## MP Board Class 9th Science Solutions Chapter 10 Gravitation

### Gravitation Intext Questions

**Gravitation Intext Questions Page No. 134**

Question 1.

State the universal law of gravitation.

Answer:

Suppose there are two objects having mass M and m respectively.

The distance between their centres is equal to d.

The force of attraction is F.

Thus, F ∝ M . m … (i)

and, F ∝ 1/d^{2} … (ii)

Joining equation (i) and (ii)

we get F ∝ M . m/d^{2}

⇒ F = G . M . m/d^{2} … (iii)

where, G is the proportionality constant and called Universal Gravitation Constant.

The expression (iii) is called expression for Universal Law of Gravitation.

The universal law of gravitation is represented as:

\(F=\frac{G m_{1} m_{2}}{r^{2}}\)

Where, G is the universal gravitation constant given by:

G = 6.67 × 10^{-11} Nm^{2} kg^{-2}.

Question 2.

Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the earth.

Answer:

Let M_{e} be the mass of the Earth and m be the mass of an object on its surface.

And say R is the radius of the Earth,

then according to the universal law of gravitation,

the gravitational force (F) acting between the Earth and the object is given by the relation:

\(F=\frac{G m_{1} m_{2}}{r^{2}}\)

F = GM_{e}m/R

**Gravitation Intext Questions Page No. 136**

Question 1.

What do you mean by free fall?

Answer:

Its a phenomenon of gravity. When an object falls from any height under the influence of gravitational force only, it is said to have a free fall. In the case of free fall, no change in direction takes place but the magnitude of velocity changes because of acceleration.

Question 2.

What do you mean by acceleration due to gravity?

Answer:

Change in velocity due to variation in height produces acceleration which is due to gravity in the object and is known as acceleration due to gravity denoted by letter g. The value of acceleration due to gravity is g = 9.8 m/s^{2}.

**Gravitation Intext Questions Page No. 138**

Question 1.

What are the differences between the mass of an object and its weight?

Answer:

Mass | Weight |

Mass is a measurement of the amount of matter something has. | Weight is the measurement of the pull of gravity on an object. |

Mass is a constant quantity. | Weight is not a constant quantity. It is different at different places. |

It is a scalar quantity. | It is a vector quantity. |

Its SI unit is kilogram (kg). | Its SI unit is the same as the SI unit of force, i.e., Newton (N). |

Question 2.

Why is the weight of an object on the moon \(\frac { 1 }{ 6 }\)th its weight on the earth?

Answer:

The mass of moon is \(\frac { 1 }{ 100 }\) times and its radius \(\frac { 1 }{ 4 }\) times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is \(\frac { 1 }{ 6 }\)th of its weight on the earth.

**Gravitation Intext Questions Page No. 141**

Question 1.

Why is it difficult to hold a school bag having a strap made of a thin and strong string?

Answer:

It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.

Question 2.

What do you mean by buoyancy?

Answer:

The upward force exerted by a liquid on an object that is immersed in it is known as buoyancy.

Question 3.

Why does an object float or sink when placed on the surface of water?

Answer:

- An object sinks in water if its density is greater than that of water.
- An object floats in water if its density is less than that of water.

**Gravitation Intext Questions Page No. 142**

Question 1.

You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

Answer:

When we weigh our body, an upward force acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value.

Question 2.

You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?

Answer:

The cotton bag is heavier than the iron bar. The cotton bag experiences larger up – thrust of air than the iron bar. So, the weighing machine indicates a smaller mass for cotton bag than its actual mass.

### Gravitation NCERT Textbook Exercises

Question 1.

How does the force of gravitation between two objects change when the distance between them is reduced to half?

Answer:

According to Universal Law of gravitation, the gravitational force of attraction between any two objects of mass M and m is proportional to the product of their masses and inversely proportional to the square of distance r between them. So, force F is given by

F = G\(\frac { M\times m }{ { r }^{ 2 } } \)

Now, when the distance ‘r’ is reduced to half then force between two masses becomes

F’ = G\(\frac { M\times m }{ { (\frac { r }{ 2 } ) }^{ 2 } } \)

Or

F’ = 4F

Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.

Question 2.

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

Answer:

All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

Question 3.

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10^{24} kg and radius of the earth is 6.4 × 10^{6 }m).

