# Gravitation : NCERT Exercise Questions

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**Q.1 Â Â Â How does the force of gravitation between two objects change when the distance between them is reduced to half
**

*Â Â According to the law of gravitation , the force of attraction between any two objects of mass m*

**Sol.****Â Â**_{1}and m

_{2}is proportional to the product of their masses and inversely proportional to the square of the distance 'R' between them.

Â Â Â Â Â Â As given hereÂ

Â Â Â Â Â Â Here G is the gravitational constant. When the distance (R) is reduced to half .

Â Â Â Â Â Â ThenÂ

Â Â Â Â Â Â OrÂ

Â Â Â Â Â Â Clearly, as the distance between the objects is reduced to half the force of gravitation becomes four times the original force.

**Q.2 Â Â Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object ?
**

*Weight of an object on surface of earth = mg*

**Sol. Â Â Â**Â Â Â Â Â Â where 'm' is mass and 'g' is acceleration due to gravity .

Â Â Â Â Â Â Gravitational force acting on the object F =

Â Â Â Â Â Â Here M is mass of earth, R is radius of earth i.e. distance between the objects and G is gravitational constant

Â Â Â Â Â Â Weight of object 'mg' = Â Gravitational force F acting on it

Â Â Â Â Â Â mg =Â

Â Â Â Â Â Â g =Â

Â Â Â Â Â Â From above expression it is clear , 'g' acceleration due to gravity is independent of mass of an object and all the objects- irrespective of being heavy or light experience the same acceleration due to gravity, hence fall with same speed from a given height.Â

**Q.3 Â Â What is magnitude of gravitational force between the earth and a 1 kg object on its surface ? Take mass of earth to be 6 Ã— 10 ^{24}Â kg and radius of the earth is 6.4 Ã— 10^{6Â }m. G = 6.67 Ã—10^{â€“11}Â Nm^{2}Â kg^{â€“2}.**

*Â Â Â As given in the statement :*

**Sol.**Â Â Â Â Â Â Gravitational Constant G = 6.67 Ã— 10â€“11Â N m2Â kg

^{â€“2}Â

Â Â Â Â Â Â Mass of the object m

_{1}Â = 1kgÂ

Â Â Â Â Â Â Mass of the Earth m

_{2}= 6 Ã— 1024Â kg ;

Â Â Â Â Â Â Radius of the earth R = 6.4 Ã—106Â m

Â Â Â Â Â Â As we know F =Â

Â Â Â Â Â Â

Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â or F = 9.8 N Approximately

**Q.4. Â Â The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth ? Why ?
**

**Â Â Â Â The universal law of gravitation states that the force of attraction between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them.**

*Sol.*Â Â Â Â Â Â As given here

Â Â Â Â Â Â Where M

_{1}Â = Mass of earth, M

_{2}Â = Mass of moon, R = Distance between earth and moon.Â Â G is the gravitational constant.Â The above law applies to all objects anywhere in the universe. This is also true in case of force of attraction between Earth and moon. The magnitude of force (F) of attraction exerted by Earth on the moon, due to gravitation, is the same as that exerted by moon (F) on earth.These forces being equal and opposite, also in accordance with Newton's Third Law of Motion, which states that to every action there is an equal and opposite reaction.Â

**Q.5 Â Â If the moon attracts the earth, why does the earth not move towards the moon ?**

* Sol.*Â Â Â As per Newtonâ€™s third law of motion, the mutual forces of attraction due to gravitation, between earth and moon is same. The mass of the earth is 5.97 x 1024Â kg and that of moon is 7.349 x 1022Â kg. Clearly the mass of earth is 82.23 times that of moon.Â Let FeÂ be the force of gravitational pull of earth on moon; FmÂ be the force of gravitational pull of moon on earth; meÂ mass of earth ;mmÂ mass of moon ; geÂ be the acceleration caused by earth on moon and gmÂ be the acceleration caused by moon on earth .

