NCERT Exemplar Class 9 Science Solutions Chapter 10 Gravitation

Question:2

The value of acceleration due to gravity
(a) is same on equator and poles
(b) is least on poles
(c) is least on the equator
(d) increases from pole to equator

Answer:

Ans. C
By universal law of gravitation, we know that the acceleration due to gravity will be inversely proportional to the square of distance of point from the centre of earth.
$F=\frac{Gm_{1}m_{2}}{r^{2}}$
The surface near pole is closer to the centre of earth than the distance between the centre of the earth and the equatorial surface.
Therefore, the acceleration due to gravity will be minimum at equator.
Due to the rotation of earth, the acceleration of gravity further decreases at the surface of equator.
Hence, the correct answer to this question will be option C

Question: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) $\frac{F}{4}$
(b) $\frac{F}{2}$
(c) F
(d) 2 F

Answer:

Ans. A
By universal law of gravitation, we know that the force between two particles kept at a fixed separation will be proportional to product of their masses.
$F=\frac{Gm_{1}m_{2}}{r^{2}}\\ \Rightarrow F \propto \: m_{1}m_{2}$
If masses of both the blocks are halved, then the net force of gravitation between them will become one fourth of the original force.
Hence, the correct option to this question is option A.

Question:4

A boy is whirling a stone tied with a string in a 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

Answer:

Ans. C
If the boy is whirling a stone tied with a string in a horizontal circular path, the force by string is changing the direction of velocity. This force is called Centripetal force.
If the string breaks, there will be no force to change the direction of velocity.
Therefore, the particle will move on a straight line, tangent to the circle with speed it had just before the breaking of string.

Question: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

Answer:

Ans. D
If an object is floating in liquid, that means the force of buoyancy is equal to the object's weight.
F_b=mg
Weight of the object in all three liquids will be the same; hence the force of buoyancy by all three liquids will be same.
Force of buoyancy in any liquid is equal to weight of displaced fluid.
F_b=dV_g
Therefore, the weight of displaced fluid will be same in all three cases.
Weight of displaced fluid will be proportional to mass of displaced fluid.
Therefore, mass of displaced liquid will be same in all three cases.
We know that mass is equal to the product of density and volume of displaced liquid.
Therefore, density will be inversely proportional to displaced liquid volume.
In the first case, displaced volume is most, hence density will be least.
In the third case, displaced volume is least, hence density will be most.
So, the correct option for this question is option D.

Question:6

In the relation $F= \frac{GmM}{d^{2}}$, 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

Answer:

D
According to the universal law of gravitation, gravitational force is proportional to product of masses and inversely proportional to square of distance between them.
$F= \frac{Gm_{1}m_{2}}{r^{2}}$
G is the universal gravitational constant, which does not depend on any other physical quantity.
The correct answer is option D.

Question: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

Answer:

C
Gravitational force is a fundamental force, and it comes because of property of mass.
According to Newton’s gravitational law, any two bodies having mass will attract each other by gravitational force.
Gravitational force is given as:
$F= \frac{Gm_{1}m_{2}}{r^{2}}$
Hence the correct answer is option C

Question: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 the earth
(d) is independent of mass and radius of the earth

Answer:

D Gravitational force is a fundamental force, and it comes because of the property of mass.
According to Newton’s gravitational law, any two bodies having mass will attract each other by gravitational force.
According to the universal law of gravitation, gravitational force is proportional to the product of masses and inversely proportional to the square of distance between them.
Gravitational force is given as:
$F= \frac{Gm_{1}m_{2}}{r^{2}}$
G is the universal gravitational constant, which does not depend on any other physical quantity.
G is independent of mass and radius of the earth.
The correct answer of this question is option D.

Question: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) $\frac{1}{4}$times (b) 4 times (c) $\frac{1}{2}$ times (d) unchanged

Answer:

B
By universal law of gravitation, we know that the force between two particles kept at a fixed separation will be proportional to the product of their masses.
$F= \frac{Gm_{1}m_{2}}{r^{2}}\\ \Rightarrow F\: \propto \: m_{1}m_{2}$
If masses of both the blocks are doubled, then the net force of gravitation between them will become four times of the original force.
Hence, the correct answer to this question is option B.

Question:10

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

Answer:

A
The atmosphere around earth has gases in it.
These gases have mass; hence they are attracted by the gravitational force of earth.
Gravitational force is given as:
$F= \frac{Gm_{1}m_{2}}{r^{2}}$
Therefore, we can say that atmosphere is held to the earth by its gravity.
The correct answer for this question is option A.

