# NCERT Solutions Class 9 Science Chapter 10 Gravitation

## NCERT Solutions Class 9 Science Chapter 10 Gravitation

NCERT Solutions Class 9 Science Chapter 10 Gravitation: National Council of Educational Research and Training Class 9 Science Chapter 10 Solutions – Gravitation. NCERT Solutions Class 9 Science Chapter 10 PDF Download.

### NCERT Solutions Class 9 Science Chapter 10: Overview

 Board NCERT Class 9 Subject Science Chapter 10 Chapter Name Gravitation Topic Exercise Solutions

### NCERT Solutions Class 9 Science Chapter 10 – Gravitation

Part 1:

1) State the universal law of gravitation.

Newton’s gave the law of universal gravitation. This law states that,

“Every object in the universe attract other object with a force which is directly proportional to product of mass of each object and square of distance between them.”

Mathematically it expressed as,

F ∝ m1 m2 ——1

F ∝ 1/r² ——–2

Combine equation 1 and 2,

F ∝m1 m2/r²

Take , proportionality constant = G = 6.67 × 10-11 N m²/kg²

F = G m1 m2/r²

Where

m1– Mass of first object,

m2 – Mass of second object,

r- distance between them,

G – gravitational constant.

2) Write the formula to find the magnitude of the gravitationalforce between theearth and anobject on the surface of the earth.

Answer: We can calculate gravitational force exerted by earth on the object which is on surface of earth using universal law of gravitation.

According to law of gravitation,

Force = G mM/R²

Where,

M- mass of earth =

m- mass of object,

G = 67 × 10-11 N m²/kg²

Part 2:

1) What do you mean by free fall?

Answer:  If an object falls under gravitational force only then such motion is called free fall. In free fall,  we ignore all opposing forces on the objects like air resistance.

Characteristics of free fall:

• The free motion takes place under constant acceleration.
• Kinetic energy of the object continuously increases in free fall.

Example: A 8 kg stone released from top pf tower.

2) What do you mean by acceleration due to gravity?

Answer: According to second law of motion, force produces acceleration. As we know that universal law of gravitation, according to this law earth exerts force on the object which lies in the field of gravitation of earth. The acceleration produced in the object because of gravitational force of earth is called acceleration due to gravity.The value of acceleration due to gravity is constant and it’s value is 9.8 m/s².

At the pole of earth this value increases upto 10 m/s².

Part 3:

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

Answer: Differences between mass of the object and its weight are as follows.

• Mass of the object is defined as total matter contained in the object. Weight of the object is defined as gravitational force exerting on that object.
• The mass of object is constant. It does not change. Weight of the object is not constant. It depend on gravitational force. When we measure weight at pole of the earth then value of weight become large.
• SI unit of mass is Kg. Si unit of weight is kg m/s² or Newton.
• Examples – mass: mass of man is 100 kg.

Weight: Weight of a man having mass 100 kg is 980 N

2) Why is the weight of an object on the moon 1/6th its weight on the earth?

Answer: Weight is defined as gravitational force exerting on the object. It depend on planet which exerts force. Mass of earth is greater than mass of moon so earth exerts much force on object than moon.

According to second law of motion, this force produces acceleration in the object. But acceleration due to gravity by earth is 6th time greater than acceleration produced by moon. Therefore the weight of an object on the moon 1/6th its weight on the earth.

Part 4:

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

Answer:  We know that, pressure is directly proportional to force and inversely proportional to area. If the area of strap of school bag is thin then pressure increases. So it hearts to soldiers.

2) What do you mean by buoyancy?

Answer: The upward force exerted on object by water or liquid is called buoyancy. This force is responsible for floating objects on water. Ships floats on water because of buoyancy force.

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

• The condition for floating objects on water-

If the buoyant force is greater than gravitational force acting on that object then object float on water.

Or

Relative density of object is less than water then it float on water.

• The condition for sink objects in water-

If the buoyant force is less than gravitational force acting on that object then object sink in water.

Or

Relative density of object is greater than water then it sinkin water.

Part 5:

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

Answer: The weighing machine is a type transducer which converts pressure due to gravitational force on the object into electrical signal and shows the mass of that body.