Â Â Â Â Â

Â Â Â Â Â

Â Â Â Â Â As per, Newton's third law of motion,

Â Â Â Â Â

Â Â Â Â Â

Â Â Â Â Â As given,Â = 83.23 Ã—Â

Â Â Â Â Â

Â Â Â Â Â Therefore,Â

Â Â Â Â Â OrÂ

Â Â Â Â Â Here clearly acceleration experienced by the earth due to gravitational pull of moon is very small, just .012 times ( 1.2 % ) of that experienced by the moon due to earth , which is not as effective to move the earth towards the moon.

**Q.6 Â Â What happens to the force between two objects, if
**

**Â Â Â Â Â Â (i) The mass of one object is doubled ?**

**Â Â Â Â Â Â (ii) The distance between the objects is doubled and tripled ?**

**Â Â Â Â Â Â (iii) The masses of both objects are doubled ?**

**Sol. Â Â***According to the law of gravitation , the force of attraction between any two objects of mass m*

**Â**_{1}and m

_{2}is proportional to the product of their masses and inversely proportional to the square of the distance 'R' between them.

Â Â Â Â Â Â i.e

Â Â Â Â Â Â Here G is the gravitational constant. When the distance (R) is reduced to half .

Â Â Â Â Â Â (i) When the mass of one object say m1 is doubled, then

Â Â Â Â Â Â i.e.Â

Â Â Â Â Â Â i.e.Â = 2F

Â Â Â Â Â Â âˆ´ As the mass of one object is doubled the force becomes 2 times.Â

Â Â Â Â Â Â (ii) When the distance between the bodies is doubled and tripled

Â Â Â Â Â Â when the distance is doubled

Â Â Â Â Â Â i.e.Â

Â Â Â Â Â Â The force is reduced to one fourth of the original force.Â when the distance is tripled :

Â Â Â Â Â Â i.e.Â

Â Â Â Â Â Â The force is reduced to one ninth of the original force.

Â Â Â Â Â Â (iii) When the masses of both the objects are doubled, thenÂ

Â Â Â Â Â Â

Â Â Â Â Â Â âˆ´ When the masses of both the objects are doubled, then the force becomes four times the original force.

**Q.7 Â Â What is importance of universal laws of gravitation ?**

* Sol.*Â Â Â The Importance of Universal law of gravitation lies in the fact, that it was successful in explaining many phenomena such as.

Â Â Â Â Â Â (i) That, how does the different objectsÂ in this universe,Â affectÂ others.

Â Â Â Â Â Â (ii) That, howÂ gravity; the force of gravitation due to earth, isÂ responsible for the weight of a body and keeps us on the ground. And that why force of gravity decreases with altitude.Â

Â Â Â Â Â Â (iii) That, how does the lunar motion around the earth occur.

Â Â Â Â Â Â (v) That, how does the planetary motion of planets in our solar system as well as that of all other celestial objects take place

Â Â Â Â Â Â (vi) That how do the tidal waves originate, due to the gravitational pull of moon and the sun.

**Q.8 Â Â What is the acceleration of free fall ?
**

*Â Â Â Acceleration of free fall is the acceleration experienced by body falling freely towards earth under the influence of gravitation force of earth alone. It is denoted by g and its value on the surface of earth is 9.8 m s*

**Sol.**^{â€“2}.

**Q.9 Â Â What do we call the gravitational force between an earth and an object ?
**

*The gravitational force between an earth and an object Weight and it is equal to product of mass(m) and acceleration due to gravity(g).*

**Sol. Â Â Â****Q.10 Â Â Amit buys a few gram (force) of gold at the poles as per instruction of one of his friends. He hands over the same when he meets him at equator. Will the friend agree with the weight of gold bought ? If not, why ?*** Sol.*Â Â Â Â The force of gravity decreases with altitude. It also varies on the surface of the earth, decreasing from poles to the equator.Â Also, The weight is equal to the product of mass and acceleration due to gravity.Â

Â Â Â Â Â Â Â i.e. W = mg,

Â Â Â Â Â Â Â where â€˜mâ€™ is mass of the object â€˜gâ€™ is the acceleration due to gravity.Â As value of g is higher at polar region than the equator, the weight of an object will also be more at polar region than at equtor. Therefore, his friend will not agree with weight of the gold bought at the poles when measured at equator.