Question: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

Answer: D

By universal law of gravitation if two particles of mass m1 and m2 are kept at separation, then the gravitational force between them will be given as:
$F= \frac{Gm_{1}m_{2}}{r^{2}}$
Here, G is the universal gravitational constant.
If both masses have unit values and their separation is also unit, then the magnitude of force will be equal to universal gravitational constant G.
$F= \frac{Gm_{1}m_{2}}{r^{2}}\\ $
if,$\;m_{1}=1;m_{2}=1;r=1;\\ F=G$
Hence, the correct answer to this question is option D.

Question: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) $\frac{1}{R^{2}}$ times the weight at surface of the earth

Answer: A

When we go towards the centre of earth from the surface, the acceleration due to gravity decreases.
Hence, the weight of any object decreases as we go towards the centre.
This decrease will be linear, and the weight of the body becomes zero at the centre of earth.
At the centre of earth, there will be no gravitational force.
Hence, the correct answer to this question is option A.

Question: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

Answer: A

The weight of an object decreases when it is immersed in any fluid. This decrease in weight is caused due to buoyancy.
In any liquid the apparent weight (decreased weight) is given as:
Apparent weight = weight in air - buoyancy force
The force of buoyancy is equal to the weight of displaced liquid.
$F_{b}=dVg$
The given question, has an apparent weight equal to 8 kg and weight in air is equal to 10 kg.
Therefore, the weight of displaced liquid will be 2 kg.
Hence, the correct answer to this question is option A.

Question: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

Answer: B

The pressure is defined as normal force per unit area.
Force applied by box will be equal to weight of the box in all orientations.
Therefore, pressure exerted by it will be inversely proportional to the area.
To maximize the value of pressure, we have to decrease the surface area.
The least surface area will be in the case when breadth and width form the base.
Correct answer is option B.

Question: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

Answer: D

Newton’s third law states that every action has an equal and opposite reaction.
If we assume gravitational attraction by earth on the Apple as action, then Apple's gravitational attraction on the earth will be reaction.
As action and reaction have equal magnitude, therefore F₁ and F₂ will have same magnitude.
Hence the correct answer to this question is option D

Question: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?

Answer:

The motion in the vertical direction, will not be affected by horizontal motion.
If a stone is dropped or thrown horizontally, its initial vertical speed is zero in both cases.
Both of them will experience same acceleration due to gravity.
Hence, they will take some time to reach the ground if they are thrown from the same height.

Question: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?

Answer:

The moon revolves around the earth.
To revolve, it requires a centripetal force which is given by earth’s gravitational force.
If the gravity of earth suddenly disappeared, the force on moon will become zero.
In the presence of zero force, the moon will continue moving in a straight line with the same speed.
Can be understood by Newton’s first law:
“In the absence of any force, the body will move on straight with constant speed, if it is not at rest.”

Question: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

Answer:

The value of gravity near the equator is less than the value of gravity near the pole.
Two packets are dropped from two airplanes, with identical conditions from pole and equator.
The time to reach the surface will be inversely proportional to the square root of acceleration which means more acceleration ensures lesser time.
Therefore, the packet near the pole will take less time.

Question: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?

Answer:

It is given that the weight of any person on the moon is about 1/6^th that on the earth.
It means, acceleration due to gravity is also 1/6th on moon in comparison with earth.
A person has to apply force equal to the weight of any body to hold it.
As the gravity becomes 1/6 on the moon, its mass must be six times for the same weight.
F = mg
Therefore, the man can lift 90 kg on the surface of moon if he can hold 15 kg on earth.

Question:21

Calculate the average density of the earth in terms of g, G and R.

Answer:

The weight of an object is nothing but the gravitational force by the earth on that body.
By universal law of gravitation, if two particles of mass m1 and m2 are kept at a separation then the gravitational force between them will be given as:
$F= \frac{Gm_{1}m_{2}}{r^{2}}$
The weight is generally given as mg, therefore:
$mg= \frac{GM_{e}m}{R_{e}^{2}}\\ \Rightarrow g= \frac{GM_{e}}{R_{e}^{2}}\\$
If density of earth is d, its mass can be calculated by using the fact: mass is product of density and volume.
$M_{e}=d \times \frac{4\pi\left ( R_{e} \right )^{3} }{3}$
By putting the value of mass:
$\begin{aligned} \Rightarrow & g=\frac{G}{R_e^2}\left(d \times \frac{4 \pi\left(R_e\right)^3}{3}\right) \\ \Rightarrow & g=\frac{G}{1}\left(d \times \frac{4 \pi\left(R_e\right)}{3}\right) \\ \Rightarrow & g=\frac{4 \pi d G\left(R_e\right)}{3} \\ \Rightarrow & d=\frac{3 g}{4 \pi G\left(R_e\right)}\end{aligned}$

Question:22

The earth is acted upon by gravitation of Sun, even though it does not fall into the Sun. Why?

