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NCERT Solution for Class 9 Physics Chapter 10 - Gravitation

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NCERT Textbook Solutions are considered extremely helpful when preparing for your CBSE Class 9 Physics exams. TopperLearning study resources infuse profound knowledge, and our Textbook Solutions compiled by our subject experts are no different. Here you will find all the answers to the NCERT textbook questions of Chapter 10 - Gravitation.

All our solutions for Chapter 10 - Gravitation are prepared considering the latest CBSE syllabus, and they are amended from time to time. Our free NCERT Textbook Solutions for CBSE Class 9 Physics will strengthen your fundamentals in this chapter and can help you to score more marks in the examination. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam.

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NCERT Solution for Class 9 Physics Chapter 10 - Gravitation Page/Excercise 134

Solution 1

The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The gravitational force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Consider two objects of masses m1 and m2 and let the distance between their centres be r. The gravitational force of attraction (F) acting between them is given by the universal law of gravitation as:

Concept Insight -             

where, G is the universal gravitation constant given by:

Solution 2

Let M  be the mass of the Earth and m be the mass of an object on its surface. If 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:

Concept Insight -
begin mathsize 14px style straight F equals GMm over straight R squared end style
 

NCERT Solution for Class 9 Physics Chapter 10 - Gravitation Page/Excercise 136

Solution 1

Gravity of the Earth attracts every object towards its centre. When an object is released from a height such that it falls towards the surface of the Earth under the influence of gravitational force alone, the motion of the object is called free fall.

Solution 2

During free fall of an object towards the earth, the magnitude of velocity of the falling object goes on increasing. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s2.

  Concept Insight - 

The value of acceleration due to gravity changes from place to place but for most of the purposes it is taken to be 9.8 m/s2.

NCERT Solution for Class 9 Physics Chapter 10 - Gravitation Page/Excercise 138

Solution 1

S. No.

Mass

Weight

I

Mass is the quantity of matter contained in a body.

Weight is the force of gravity acting on a body.

II

Mass of an object is a constant quantity.

Weight of an object is not a constant quantity. It is different at different places.

III

It only has magnitude.

It has magnitude as well as direction.

IV

Its SI unit is kilogram (kg).

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

V

Mass of an object is never zero.

Weight of an object is zero at the centre of Earth.

 

Concept Insight  Weight is a force but mass is not.

Solution 2

Let ME be the mass of the Earth and m be the mass of an object on the surface of the Earth. Let RE be the radius of the Earth. According to the universal law of gravitation, weight WE of the object on the surface of the Earth is given by,

begin mathsize 14px style straight W subscript straight E equals fraction numerator GM subscript straight E straight m over denominator straight R subscript straight E superscript 2 end fraction end style 

Let MM and RM be the mass and radius of the Moon. Then, according to the universal law of gravitation, weight WM of the same object on the surface of the Moon is given by:

begin mathsize 14px style straight W subscript straight M equals fraction numerator GM subscript straight M straight m over denominator straight R subscript straight M superscript 2 end fraction
straight W subscript straight M over straight W subscript straight E equals fraction numerator straight M subscript straight M straight R subscript straight E superscript 2 over denominator straight M subscript straight E straight R subscript straight M superscript 2 end fraction end style

 Therefore, weight of an object on the moon is th of its weight on the Earth.

Concept Insight Gravitational force of the moon is th that of the earth.

NCERT Solution for Class 9 Physics Chapter 10 - Gravitation Page/Excercise 141

Solution 1

It is difficult to hold a school bag having a thin strap because the pressure exerted 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 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 shoulders is very large.

  Concept Insight - Pressure = Force/Area  

Solution 2

Whenever an object is immersed in a fluid, either partially or fully, it experiences an upward force by the fluid. This force is called buoyant force or upthrust and the property due to which a fluid exerts this upthrust on an object placed in it, is called buoyancy.

Solution 3

If the density of an object is more than the density of water, then it sinks in water. This is because the buoyant force acting on the object is less than the force of gravity acting on it. On the other hand, if the density of an object is less than the density of water, then it floats on the surface of water. This is because the buoyant force acting on the object is greater than the force of gravity acting on it.