**Q.11 Â Â Why will a sheet of paper fall slower than one that is crumpled into a ball ?**

* Sol.*Â Â Â Â Surface area of a Sheet which is crumpled into a ball, is much smaller than the surface area of a plain or flat sheet. Therefore, despite both experince same force of gravity, the plain or flat sheet of paper will have to face more air resistance than the crumpled ball, so it will fall slower than the sheet crumpled into a ball.

**Q.12 Â Â Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newton of a 10 kg object on the moon and on the earth ?
**

*Â Â Â Â The mass of an object always being same.*

**Sol.**Â Â Â Â Â Â Â Therefore,The mass of the object on the moon will be the same, as that on earth, m = 10 kg.Â

Â Â Â Â Â Â Â We know that, Acceleration due to gravity on earth (ge) = 9.81 ms

^{â€“2}

Â Â Â Â Â Â Â Acc. due to gravity on moon (gm) =Â

Â Â Â Â Â Â Â Therefore, Weight of the object on the surface of moon = mgm

Â Â Â Â Â Â Â = 10 Ã—Â Â and Weight of the object on the surface of Earth = mge

Â Â Â Â Â Â Â = 10 Ã— 9.81 = 98.1 N

**Q.13 Â Â A ball is thrown vertically upwards with a velocity of 49 ms ^{â€“1}. Calculate
Â Â Â Â Â Â Â (i) the maximum height to which it rises.
Â Â Â Â Â Â Â (ii) the total time it takes to return to the surface of the earth.**

*Â Â Â Â (i)Â Â As per the statement given*

**Sol.**Â Â Â Â Â Â Â Â Â Initial velocity of the ball (u) = 49 ms

^{âˆ’1}Â

Â Â Â Â Â Â Â Final velocity of the ballÂ (v) = 0 ms

^{âˆ’1}Â

Â Â Â Â Â Â Â Downward gravity (Â gÂ ) =Â 9.8 ms

^{âˆ’2 }Â Â Â Â Â Â Â Upward gravityÂ Â (g) =Â âˆ’Â 9.8 ms

^{âˆ’2 }Â Â Â Â Â Â Â Max. Height attained by the ballÂ Â (sÂ ) = ?

Â Â Â Â Â Â Â As we know :

Â Â Â Â Â Â Â v

^{2}Â âˆ’Â u

^{2}Â =Â 2gsÂ =Â >Â (0)

^{2}Â âˆ’Â (Â 49Â )

^{2}Â =Â 2Â Ã—Â (âˆ’Â 9.8Â )Â Ã—Â sÂ =Â > sÂ =Â âˆ’Â 49Â Ã—Â 49Â âˆ’Â 2Â Ã—Â 9.8

Â Â Â Â Â Â Â Max. Height attained by the ball (s) =Â 122.5 m

Â Â Â Â Â Â Â (ii) Also, as we know :

Â Â Â Â Â Â Â v =Â uÂ Â +Â Â gtÂ =Â >Â 0Â Â =Â 49Â âˆ’Â 9.8Â Ã—Â tÂ =Â > t = 49Â 9.8Â =5 s

Â Â Â Â Â Â Â Time for upward journey of the ball will be the sameÂ as Â time for downward journey i.e.

tÂ =Â Â 5 s.

Â Â Â Â Â Â Â Total time taken by the ball to returnÂ to the surface of earthÂ Â =Â 2Â Ã—Â tÂ =Â 2Â Ã—Â 5Â =Â 10 s

Â **Q.14 Â Â A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.Â **

** Sol.Â Â Â **As per given statement, initial velocity u = 0, g = â€“ 9.8 ms

^{â€“2}, height,Â s = â€“19.6 m

Â Â Â Â Â Â As we know that u

^{2}â€“ u

^{2}= 2 gs

Â Â Â Â Â Â Therefore, u

^{2}â€“ 02 = 2 Ã— (â€“ 9.8) Ã— (â€“19.6)

Â Â Â Â Â Â or v

^{2}= (19.6)

^{2}

Â Â Â Â Â Â v = â€“19.6 ms

^{â€“1}

Â Â Â Â Â Â The negative sign indicates that the velocity is in the downward direction.