Answer:

The earth is acted upon by the gravitation of sun, this gravitational force provides needed centripetal force to the earth.
Centripetal force is required to change the direction of velocity of the earth.
Because of this force the earth maintains its separation from sun; otherwise, it will move in a straight-line tangent to the current orbit.
The gravitational pull of the sun forbids earth to go away from it.

Question: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.

Answer:

By universal law of gravitation, we know that the force between two particles kept at a fixed separation will be proportional to product of their masses.
$F= \frac{Gm_{1}m_{2}}{r^{2}} \\ \Rightarrow F\: \propto \: m_{1}m_{2}\\ \Rightarrow F\: \propto \: \frac{1}{r^{2}}$
The weight of an object is nothing but the gravitational force by earth on that body.

The weight is generally given as mg, therefore,
$mg= \frac{GM_{e}m}{R_{e}^{2}}\\ \Rightarrow g=\frac{GM_{e}}{R_{e}^{2}}$
Hence, the acceleration due to gravity does not depend on mass of the object.
Whether you drop a single body or two bodies tied with the string, their acceleration under gravity will be same in free fall.
Therefore, they will take the same time to reach the ground.
Hence, the hypothesis is wrong.

Question: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.

Answer:

The weight of an object is nothing but the gravitational force by earth on that body.
The weight is generally given as mg, therefore,
$mg= \frac{GM_{e}m}{R_{e}^{2}}\\ \Rightarrow g=\frac{GM_{e}}{R_{e}^{2}}$
Hence, the acceleration due to gravity does not depend on mass of object.
If we neglect, the friction force of air, mass does not depend on size also.
Whether you drop a solid body or hollow body, their acceleration under gravity will be same in free fall.
Therefore, they will take same time to reach the ground.
The time to fall can be calculated by equation of motion.
$s=ut+\frac{1}{2}at^{2}$
$ \Rightarrow h=0t+\frac{1}{2}gt^{2}$
$\Rightarrow t=\sqrt{\frac{2h}{g}}$
For the given objects, time will be:
$\Rightarrow t_{1}=\sqrt{\frac{2h_{1}}{g}}\; \; and\; \; t_{2}=\sqrt{\frac{2h_{2}}{g}}$
$\Rightarrow \frac{t_{1}}{t_{2}}=\sqrt{\frac{h_{1}}{h_2}}\;$
The ratio will not change in either case because acceleration will remain same. In case of free-fall acceleration does not depend upon mass and size of the body

Question: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 kgm^-3 is completely immersed in water of density 1000 kgm^-3 .Find the force of buoyancy on it.

Answer:

(a) Force of buoyancy in any liquid is equal to weight of displaced fluid.
$F_{b}=dVg$
As the cube is completely immersed, displaced volume will be equal to volume of cube.
In both cases, volume and acceleration due to gravity is same.
Force of buoyancy will be only affected by density.
Density of saline solution will be more than density of water, hence force in water will be less.
If we change the side length of cube, its volume will be changed.
The force of buoyancy will be changed in same ratio as of volume.
Old volume = (5cm)³ = 125cm³
New volume = (4cm)³= 64cm³
Hence, the new force will be 64/125 times of old buoyancy force.
(b) Force of buoyancy in any liquid is equal to weight of displaced fluid.
$F_{b}=dVg$
As the density of solid is more than density of water, it will drown in it. Therefore, the displaced volume will be equal to the volume of solid.
We know that mass is product of density and volume.

$\begin{gathered}
M=d \times V \\
V=\frac{M}{d}=\frac{4}{4000} m^3
\end{gathered}$

Now, we know the displaced volume, we can calculate force of buoyancy.

$\begin{aligned}
& F_b=d V g \\
& \text { where } V=\frac{4}{4000} m^3 \\
& \Rightarrow F_b=1000 \times \frac{4}{4000} \times 10=10 \mathrm{~N}
\end{aligned}$