Concept Insight -Density of an object is a key factor to decide whether it'll float or sink in a liquid.

NCERT Solution for Class 9 Physics Chapter 10 - Gravitation Page/Excercise 142

Solution 1

When you weigh your body, an upward force (due to air) 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.


Concept Insight -The upthrust of air acting on the body reduces the weight of the body.

Solution 2

We know that true weight = apparent weight + upthrust. The cotton bag is heavier than the iron bar. This is due to the reason, that the bag of cotton which has more volume (as it has less density) than the iron bar (which has more density), experiences more upthrust due to air.

Concept Insight - Same mass does not necessarily means same buoyant force.

NCERT Solution for Class 9 Physics Chapter 10 - Gravitation Page/Excercise 143

Solution 1

According to the universal law of gravitation, gravitational force (F) acting between two objects is inversely proportional to the square of the distance (r) between them, i.e.

Concept Insight - If distance r becomes r /2, then the gravitational force will be proportional to  

Hence, if the distance is reduced to half, then the gravitational force would become four times.

Solution 2

It is true that gravitational force acts on all objects in proportion to their masses. But a heavy object does not fall faster than a light object. This is because of the reason that

As Force Mass, therefore, acceleration is constant for a body of any mass.

  Concept Insight - Acceleration due to gravity is constant for a heavy object as well as a light object.

Solution 3

According to the universal law of gravitation, gravitational force between two objects of masses M and m at a distance r from each other is given by:

Let mass of Earth be represented by M = 6 × 1024 kg

Let mass of the object be represented by m = 1 kg

Universal gravitational constant, G = 6.7 × 10-11 Nm2 kg-2

Since the object is on the surface of the Earth, r = radius of the Earth (R)

r = R = 6.4 × 106 m

 

Concept Insight -  Gravitational force,   begin mathsize 14px style fraction numerator 6.7 cross times 10 to the power of negative 11 end exponent cross times 6 cross times 10 to the power of 24 cross times 1 over denominator left parenthesis 6.4 cross times 10 to the power of 6 right parenthesis squared end fraction equals 9.8 space straight N end style

Solution 4

The Earth attracts the Moon with a force which is same as the force with which the moon attracts the Earth because according to Newton's third law of motion, force of action and reaction are always equal and opposite. So, the force of attraction of Earth on Moon is equal and opposite to the force of attraction of Moon on Earth.

Concept Insight - Action and reaction forces are always equal and opposite.

Solution 5

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 much lesser than the acceleration rate of the Moon towards the Earth. For this reason, we do not see the Earth moving towards the Moon.


Concept Insight - For the same force, an object with greater mass obtains less acceleration as compared to an object with lesser mass.

NCERT Solution for Class 9 Physics Chapter 10 - Gravitation Page/Excercise 144

Solution 6

According to the universal law of gravitation, the force of gravitation between two objects is given by:   (i) F is directly proportional to the product of masses of the objects. If the mass of one object is doubled, then the gravitational force will also get doubled.

(ii) F is inversely proportional to the square of the distance between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value.

Similarly, if the distance is tripled, then the gravitational force becomes one-ninth of its original value.

(iii) F is directly proportional to the product of masses of the objects. If the masses of both the objects are doubled, then the gravitational force becomes four times the original value.

Solution 7

The universal law of gravitation is important because it accounts for the force that binds us to the Earth, motion of planets around the Sun, motion of the Moon and other artificial satellites around the Earth, tides due to the Moon and the Sun and many other phenomena. 

Solution 8

When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall.  This force produces a uniform acceleration in the objects, which is called acceleration of free fall or acceleration due to gravity. Its value is 9.8 m/s2.

Concept Insight - The value of acceleration of free fall is 9.8 m/s2, which is constant for all objects (irrespective of their masses).

Solution 9

Gravitational force between the Earth and an object is known as Earth's gravity or weight of the object.

Solution 10

Weight of a body on the earth is given by:

W = m g

where,

m = Mass of the body

g = Acceleration due to gravity

The value of g is greater at poles than at the equator. Therefore, the same mass of gold weighs lesser at the equator than at the poles. Hence, Amit's friend will not agree with the weight of the gold bought.