When we stand on weighing machine then there are two types of forces acting on us. 1stforce is gravitational force acting on us in downward direction and 2nd is air resistances which exerts in upward direction. So weighing machine measure mass of the object on the basis of gravitational force only. Thus it shows less mass of the object than original mass.

If the weighing machine shows 42 kg then actual mass will be greater than 42 kg.

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:  As we know that weighing machine is work on gravitational force acting on object only. It does not include upthrust due to air resistance. So the actual mass of objects appears less than original mass.

The weighing machines show mass of bag of cotton and iron bar is 100 kg. Air resistance in upward direction for cotton bag is more because its density is greater than density of iron. Thus the actual mass of cotton bag is more than iron bar.

Gravitation Exercise

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

Answer: Universal law of gravitation states that, force of attraction between two objects is inversely proportional to square of distance between them.Gravitational force also depend on mass but we have to focus on distance in this problem.

If d is the distance between them then the magnitude of force is,

F ∝ 1/d²………..1

Now the distance becomes half then,

F’∝[1/(d/2)²]

F’ ∝4/d² ………..2

Divide equation 2 by equation 1 we get,

F’/F∝[(4/d²)/1/d²]

F’ ∝4F

Final force becomes 4 times initial force.

If distance reduced to half then force increases by 4 times.

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

Answer: According to second law of motion force is directly proportional to change in momentum. When objects are at certain height then they moves towards earth because gravitational force. This gravitational force produces constant acceleration of 9.8 m/s². This is not depend on mass of that object. Thus a heavy object does not fall faster than alight object.

3.) What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is6 × 1024 kg and radius of the earth is 6.4 × 106 m.)

Answer: Given, mass of earth = M = 6 × 1024 kg, radius of earth = r =  6.4 × 106 m, mass of object = m = 1 kg,

G= 6.67 × 10-11N m²/kg².

According to universal law of gravitation,

F = G mM/r²

F = 6.67 × 10-11 × 1×6× 1024/(6.4 × 106

F = 40.02× 1013/(6.4 × 106)

F = 40.02× 1013/(40.96 × 1012)

F = 0.98 × 10

F = 9.8 N.

The gravitational force exerted on that object is 9.8 N.

Note- we can also solve this using second law of motion.

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 third law of motion,  magnitude of force exerted by first object on second object  is equal to the magnitude of force exerted by second object on first.

So the magnitude of force exerted by earth on moon is always equal to magnitude of force exerted by moon on earth but direction of these two forces are always opposite.

5.) If the moon attracts the earth,why does the earth not movetowards the moon?

Answer: According to third law of motion, the magnitude of  force exerted by earth on moon is always equal to magnitude of force exerted by moon on earth but direction of these two forces are always opposite.

But the magnitude of that force is not sufficient to move earth towards moon. So earth does not moves towards moon.

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) as we know that,

F ∝ m ×M

If one mass is doubled then force also becomes doubled.

(ii) As we know that,

F ∝ 1/d²

If distance becomes doubled then

Force increases by 4 times .

If distance becomes triple then force increases by 9 times.

(iii) As we know that,

F ∝ m ×M

If mass of both object becomes doubled then,

F ∝2m ×2M

F ∝ 4m ×M

If the mass of both objects become doubled then force increases by 4 times.

7.) What is the importance ofuniversal law of gravitation?

Answer: Importance of universal law of gravitation.

• Gravitational force binds us to earth. If gravitational force vanishes suddenly then all the objects on earth throws in space.
• All geostationary satellites binds with earth and moves according to earth because of gravitational force.
• Tides, rise and fall of sea levels are due to moons gravitational force only.
• Gravitational force binds planets with earth.

8.) What is the acceleration of freefall?

Answer: We know that, when objects fall under gravitational force only then such motion is called as free fall. All object moves under constant acceleration of 9.8 m/s². This acceleration varies with altitude.

9.) What do we call the gravitational force between the earth and an object?

Answer: As we know that second law of motion,

Force = mass × acceleration.

We calculate gravitational exerted by earth on object as,

Gravitational force = mass ×acceleration due to gravity.