Answer:

Given that,

Mass of the body, m = 1 kg

Mass of the Earth, M = 6 × 10^{24} kg

Radius of the Earth, R = 6.4 × 10^{6} m

Now, magnitude of the gravitational force (F) between the Earth and the body can be given as,

F = G\(\frac { M\times m }{ { r }^{ 2 } } \) = \(\frac { 6.67 × 10 × 6 × 10 × 1 }{ (6.4 × 6.4) }\)

= \(\frac { 6.67 × 6 × 10 }{ (6.4 × 6.4) }\) = 9.8N (approx.)

Question 4.

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?

Answer:

According to the Universal Law of Gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the Moon with an equal force with which the Moon attracts the Earth.

Question 5.

If the Moon attracts the Earth, why does the Earth not move towards the Moon?

Answer:

The Earth and the Moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the Moon. Hence, it accelerates at a rate lesser than the acceleration rate of the Moon towards the Earth. For this reason, the Earth does not move towards the Moon.

Question 6.

What happens to the force between two objects, if

(i) the mass of one object is doubled?

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

(iii) the masses of both objects are doubled?

Answer:

(i) From Universal Law of Gravitation, force exerted on an object of mass at by Earth is given by

F = G\(\frac { M\times m }{ { R }^{ 2 } } \) ….1

When nws of the object say ne is doubled than

F’ = G\(\frac { M\times 2m }{ { R }^{ 2 } } \) = 2F

So as the mass of any one of the object is doubled the force is also doubled,

(ii) The force F is inversely proportional to the distance between the objects. So if the distance between two objects es doubled, then the gravitational force of attraction between them is reduced to one fourth of its original value, Similarly, if the distance between two objects is tripled. then the gravitational force of attraction becomes one ninth of its original value.

(iii) Again from Universal Law of Attraction, from equation 1, force ‘F’ is directly proportional to the product of both the masses, So, if both the masses are doubled then, the gravitational force of attraction become four times the original value.

Question 7.

What in the importance of Universal Law of Gravitation?

Answer:

Universal Law of Gravitation is important because it tells us about:

- the force that is responsible for binding us to Earth.
- the motion of Moon around the Earth.
- the motion of planets around the Sun.
- the tides formed by rising and falling of water level in the ocean are due to the gravitational force exerted by both Sun and Moon on the Earth.

Question 8.

What in the acceleration of free fall?

Answer:

Acceleration of free tall is the acceleration produced when a body falls under the influence of the force of gravitation of the Earth alone. It is denoted by ‘g’ and its value on the surface of the Earth is 9.8 ms^{-2}.

Question 9.

What do we call the gravitational force between the Earth and an object?

Answer:

Gravitational force between the Earth and an object is known as the weight of the object.

Question 10.

Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator}.

Answer:

Weight of a body on the Earth is given by:

W = mg

Where,

m = Mass of the body

g = Acceleration due to gravity

The value of g is greater at poles than at the equator.

Therefore, gold at the equator weighs less than at the poles.

Hence, Amit’s friend will not agree with the weight of the gold bought.

Question 11.

Why does a sheet of paper fell slower than one that is crumpled into a ball?

Answer:

When a sheet of paper is crumpled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper.

Question 12.

Gravitational force on the surface of the Moon is only \(\frac { 1 }{ 6 }\) as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth?

Answer:

Weight of an object on the Moon = \(\frac { 1 }{ 6 }\) × Weight of an object on the Earth.

Also,

Weight = Mass × Acceleration

Acceleration due to gravity, g = 9.8 m/s^{2}

Therefore, weight of a 10 kg object on the earth = 10 × 9.8 = 98 N

And, weight of the same object on the Moon = 1.6 × 9.8 = 16.3 N.

Question 13.

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

(i) the maximum height to which it rises.

(ii) the total time it takes to return to the surface of the earth.

Answer:

According to the equation of motion under gravity:

-u^{2 }= 2 gs

Where,

u = Initial velocity of the ball

v = Final velocity of the ball

s = Height achieved by the ball

g = Acceleration due to gravity

At maximum height, final velocity of the ball is zero, i.e., v = 0, u = 49 m/s.

During upward motion, g = – 9.8 ms^{-2}.

(i) Let ‘h’ be the maximum height attained by the ball.

Hence,

0 – 49^{2} = 2 × 9.8 × h

h = \(\frac { 49 × 49 }{ 2 × 9.8}\) = 122.5 m

(ii) Let ‘t’ be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:

v = u + gt

We get,

= 49 + t × (- 9.8)

9.8 t = 49

t = \(\frac { 49 }{ 9.8}\) = 5 s

But,

Time of ascent = Time of descent

Therefore, total time taken by the ball to return = 5 + 5 = 10 s

Question 14.

