universal gravitational constant M mass of the earth and R = radius of the earth
(3) G is known as the universal gravitational constant because its value remains the same all the time everywhere in the universe, applicable to all bodies whether celestial or terrestrial.
Question 3. (1) A person weighs 110.84 N on the moon, whose acceleration due to gravity is 1/6 of that of the earth. If the value of g on the earth is 9.8 m/s², then calculate
- g on the moon
- mass of people on the moon
- weight of the person on the earth
(2) How does the value of g on the earth is related to the mass of the earth and its radius? Derive it.
Answer:
(1) g on the moon is given by
\(g^{\prime}=\frac{g}{6}=\frac{9.8}{6}=1.63 \mathrm{~m} / \mathrm{s}^2\)(2) Mass of the person on the moon
=\(\frac{110.84}{1.63}=68 \mathrm{~kg}\)
(3) Weight of person on the earth = mg
=68 \(\times\) 9.8
=666.4 \(\mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
Weight of a person on the earth will be
w=\(\frac{G M m}{R^2}\)
where, M=mass of the earth R = radius of the earth m = mass of person and G=6.67 \(\times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\)
Question 4. Two objects of masses mx and having the same size are dropped simultaneously from heights ha and h2, respectively. Find out the ratio of time they would take to reach the ground. Will this ratio remain the same, if
- one of the objects is hollow and the other one is solid and
- both of them are hollow, size remains the same in each case. Give reason.
Answer:
Height of object A, \(h_1=\frac{1}{2} g t_1^2\)
Height of object B, \(h_2=\frac{1}{2} g t_2^2\)
⇒ \(h_1: h_2=t_1^2: t_2^2\)
or \(t_1: t_2=\sqrt{h_1}: \sqrt{h_2}\)
(1) Acceleration due to gravity is independent of the mass of the falling body. So, the ratio remains the same.
(2) If bodies are hollow, then also ratio remains the same, \(t_1: t_2=\sqrt{h_1}: \sqrt{h_2}\)
Question 5. A stone is dropped from the edge of a roof.
- How long does it take to fall 4: 9 am
- How fast does it move at the end of that fall?
- How fast does it move at the end of 7.9 m?
- What is its acceleration after Is and after 2s?
Answer:
Given, initial velocity u=0
Acceleration g=9.8 \(\mathrm{~m} / \mathrm{s}^2\)
(1) We have, s =u t+\(\frac{1}{2} g t^2\)
4.9 =0 \(\times t+\frac{1}{2} \times 9.8 \times t^2\)
⇒ \(t^2 =\frac{9.8}{9.8}=1 \Rightarrow t=1 s\)
The stone takes 1 s to fall 4.9 m
(2) We have, \(v^2-u^2=2\) as
⇒ \(v^2-0^2 =2 \times 9.8 \times\) 4.9
v =\(\sqrt{96.04}=9.8 \mathrm{~m} / \mathrm{s}\)
At the end of 4.9 m, the stone will be moving at a speed of 9.8 m/s.
We have, \(v^2-u^2\)=2 as
⇒ \(v^2-0^2 =2 \times 9.8 \times 7.9\)
v =12.44 \(\mathrm{~m} / \mathrm{s}\)
=12.44 m/s
The stone will be moving with a speed of 12.44 m/s at the end of 7.9 m
(4) During the free fall the acceleration produced in a body remains constant.
So, acceleration after 1 s = 9.8 m/s² Acceleration after 2 s = 9.8 m/s²
Question 6. (1) A steel needle sinks in water but a steel ship floats. Explain, how.
(2) Why do you prefer a broad and thick handle for your suitcase?
Answer:
(1) The ship displaces more water than the needle as the volume of the ship is more than that of the needle. Since upthrust depends on the volume of the object (U = Vdg), so more the volume of the object, the more upthrust acts on it and the object floats.
(2) Since, pressure act on the body is inversely proportional to the surface area of contact, i.e. p \(\propto \frac{1}{A}\)
It means that the more the area of contact, the less pressure will act on the body. As the broad and thick handle of our suitcase has a large area, due to which less pressure acts on our hand and it is very easy to take from one place to another
Question 7. The radius of the earth at the poles is 6357 km and the radius at the equator is 6378 km. Calculate the percentage change in the weight of a body when it is taken from the equator to the poles.
Answer:
Let acceleration due to gravity at the equator,
⇒ \(g_e=\frac{G M_e}{R_e^2}\)
and acceleration due to gravity at poles,
⇒ \(g_p=\frac{G M_e}{R_p^2}\)
The variation of acceleration due to gravity,
⇒ \(\Delta g=g_p-g_e=G M_e\left(\frac{1}{R_p^2}-\frac{1}{R_e^2}\right)\)
e variation in g=\(\frac{G M_e\left(\frac{1}{R_p^2}-\frac{1}{R_e^2}\right)}{\frac{G M_e}{R_e^2}} \times 100\)
Percentage variation in g=\(\frac{G M_e\left(\frac{1}{R_p^2}-\frac{1}{R_e^2}\right)}{\frac{G M_e}{R_e^2}} \times 100\)
= \(\frac{R_e^2-R_p^2}{R_e^2 R_p^2} \times 100 \times R_e^2=\frac{R_e^2-R_p^2}{R_p^2} \times 100\)
= \(\frac{(6378)^2-(6357)^2}{(6357)^2} \times 100 \approx 0.7 \%\)
% variation in the weight of a body = % Change in g=0.7 \%