But thus value is also called as weight.

Thus gravitational force on object is called as weight.

10.) Amit buys few grams of gold at the poles as per the instruction of one of his friends. He handsover 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: Amit measured weight of gold at poles and his friend measured at equator. We know that, weight is depend on value of acceleration due to gravity. Thus value is greater at poles than equator.  Thus there will be some difference occurred in that two measurements.

11.) Why will a sheet of paper fall slower than one that is crumpledinto a ball?

Answer: Any object falls under gravitational force then air exerts force in upward direction. This force exerts on the area of that object. If the area is greater then the value of opposing forces also greater.

If a sheet of paper fall then area is greater and the value of opposing forces also greater. Thus it moves slowly.

But if we crumpled it into a ball then the value of opposition force decreases and it falls with greater acceleration.

12.) Gravitational force on the surface of the moon is only1/6 as strong as gravitational force on the earth. What is theweightin newtons of a 10 kg object on the moon and on the earth?

Answer: As given by, Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth.

Thus acceleration due to gravity on earth  is 6 times greater than acceleration due to gravity on moon.

We know that value of acceleration due to gravity on earth = 9.8 m/s².

Thus acceleration due to gravity on moon becomes 9.8/6 m/s² = 1.633 m/s².

Now we calculate weight of the object on earth.

Given, mass of the object = 10 kg.

According to second law of motion,

Weight = mass × acceleration due to gravity.

Weight = 10 × 9.8

Weight = 98 N.

Now we calculate weight of the object on moon.

Given, mass = 10 mass

As,

Weight = mass × acceleration due to gravity on moon.

Weight = 10 × 1.633

Weight = 16.33 N.

The weight of the object on earth will be 98 N and moon will be 16.33 N.

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: Given, initial velocity =u = 49 m/s,

The value of acceleration in upwards journey  = -9.8 m/s².

(i) We know that the velocity of ball becomes zero when it reach at maximum height.

v = 0 m/s.

The value of acceleration in upwards journey  = -9.8 m/s².

We can calculate maximum height using third equation of motion,

v² = u² +2as

0² = 49² + 2 × (-9.8 )× s

19.6 s = 2401

s = 2401/19.6

s = 122.5 m.

The maximum height gained by ball is 122.5 m.

(ii) Firstly we calculate time required to gain maximum height.

According to 1st equation of motion,

v = u + at

0 = 49 + (-9.8)× t

9.8t = 49

t = 49/9.8

t = 5 seconds

Similarly we calculate time for downward journey.

Thus, total time = Time required for upward journey + Time required for downward journey.

Total time = 5 + 5

Total time = 10.s.

Total time required for journey is 10 seconds.

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: Given, a stone is released so initial velocity= u = 0 m/s.

Final velocity= v m/s.

Height = s = 19.6 m, acceleration= 9.8 m/s².

As we know third equation of motion,

v² = u² +2as

v²= 0² + 2 ×9.8× 19.6

v² = 0 + 384

Taking square root on both side,

v = 19.6 m/s.

The final velocity becomes 19.6 m/s.

15.) A stone is thrown vertically upward with an initial velocity of40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the totaldistance covered by the stone?

Answer: Given, initial velocity= u = 40 m/s, final velocity becomes zero when it reach it maximum height,

g = 10 m/s²,  acceleration opposes the motion so we take negative sign for acceleration.

g = -10 m/s².

As we know third equation of motion,

v² = u² +2as

0² = 40² + 2 × (-10)s

0= 1600 -20s

20s = 1600

s = 1600/20

s = 80 m

The maximum height gained by that stone is 80 m.

If the stone is thrown vertically upward then it reach at same point.

So initial and final position become lies at same point .

Thus displacement becomes zero.

The stones covers 80 m in upward direction and 80 m in downward direction. Thus distance covered by stone is 160 m.

16.) Calculate the force of gravitation between the earth and theSun, given that the mass of the earth = 6 × 1024 kg and of theSun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011 m.

Answer: Given, mass of earth = m = 6 × 1024 kg,

Separation between earth and sun  = r = 1.5 × 1011 m,

mass of Sun = M = 2 × 1030 kg,

G= 6.67 × 10-11N m²/kg².