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

Answer:

According to the equation of motion under gravity: v^{2 }– u^{2} = 2 gs

Where,

u = Initial velocity of the stone = 0

v = Final velocity of the stone

s = Height of the stone = 19.6 m

g = Acceleration due to gravity = 9.8 ms^{-2}

∴ v^{2} – 0^{2} = 2 × 9.8 × 19.6

v^{2} = 2 × 9.8 × 19.6 = (19.6)^{2}

v = 19.6 ms^{-1}

Hence, the velocity of the stone just before touching the ground is 19.6 m s^{-1}.

Question 15.

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s^{2}, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Answer:

According to the equation of motion under gravity:

v^{2} – u^{2} = 2 gs

Where,

u = Initial velocity of the stone = 40 m/s

v = Final velocity of the stone = 0

s = Height of the stone

g = Acceleration due to gravity = -10 ms^{-2}

Let h be the maximum height attained by the stone.

Therefore,

0 – (40)^{2} = 2 × h × (-10)

h = \(\frac { 40 × 40 }{ 20 }\) = 80 m

Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m

Net displacement of the stone during its upward and downward journey

= 80 + (-80) = 0.

Question 16.

Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10^{24} kg and of the Sun = 2 × 10^{30 }kg. The average distance between the two is 1.5 × 10^{11 }m.

Answer:

According to question,

M_{Sun} = Mass of the Sun = 2 × 10^{30} kg

M_{Earth }= Mass of the Earth = 6 × 10^{24} kg

R = Average distance between the Earth and the Sun = 1.5 × 10^{11 }m.

From Universal Law of Gravitation,

F = G\(\frac { M\times m }{ { R }^{ 2 } } \)

Therefore, putting all the values given in question in above equation we get

F = 6.67 × 10^{-11} \(\frac { (6\times { 10 }^{ 24 })\times (2\times { 10 }^{ 30 }) }{ { (1.5\times { 10 }^{ 11 }) }^{ 2 } } \) = 3.56 × 10^{22} N.

Question 17.

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.

Answer:

Let be the point at which two stones meet and let ‘h’ be their height from the ground. It is given in the question that height of the tower is H = 100 m

Now, first consider the stone which falls from the top of the tower.

So, distance covered by this stone at time ‘t’ can be calculated using equation of motion:

x – x_{0} = u_{0}t + \(\frac { 1 }{ 2 }\)gt^{2}

Since, initial velocity u = 0,

so we get

100 = x \(\frac { 1 }{ 2 }\)gt^{2 }………… (1)

The distance covered by the same stone that is thrown in upward direction from ground is

x = 25t –\(\frac { 1 }{ 2 }\)gt^{2}

In this case intitial velocity is 25 m/s.

So, x = 25t – \(\frac { 1 }{ 2 }\)gt^{2 }………… (2)

Adding equations (1) and (2) we get,

100=25t

or,

t = 4s

Putting value in equation (2).

x = 25 × 4 – \(\frac { 1 }{ 2 }\) × 9.8 × (4)^{2
}= 100 – 78.4

= 21.6 m.

Question 18.

A ball thrown up vertically returns to the thrower after 6 s. Find:

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

(c) its position after 4 s.

Answer:

(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey. Hence, it has taken 3 s to attain the maximum height. Final velocity of the ball at the maximum height, v = 0 Acceleration due to gravity, g = -9.8 m s^{-2}

Equation of motion, v = u + gt

will give,

0 = u + (-9.8 × 3)

u = 9.8 × 3

= 29.4 ms^{-1}

Hence, the ball was thrown upwards with a velocity of 29.4 ms^{-1}.

(b) Let the maximum height attained by the ball be ‘h’

Initial velocity during the upward journey, u = 29.4 ms^{-1}

Final velocity, v = 0

Acceleration due to gravity, g = -9.8 ms^{-2}

From the equation of motion, s = ut + \(\frac { 1 }{ 2 }\) at^{2}

h = 29.4 × 3 + \(\frac { 1 }{ 2 }\) × – 9.8 × (3)^{2} = 44.1 m.

(c) Ball attains the maximum height after 3 s.

After attaining this height, it will start falling downwards.

In this case, Initial velocity, u = 0

Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in

4 s – 3 s = 1 s.