**Q.15 Â Â A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/sÂ², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone ?
**

*Â Â Here,initial velocity u = 40 msâ€“1Â and g =Â â€“ 10 ms*

**Sol. Â Â**^{â€“2}

Â Â Â Â Â Â Â At the maximum height reached S , final velocity, v = 0

Â Â Â Â Â Â Â AS we know the v

^{2}â€“ u

^{2}= 2gsÂ

Â Â Â Â Â Â Â S =Â

Â Â Â Â Â Â Â Also, the total distance covered = 2 Ã— S = 2 Ã—= 160 m

Â Â Â Â Â Â Â As the stone returns back to the original position.

Â Â Â Â Â Â Â Therefore, Net= S â€“ (S)

Â Â Â Â Â Â Â = 80 â€“ 80 = 0

**Q.16 Â Â Calculate the force of gravitation between the earth and the sun, given that the mass of the earth = 6 Ã— 10 ^{24}Â kg and of the sun = 2 Ã— 10^{30}Â kg. The average distance between the two is 1.5 Ã— 10^{11}Â m.**

**As given in the statement, M**

*Sol. Â Â Â Â*_{e}Â = 6 Ã— 10

^{24}kg, M

_{s}= 2 Ã— 10

^{30 }Â Â Â Â Â Â Â r = 1.5 Ã— 10

^{11}m

Â Â Â Â Â Â Â As we know that F = G

Â Â Â Â Â Â Â Therefore,Â {{10}^{30}}} \over {{{\left( {1.5 \times {{10}^{11}}} \right)}^2}}}N" />

Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Therefore, F = 3.56 Ã— 10

^{22}N

**Q.17 Â Â A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet ?
**

*Let the point at which, two stones meet after time*

**Sol.**Â Â Â Â Â*Â t*Â from the start, be at a height xÂ from the ground.Â Height of the tower = 100 m

Â Â Â Â Â Â Â Â Â Then distance covered by the stone allowed to fall from the top of the tower,

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â ...(1)

Â Â Â Â Â Â Â The distance covered by the stone thrown from the ground,

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â ...(2)

Â Â Â Â Â Â Â Combining eq. (1) and (2), we get

Â Â Â Â Â Â Â 100 = 25t

Â Â Â Â Â Â Â t = 4s

Â Â Â Â Â Â Â Therefore, x = 25 Ã— 4 â€“ Ã— 9.8 Ã— 4

^{2 }Â Â Â Â Â Â Â = 100 â€“ 78.4

Â Â Â Â Â Â Â = 21.6 m

**Q.18 Â Â Â A ball thrown up vertically returns to the thrower after 6s. Find:
Â Â Â Â Â Â Â (a) the velocity with which it was thrown up.
Â Â Â Â Â Â Â (b) the maximum height it reaches, and
Â Â Â Â Â Â Â (c) its position after 4s.
**Â

*Â Â Â Â (a) The velocity with which ball was thrown up :*

**Sol.**Â Â Â Â Â Â Â Acceleration due to gravity, g = â€“ 9.8 ms

^{â€“2}Â

Â Â Â Â Â Â Â As the total time taken in upward and return journey by the ball is 6 s.

Â Â Â Â Â Â Â Therefore, The upward journey, t = 6/2 s = 3 s

Â Â Â Â Â Â Â Final velocity, u = 0 ms

^{â€“1}

Â Â Â Â Â Â Â Initial velocity, u = ?

Â Â Â Â Â Â Â As we know, by the first equation of motion,

Â Â Â Â Â Â Â u = u + gt

Â Â Â Â Â Â Â Â 0 = u + (9.8) Ã— 3

Â Â Â Â Â Â Â 0 = 4 â€“ 29.4

Â Â Â Â Â Â Â u = 29.4 msâ€“

Â Â Â Â Â Â Â Therefore, The velocity with which ball was thrown up = 29.4 ms

^{â€“1}Â

Â Â Â Â Â Â Â T (b) Max. Height the ball reaches (h) = Distance (s) = ?

