NCERT Exemplar: Gravitation | Science Class 9 PDF Download

Multiple Choice Questions

Q.1. Two objects of different masses falling freely near the surface of moon would
(a) have same velocities at any instant
(b) have different accelerations
(c) experience forces of same magnitude
(d) undergo a change in their inertia
Ans: (a)
Explanation: 

• The same acceleration due to gravity of Moon is applied on both the objects.
• In a free fall, velocity depends only on the acceleration produced by gravity. 
• The acceleration due to gravity is independent of the mass of the body. Hence, the same acceleration due to gravity of Moon is applied on both objects. So, they will have the same velocities at any instant irrespective of their masses. 
• The two will not experience the same force as force is dependent on mass and mass of the body is different here.

Q.2. The value of acceleration due to gravity
(a) is same on equator and poles
(b) is least on poles
(c) is least on equator
(d) increases from pole to equator
Ans: (c)
Explanation:  

• The acceleration due to gravity is given by g=GM_E / R_E where G is the universal gravitational constant, M_E is the mass of the Earth, and R_E is the radius of the Earth.
•  Acceleration due to gravity is inversely proportional to the radius of the Earth. 
• At the equator, the radius of the Earth is maximum. Hence, the acceleration due to gravity is least at the equator.

Q.3. The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become
(a) F/4
(b) F/2
(c) F
(d) 2 F
Ans: (a)

Explanation:

•  Gravitational force between two objects varies directly as their masses and inversely as the square of the distance between them. 
• So, when the masses of both objects are halved without changing the distance, the gravitational force between them would become one-fourth of the original value.

Q.4. A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone
(a) will continue to move in the circular path
(b) will move along a straight line towards the centre of the circular path
(c) will move along a straight line tangential to the circular path
(d) will move along a straight line perpendicular to the circular path away from the boy
Ans: (c)
Explanation: At any instance of time object in circular motion tend to be in rectilinear motion. Object keeps on moving due to centripetal force and it moves along a straight line tangential to the circular path when strings breaks.

Q.5. An object is put one by one in three liquids having different densities. The object floats with 1/9 , 2/11 and 3/7 parts of their volumes outside the liquid surface in liquids of densities d₁, d₂ and d₃ respectively. Which of the following statement is correct?
(a) d1> d2> d3
(b) d1> d2< d3
(c) d1< d2> d3
(d) d1< d2< d3
Ans: (d) 

Explanation: 

• Buoyant Force is when a rigid object is submerged in a fluid (completely or partially), there exits an upward force on the object that is equal to the weight of the fluid that is displaced by the object.
• In this case, there are 3 liquids with varying densities. 9 / 1=0.11, 2/11=0.18and 3/7=0.43. These are the volume outside of liquids of densities d₁,d₂ and d_e respectively. This means that the buoyant force is maximum for d₃ and minimum for d₁. As the buoyant force is proportional to density, the densities of the liquids are in the order d₁<d₂<d₃.

Q.6. In the relation F = G M m/d², the quantity G
(a) depends on the value of g at the place of observation
(b) is used only when the earth is one of the two masses
(c) is greatest at the surface of the earth
(d) is universal constant of nature
Ans: (d)
Explanation: 

• G is called as Newton’s constant.  Value of G is 6.66x 10^-11 Nm²kg^-2.  
• G is the universal gravitational constant which remains constant at all places in the universe. 
• G is equivalent to the force of attraction between two bodies of unit mass and unit distance apart.

Q.7. Law of gravitation gives the gravitational force between
(a) the earth and a point mass only
(b) the earth and Sun only
(c) any two bodies having some mass
(d) two charged bodies only
Ans: (c)

Explanation: Newton's law of gravitation can be used to calculate the gravitational force that exists between any two bodies of a certain mass. The force of attraction between two bodies is inversely proportional to the square of the distance between them and directly proportional to the product of their masses.

Q.8. The value of quantity G in the law of gravitation
(a) depends on mass of earth only
(b) depends on radius of earth only
(c) depends on both mass and radius of earth
(d) is independent of mass and radius of the earth
Ans: (d)
Explanation: G is an universal constant hence it is independent of mass and radius of the earth.