Equation of motion, s = ut + \(\frac { 1 }{ 2 }\) g^{2}

will give,

s = 0 x t + \(\frac { 1 }{ 2 }\) x 9.8 x 12 = 4.9 m

Total height = 44.1 m.

This means that the ball is 39.2 m (44.1 m – 4.9 m) above the ground after 4 seconds.

Question 19.

In what direction does the buoyant force on an object immersed in a liquid act?

Answer:

An object immersed in a liquid experiences buoyant force in the upward direction.

Question 20.

Why does a block of plastic released under water come up to the surface of water?

Answer:

For an object immersed in water two forces act on it:

- Gravitational force which tends to pull object in downward direction
- Buoyant force that pushes the object in upward direction.

Here, in this case buoyant force is greater than the gravitational pull on the plastic block. This is the reason the plastic block comes up to the surface of the water as soon as it is released under water.

Question 21.

The volume of 50 g of a substance is 20 cm^{3}. If the density of water is 1 g cm^{-3}, will the substance float or sink?

Answer:

If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.

= \(\frac { 50 }{ 20 }\)

= 2.5 g cm^{-3}

The density of the substance is more than the density of water (1 g cm^{-3}).

Hence, the substance will sink in water.

Question 22.

The volume of a 500 g sealed packet is 350 cm^{3}. Will the packet float or sink in water if the density of water is 1 g cm^{-3}? What will be the mass of the water displaced by this packet?

Answer:

Density of the 500 g sealed packet

= \(\frac { 500 }{ 350 }\)

= 1.428 g cm^{-3}

The density of the substance is more than the density of water (1 g cm^{-3}). Hence, it will sink in water.

The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 g.

### Gravitation Additional Questions

**Gravitation Multiple Choice Questions**

Question 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.

Answer:

(c) experience forces of same magnitude.

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 equator.

(d) increases from pole to equator.

Answer:

(c) is least on equator.

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:

(a) \(\frac { F }{ 4 }\)

Question 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.

Answer:

(c) will move along a straight line tangential to the circular path.

Question 5.

An object is put one by one in three liquids having different densities. The object floats with 1 2 3, and 9 11 7 parts of their volumes outside the liquid surface in liquids of densities d_{1}, d_{2} and d_{3} respectively. Which of the following statement is correct?

(a) d_{1 }> d_{2 }> d_{3}

(b) d_{21 }> d_{2 }< d_{3}

(c) d_{1 }< d_{2 }> d_{3}

(d) d_{1 }< d_{2 }< d_{3}.

Answer:

(d) d_{1 }< d_{2 }< d_{3}.

Question 6.

In the relation F = GM m/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) is universal constant of nature.

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) any two bodies having some mass.

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 Earth.

(d) is independent of mass and radius of the Earth.

Answer:

(d) is independent of mass and radius of the Earth.

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) 14 times

(b) 4 times

(c) 12 times

(d) unchanged.

Answer:

(b) 4 times

Question 10.

The atmosphere is held to the earth by __________ .

(a) gravity

(b) wind

(c) clouds

(d) Earth’s magnetic field.

Answer:

(a) gravity

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) universal gravitational constant.

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) zero.

**Gravitation Very Short Answer Type Questions**

Question 1.

Why Moon revolves around the Earth?

Answer:

Gravitational force of Earth.

Question 2.

What is the SI unit of gravitational force?

Answer:

Newton (N).

Question 3.

Which law of physics is represented by the statement every object attract other object in universe towards itself’?

Answer:

Universal Law of Gravitation.

Question 4.

State the relation between gravitational force and distance among objects.

Answer:

Inversely proportional.

Question 5.

In which conditions free fall of an object occur?

Answer:

When an object falls from a height under the influence of gravity and no other force, it is said to have a free fall.

Question 6.

What is the SI unit of gravitational constant?

Answer:

Nm^{2}kg^{-2}.

Question 7.

Express the relation between thrust and pressure.

Answer:

Pressure = thrust / area.

Question 8.

What kind of force is exerted by a liquid?

Answer:

Equal and unidirectional.

Question 9.

In what condition an object sinks?

Answer:

If the weight of the object is more than 9.8 N, then the object will sink.

Question 10.

Which material is taken as standard to calculate any object’s relative density?

Answer:

Water.

**Gravitation Short Answer Type Questions**

Question 1.

What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?

Answer:

Gravitational force. This force depends on the product of the masses of the planet and Sun, and the distance between them.