Concept Insight - Acceleration due to gravity changes from place to place.  

Solution 11

A sheet of paper has more surface area than a crumpled ball of paper. So, the resistance offered by air to a sheet of paper falling through it is more than the resistance offered to a falling crumpled ball of paper. This decreases the speed of the sheet of paper and hence it falls slower than the crumpled ball.

Concept Insight - Amount of air resistance depends on the cross-sectional area of the object.

Solution 12

Weight = Mass × Acceleration

Acceleration due to gravity on earth, ge = 9.8 m/s2

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

Acceleration due to gravity on Moon,

Therefore, weight of the same object on the Moon =

Concept Insight Weight of an object on Moon =   of the weight of the object on Earth

Solution 13

(i) For the upward motion of the ball, we use the equation:

v2 - u2 = 2 g h

where,

u = Initial velocity of the ball = 49 m/s (Given)

v = Final velocity of the ball = 0 (At the highest point)

h = Maximum height attained by the ball

g = Acceleration due to gravity = -9.8 m/s2 (Ball goes up)

Putting the values, we get

0 - (49)2 = 2 × (-9.8) × h

begin mathsize 14px style straight h equals fraction numerator 49 cross times 49 over denominator 2 cross times 9.8 end fraction equals 122.5 space straight m end style

(ii) Let t be the time taken by the ball to reach the height 122.5 m, then to calculate t we use the following equation of motion:

v = u + g t

Putting the values, we get 

0 = 49 + (-9.8) × t

0 = 49 - 9.8 t

49 = 9.8 t

begin mathsize 14px style straight t equals fraction numerator 49 over denominator 9.8 end fraction equals 5 space straight s end style

But, Time of ascent = Time of descent   Therefore, the total time taken by the ball to return = 5 + 5 = 10 s

Concept Insight - When a body is thrown vertically upwards, its velocity decreases, so the acceleration due to gravity g is taken as negative.

Solution 14

Initial velocity of the stone, u = 0

Final velocity of the stone, v = ?

Height of the tower, h = 19.6 m

Acceleration due to gravity, g = 9.8 ms-2

For a freely falling body:

v2 - u2 = 2 g h

v2 - 02 = 2 × 9.8 × 19.6

v2 = 19.6 × 19.6 = (19.6)

v2 = 384.16

begin mathsize 14px style straight v equals square root of 384.16 end root end style 

v = 19.6 ms-1

Hence, the velocity of the stone just before touching the ground is 19.6 ms-1.

 

Concept Insight When a body is falling vertically downwards, its velocity increases, so the acceleration due to gravity g is taken as positive.

Solution 15

Given: 

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

Final velocity of the stone, v = 0 (At the highest point)

Maximum height reached by the stone, h = ?

Acceleration due to gravity, g = -10 ms-2 (Stone goes up)

Using the equation of motion:

v2 - u2 = 2 g h

0 - (40)2 = 2 × (-10) × h

0 - 1600 = - 20 h

-1600 = -20 h

1600 = 20 h

begin mathsize 14px style 1600 over 20 equals straight h
therefore straight h equals 80 space straight m end style

Thus, 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 = 0 (since final position coincides with the initial position)

Concept Insight Distance is the length of the actual path covered by an object, while displacement is the shortest distance between the initial and final position of the object.

Solution 16

According to the universal law of gravitation, the force of attraction between the Earth and the Sun is given by:

  Concept Insight-

where,

MSun = Mass of the Sun = 2 × 1030 kg

MEarth = Mass of the Earth = 6 × 1024 kg

R = Average distance between the Earth and the Sun = 1.5 × 1011 m

G = Universal gravitational constant = 6.7 × 10-11 Nm2 kg-2

Substituting the values, we get

begin mathsize 14px style straight F equals fraction numerator 6.7 cross times 10 to the power of negative 11 end exponent cross times 2 cross times 10 to the power of 30 cross times 6 cross times 10 to the power of 24 over denominator left parenthesis 1.5 cross times 10 to the power of 11 right parenthesis squared end fraction equals 3.57 cross times 10 to the power of 22 space straight N end style 

Solution 17

Let the two stones meet after a time t.