Q.9. Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be
(a) 14 times
(b) 4 times
(c) 12 times
(d) unchanged
Ans: (b)

Explanation: The gravitational force between two bodies is inversely proportional to the square of the distance between them and directly proportional to the product of their masses. Therefore, the gravitational force between the two objects would increase to four times its initial value when their masses are both doubled while remaining at the same distance.

Q.10. The atmosphere is held to the earth by
(a) gravity
(b) wind
(c) clouds
(d) earth’s magnetic field
Ans: (a)

Explanation: Due to gravity, everything is drawn toward the earth.  In fact, the air encircling our planet is kept in place by gravity, and this protective layer of air is referred to as the atmosphere.

Q.11. The force of attraction between two unit point masses separated by a unit distance is called
(a) gravitational potential
(b) acceleration due to gravity
(c) gravitational field
(d) universal gravitational constant 
Ans: (d)
Explanation:
 
Here point masses are separated by unit distance
Hence m₁, m₂ and r = 1
Hence F = G which is a universal constant. Hence answer is universal gravitational constant

Q.12. The weight of an object at the centre of the earth of radius R is
(a) zero
(b) infinite
(c) R times the weight at the surface of the earth
(d) 1/R2 times the weight at surface of the earth
Ans: (a)
Explanation: At the centre of the earth acceleration due to gravity is zero. Since weight is the product of mass and acceleration due to gravity. Weight of the object at the centre of the earth will be zero.

Q.13. An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be
(a) 2 N
(b) 8 N
(c) 10 N
(d) 12 N
Ans: (a)
Explanation:According to Archimedes principle,

 Weight of displaced liquid = weight of object in air- weight of object in liquid
= 10N - 8N
= 2N

Q.14. A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be
(a) maximum when length and breadth form the base
(b) maximum when breadth and width form the base
(c) maximum when width and length form the base
(d) the same in all the above three cases
Ans: (b)
Explanation: Surface area and pressure are inversely proportional to each other. Pressure will be maximum when surface area is minimum. Hence, the answer is maximum when breadth and width form the base because surface area is going to be minimum when breadth and width form the base. 

• Length = 60cm
• Breadth = 40cm
• Width = 20cm

Area of the base formed by length and breadth = 60 × 40 = 2400 cm²

Area of the base formed by breadth and width = 40 × 20 = 800 cm²

Area of the base formed by width and length = 20 × 60 = 1200cm²

Q.15. An apple falls from a tree because of gravitational attraction between the earth and apple. If F₁ is the magnitude of force exerted by the earth on the apple and _F2 is the magnitude of force exerted by apple on earth, then
(a) F₁ is very much greater than F₂
(b) F₂ is very much greater than F₁
(c) F₁ is only a little greater than F₂
(d) F₁ and F₂ are equal
Ans: (d)
Explanation: Netwon’s third law of motion states that for every action there is equal and opposite reaction. Hence F₁ and F₂ are equal.

Short Answer Questions

Q.16. What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?
Ans: Gravitational force is the source of centripetal force required to revolve around the sun. This force depends on the distance between the planet and sun along with their masses. If this force becomes zero as a result of the absence of centripetal force, the planet would shift to moving tangentially outwards to the circular route.

Q.17. 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?
Ans: Both the stones reach the ground simultaneously as they are dropped from the same height and their initial velocity will be the same.

Q.18. Suppose gravity of earth suddenly becomes zero, then in which direction will the moon begin to move if no other celestial body affects it?
Ans: If there is no gravitational pull from the earth, the moon starts to move in a straight line tangent to its circular path.

Q.19. 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 same time to reach the surface of earth. Justify your answer.
Ans: 

• The value of ‘g’ – acceleration due to gravity is constant, but depending upon the surface of the earth it varies from place to place as earth is not completely spherical. 
• As it is flattened at the poles, the value of ‘g’ is maximum at the poles and the bulging at the equator causes the ‘g’ value to be minimum at the equator. 
• The ‘g’ value increases as we move towards the poles. 
• Hence, the packets falls gradually at the equator in comparison to the poles. Thereby, the packets stay in air for longer when dropped at the equator.