According to universal law of gravitation,

F = G mM/r²

F = 6.67 × 10-11 ×6 × 1024×2 ×1030/(1.5 × 1011

F = 80.04× 1043/(2.25× 1022)

F = 35.57 × 1021

F = 3.557 × 1022  N

The gravitational force between earth and sun is 3.557 × 1022  N.

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.

• When a stone is falling from the top:

When stone is falling from top its initial velocity u= 0

Acceleration due to gravity g= 9.8m/s².

Let us suppose, s be the displacement of the stone in time t when it is falling from the top,

According to second law of motion,

s= ut + ½ gt²

But, here u= 0

Hence, s= ½ × 9.8 × t²

Thus, s= 4.9 × t²

• When stone is thrown vertically upwards:

Given that, initial velocity u= 25m/s.

Again we suppose, s’ be the displacement of the stone in time t when vertically thrown upwards.

Here, stone is moving upwards against the acceleration due to gravity, hence g= -9.8m/s²

From equation of motion, we can write,

s’ = ut + ½ gt²

s’ = 25t -4.9× t²

Now, the total displacement of the both stones at the point where they meet is nothing but the height of tower.

Given that, height of tower = 100 m

Hence, we can write,

s + s’ = 100

½ gt² + 25t – ½ gt² = 100

Thus, 25t = 100

Hence, t= 4 second

Hence, in 4 s the distance covered by the falling stone will be given by,

s= ½ g × t² = ½ × 10×16

Thus, s= 80 m

Thus, the height at which both stone will meet after 4 s will be (100 – 80)= 20 m

Thus, the both stones will meet at a distance 20 m from the ground..

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.

The ball goes up and returned to thrower in 6 seconds.

time of journey = 6 seconds.

We split this problem in 2 parts.

First part- ball moves from thrower to maximum height.

Second part-ball moves from maximum height to thrower.

Time required to gain maximum height = 3 s.

Also time required to react at thrower from maximum height = 3 seconds.

(a) We can find initial velocity using first part of journey.

Let, initial velocity of upward journey = u m/s.

The velocity of ball becomes zero at maximum height. Final velocity = v = 0.m/s.

Time of upward journey = 3 m/s.

Acceleration due to gravity opposes the motion. Value of acceleration due to gravity = -9.8 m/s².

According to first equation of motion,

v = u +at

0 = u +(-9.8)3

u = 29.4 m/s.

The ball is grown with 29.4 m/s.

(b) Now we calculate maximum height using first part of journey.

According to third law of motion,

v² = u² +2as

0² = 29.4² + 2 × (-9.8)s

0 = 864.36- 19.6 s

s = 864.36/19.6

s = 44.1 m

The maximum height gained by ball is 44.1 m.

(c)

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

Answer: Buoyancy force always exerts in vertically upward direction which immersed in liquid.

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

Answer: The mass of block of equal volume of plastic is less than mass of equal volume of water. In simple way, we conclude that the relative density of plastic is less than water. We know that if the density of object is less than density of water then it float on water.

Therefore plastic float on surface of water.

21.) The volume of 50 g of a substance is 20 cm³. If the density ofwater is 1 g/cm³, will the substance float or sink?

volume of substance = 20 cm³

mass = 50 g ,

Density of water = 1 g /cm³.

Now we calculate density of substance.

As we know that,

Density = mass /volume

Density= 50/20

Density = 2.5 g/cm³

The density of substance is greater than density of water thus the substance sink into water.

22.) The volume of a 500 g sealed packet is 350 cm³. Will the packetfloat orsink in water if the density of water is 1 g/cm³? Whatwill be the mass of the water displaced by this packet?

volume of pocket = 350 cm³

mass = 500 g ,

Density of water = 1 g /cm³.

As we know that,

Density = mass /volume

Density= 500/350

Density = 1.43 g/cm³

The density of pocket is greater than density of water thus the pocket sink into water.

According to Archimedes principle, the displacement of water is equal to mass of the object sink into it.

Therefore amount of water displacement of water is 350 g.

Updated: August 24, 2021 — 1:53 pm