Question 2.

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:

Both the stones will take the same time to reach the ground because the two stones fall from the same height.

Question 3.

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 will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of Moon is due to centripetal force provided by the gravitational force of Earth.

Question 4.

Two 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 ‘g’ at the equator of the Earth is less than that at poles. Therefore, the packet falls slowly at equator in comparison to the poles. Thus, the packet will remain in air for longer time interval, when it is dropped at the equator.

Question 5.

The weight of any person on the Moon is about \(\frac { 1 }{ 6 }\) 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:

The value of ‘g’ at the equator of the Earth is less than that at poles. Therefore, the packet falls slowly at equator in comparison to the poles. Thus, the packet will remain in air for longer time interval, when it is dropped at the equator.

**Gravitation Long Answer Type Questions**

Question 1.

State ‘Archimedes’ Principle and write its two applications.

Answer:

‘Archimedes’ Principle:

Force exerted by liquid on wholly or partly immersed object is equal to the weight of the fluid displaced by the object. Applications based on ‘Archimedes’ principle are:

- designing of water transport vehicles.
- hydrometers used for determining the density of liquids

Question 2.

What is relative density? What is the density of water?

Answer:

Relative Density (RD) or Specific Gravity (SG) is the ratio of either densities or weights. Hence, when we compare or divide value of an objects’ density with water’s density, it is called Relative Density of a substance to water.

- In SI units, the density of water is (approximately) 1000 kg/m
^{3}Or 1 g/cm^{3}.

Question 3.

Mass of a rectangular copper solid piece is 300 g. With dimensions 5 × 2 × 5 cm^{3}, what should be its specific gravity, calculate? Will the bar float or sink in water?

Answer:

Given:

Mass of copper = 300 g

5 × 2 × 5 = 50 cm^{3}

Density of copper = mass / volume

\(\frac { 300 }{ 50 }\) = 6 g / cm^{3}

Density of water, = 1 g/cm^{3}

Specific gravity of iron = \(\frac { 6 }{ 1 }\) = 6.

Hence the bar will sink.

Question 4.

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?

Answer:

We know, 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 a body ∝ \(\frac { M }{ { R }^{ 2 } } \)

Original weight, W_{0} = mg = mG\(\frac { M }{ { R }^{ 2 } } \)

When hypothetically M becomes 4 M and R becomes \(\frac { R }{ 2 }\) then weight becomes

W_{0} = mG \(\frac { 4M }{ { (\frac { R }{ 2 } ) }^{ 2 } } \) = (16 m G) M

R^{2} = 16 × W_{0}

The weight will be 16 times heavier.

Question 5.

(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 10^{3} kg m^{-3}. Find the force of buoyancy on it. (Given: g = 10 ms^{-2}.)

Answer:

(a)

- The cube will experience a greater buoyant force in the saturated salt solution because the density of the salt solution is greater than that of water.
- The smaller cube will experience lesser buoyant force as its volume is lesser than the initial cube.

(b) Buoyant force = weight of the liquid displaced = density of water x volume of water displaced xg 4

= 1000 × \(\frac { 4 }{ 4000 }\) × 10 = 10N.

4000

**Gravitation Higher Order Thinking Skills (HOTS)**

Question 1.

How will the weight of a body of mass 250 g of changes, if it is taken from equator to the poles? Give reasons.

Answer:

As we move from equator to poles, acceleration due to gravity increases. It is because radius of earth is less at poles than at equator. Therefore, its weight will increase.

Question 2.

Aman tried to immerse an empty plastic bottle in a bucket of water. But each time he fails. Why does this happen?

Answer:

When Aman tried to immerse an empty plastic bottle in a bucket of water, it comes above the surface of water. It is due to the upward force (upthrust or buoyant force). The upthrust exerted by water on the bottle is greater than its own weight.

**Gravitation Value Based Question**

Question 1.

Rashmi was wearing a high heel shoes for a beach party. Her friend told her to wear flat shoes as she will be tired soon with high heel and will not feel comfortable.

- What is the reason of one’s feeling tired with high heel shoes on a beach?
- Name the unit of pressure.
- What value of Rashmi’s friend is reflected in the above act?

Answer:

- Because the high heel shoes would exert lot of pressure on the loose sand of beach and will sink more in the soil as compared to flat shoes. Therefore, large amount of force will be required to walk with high heels.
- Pascal.
- Rashmi’s friend showed the value of being intelligent, concerned and helpful.