Q.20. The weight of any person on the moon is about 1/6 times 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?
Ans:
Weight of person on moon = 1/6^th of weight on earth 
Therefore, ‘g’ on moon = 1/6th ‘g’ on earth 
The force that is applied by the man to lift mass ‘m’ is 
F = mg = 15g    (on earth) 
If he can lift a certain mass ‘m’ by applying the same force on moon, then 
F = 15*6 = 90kg 

This proves acceleration due to gravity on the moon is 1/6^th of acceleration due to gravity on earth. Hence the person can lift a mass 6 times heavier on moon than on earth. 

Q.21. Calculate the average density of the earth in terms of g, G and R.
Ans:
Acceleration due to gravity, g = GM /R², where R is the radius of the Earth.
Mass of the Earth, M = gR²/G
Density of earth D= mass/volume
= gR²/G x Ve
= gR²/G x 4πR³/3
= 3g/4πGR

Q.22. The earth is acted upon by gravitation of Sun, even though it does not fall into the Sun. Why?
Ans: Sun provides enough centripetal force to keep the earth in its orbit and the earth provides centrifugal force due to its motion. These two forces balance each other which prevents the earth from falling into the sun.

Long Answer Questions

Q.23. How does the weight of an object vary with respect to mass and radius of the earth. In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the earth be affected?
Ans: Let R and M be the radius and mass of the earth
Then, the weight of an object is directly proportional to the mass of the earth and inversely proportional to the square of the radius of the earth. i.e. 

Weight of the object ∝ M/R²
Original weight W₀ = mg = mG M/R²
Hypothetically M becomes 4M and R becomes R/2
Then, weight = mG 4M /(R/2) ²
= (16m G) M/R²
= 10xW₀
Weight will be 16 times.

Q.24. How does the force of attraction between the two bodies depend upon their masses and distance between them? A student thought that two bricks tied together would fall faster than a single one under the action of gravity. Do you agree with his hypothesis or not? Comment.
Ans:

• The hypothesis is incorrect. Force of attraction between two masses separated by distance r is given by Newton law of gravitation where F= Gm₁m₂/R² Where G is gravitational force and it is an universal constant.
• Gravitational force is directly proportional to the product of the masses of two bodies and inversely proportional to the square of the distance between them.
• Bodies thrown from same height fall with same speed irrespective of their mass. This is due to acceleration due to gravity.g = GM/R² Where M is mass R is radius of the earth
• This equation shows that acceleration due to gravity depends on the mass of the earth and radius of the earth. Hence, two brick tied will not fall faster than the single one.

Q.25. Two objects of masses m₁ and m₂ having the same size are dropped simultaneously from heights h₁ and h₂ respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if (i) one of the objects is hollow and the other one is solid and (ii) both of them are hollow, size remaining the same in each case. Give reason.
Ans: 

We know that

v=u+at

Here v=0 and a=g

so 0=u+gt

u=−gt --- (i)

we also know

v²- u²=2as   ---(ii)

Here v=0,u=−gt from eqn (i) and S=h₁ and taking t=t₁

putting all above values in eq (ii)

g²t₁²=2gh₁

or

h₁=1/2gt₁²  ----(iii)

Similarly for S=h₂ and t=t₂

h₂=1/2gt₂²    ----(iv)

Dividing eqn (iii) & (iv) we get,

t₁/t₂=√h₁/h₂

The ratio will not change in either case because acceleration remains the same. In the case of free-fall acceleration does not depend upon mass and size of the body.

26. (a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case will it experience a greater buoyant force. If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water. Give reason for each case.
(b) A ball weighing 4 kg of density 4000 kg m^–3 is completely immersed in water of density 103 kg m^–3 Find the force of buoyancy on it. (Given g = 10 m s^–2)
Ans: (i) Buoyant force, F = Vpg
p = Density of water, V = Volume of water displaced by the body 
Volume and density of an object decides its Buoyancy. Cube will experience a greater buoyancy in a saturated solution. If the cube is reduced to 4 cm on each side, the volume of cube becomes less as the buoyancy will be reduced as buoyant force is directly proportional to volume.
(ii) The magnitude of the buoyant force given by F = Vpg
where V = Volume of body immersed in water or volume of water displaced, p = Density of liquid.
[∴ Given, mass of a ball = 4 kg, density = 4000 kgm^-3]. 
Hence Volume of solid = mass/density