Â Â Â Â Â Â Â As we know, by the second equation of motion,

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Therefore, s = 29.4 Ã— 3 + (â€“ 9.8) Ã— 32

Â Â Â Â Â Â Â = s = 88.2 m Â 44.1mÂ

Â Â Â Â Â Â Â (c) The position of the bell after Â 4 seconds

Â Â Â Â Â Â Â = Distance, s = ?Â

Â Â Â Â Â Â Â Here as given Â Â Â Â Time, t = 4 s Â

Â Â Â Â Â Â Â As we know by equation from laws of motion,

Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Therefore,Â

Â Â Â Â Â Â Â s = 117.6 m â€“ 78.4 m

Â Â Â Â Â Â Â s = 39.2 m

**Q.19 Â Â In what direction does the buoyant force on an object immersed in a liquid act?
**

*The direction of Buoyant force on an object immersed in a liquid acts is vertically upward towards the centre of buoyancy.*

**Sol. Â Â Â Â****Q.20 Â Â Why does a block of plastic released under water come up to the surface of water?
**

*Â Â Â Â All objects experience a force of buoyancy when they are immersed in a fluid. Objects having density less than that of the liquid in which they are immersed, float on the surface of the liquid. If the density of the object is more than the density of the liquid in which it is immersed then it sinks in the liquid. Here the density of plastic block is less than that of water, there for, weight of water displaced by fully immersed plastic block is more than its own weight. Thus, the upward force acting on the plastic block due to buoyancy , is much more than the downward force due to the weight of the block. Due to this upward buoyant force, the block will be forced up tilll the weight of displaced liquid is equal to the weight of plastic block.*

**Sol.****Q.21 Â Â The volume of 50 g of substance is 20 cm ^{3}. If the density of water is 1 g cm^{â€“3}, will the substance float or sink ?**

*Â Â Â Â As given in the statement: The Mass of substance = 50 gÂ*

**Sol.**Â Â Â Â Â Â Â The volume of substance = 20 cm

^{3 }Â Â Â Â Â Â Â density of water is 1 g cm

^{â€“3 }Â Â Â Â Â Â Â Density of substanceÂ Â =Â Â Â Mass of substanceÂ volume of substance Â = 50Â 20Â =Â 2.5 Â gÂ Â cm

^{âˆ’3}Â

Â Â Â Â Â Â Â Clearly As the density of substance (2.5 g cm

^{âˆ’3}Â )is more than that of water (1 g cm

^{âˆ’3}), so it will sink.Â

**Q.22 Â Â The volume of a 500 g sealed packed is 350 cm ^{3}. Will the packet float or sink in water if the density of water is 1 g cm^{â€“3}Â ? What will be the mass of water displaced by this packet ?**

*Â Â Â Â As per the statement of question :*

**Sol.**Â Â Â Â Â Â Â Mass of sealed packed M = 500g

Â Â Â Â Â Â Â Volume of sealed packed V= 350 cm

^{3}

Â Â Â Â Â Â Â Density of water = 1 g cm

^{â€“3}

Â Â Â Â Â Â Â Density of sealed packetÂ Â =Â MÂ V

Â Â Â Â Â Â Â =Â 500Â 350Â =Â 107

Â Â Â Â Â Â Â =Â 1.43 gÂ Â cm

^{âˆ’3}Â As density of packet which is 1.43 g cm

^{â€“3}Â is more than that of water

Â Â Â Â Â Â Â (1 g cm

^{â€“3}), it will sink in water.

Â Â Â Â Â Â The volume of water displaced is equal to the volume of packet V i.e., 350 cmÂ³.Â

Â Â Â Â Â Â Â Â Â Â Â âˆ´ Mass of the water displacedÂ Â =Â DensityÂ Â Ã—Â Â Volume =Â 1 gÂ cmâˆ’Â 3Â Ã— 350 cm

^{3}=Â 350 gÂ

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