(i) For the stone dropped from the top of the tower:

Initial velocity, u = 0

Let the displacement of the stone in time t from the top of the tower be s.

Acceleration due to gravity, g = 9.8 ms-2

From the equation of motion,

begin mathsize 14px style straight s equals ut plus 1 half gt squared
straight s equals 0 cross times straight t plus 1 half cross times 9.8 cross times straight t squared
therefore straight s equals 4.9 space straight t squared space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis end style

 

(ii) For the stone thrown upwards:

Initial velocity, u = 25 ms-1

Let the displacement of the stone from the ground in time t be s'.

Acceleration due to gravity, g = -9.8 ms-2

From the equation of motion,

begin mathsize 14px style straight s apostrophe equals ut plus 1 half gt squared
straight s apostrophe equals 25 cross times straight t space minus space 1 half cross times 9.8 cross times straight t squared
therefore straight s apostrophe equals 25 space straight t minus 4.9 space straight t squared space space space space space space space space space space... space left parenthesis 2 right parenthesis end style

  The combined displacement (s + s') of both the stones at the meeting point is equal to the height of the tower 100 m.   From eqs. (1) and (2), we get, s + s' = 4.9 t2 + 25 t - 4.9 t2 100 = 25 t   begin mathsize 14px style straight t equals 100 over 25 equals 4 space sec end style In 4 s, the falling stone has covered a distance given by equation (1) as s = 4.9 t2 = 4.9 × (4)2 = 78.4 m   Therefore, the stones will meet after 4 s at a height (100 - 78.4) = 21.6 m from the ground. 

Concept Insight Choose the equation of motion wisely to minimize the number of steps in calculations.

Solution 18

(a) Concept Insight - 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.

Let the initial velocity of the ball be u.

Final velocity of the ball at the maximum height, v = 0

Acceleration due to gravity, g = -9.8 ms-2

From the equation of motion, v = u + gt, we get,

0 = u + (-9.8) × 3

0 = u - 29.4

u = 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 m s -2

Concept Insight Choose the equation of motion wisely to minimize the number of steps in calculations.

From the equation of motion,  begin mathsize 14px style straight h equals 29.4 cross times 3 plus 1 half cross times left parenthesis negative 9.8 right parenthesis cross times left parenthesis 3 right parenthesis squared end style  h = 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 its position during the downward journey in 4 s - 3 s = 1 s.

Concept Insight - Choose the equation of motion wisely to minimize the number of steps in calculations.   From the equation of motion,  , we get, begin mathsize 14px style straight s equals 0 cross times 1 plus 1 half cross times 9.8 cross times left parenthesis 1 right parenthesis squared
straight s equals 0 plus 1 half cross times 9.8
straight s equals fraction numerator 9.8 over denominator 2 end fraction
therefore straight s equals 4.9 space straight m end style
  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.

NCERT Solution for Class 9 Physics Chapter 10 - Gravitation Page/Excercise 145

Solution 19

An object immersed in a liquid experiences buoyant force in the vertically upward direction, i.e., in a direction opposite to the weight of the object.

Solution 20

In case of a block of plastic, the upward buoyant force is greater than the weight of the object. The large buoyant force on the block is due to its density being smaller than that of water. Due to the larger buoyant force, the block of plastic comes up when released under water.

Concept Insight - A body floats when its weight is lesser than the upward buoyant force acting on it. 

Solution 21

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.

Here, density of the substance     The density of the substance is more than the density of water (1 g cm-3). Hence, the substance will sink in water.   Concept Insight - Density of an object is a key factor in deciding whether the object will sink or float.

Solution 22

begin mathsize 14px style Density space of space the space 500 space straight g space sealed space packet equals fraction numerator Mass space of space the space packet over denominator Volume space of space the space packet end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 500 over 350 equals space space 1.428 space straight g space cm to the power of negative 3 end exponent space space space space space space space space space space space space space space space space space space space space end style The density of the substance is more than the density of water (1 g cm-3). Hence, it will sink in water.

As the packet is fully submerged in water,

Mass of water displaced by the packet = volume of the packet x density of water

                                                    = 350 cm3 × 1 g/cm3 = 350 g

Concept Insight - Density of an object is a key factor in deciding whether the object will sink or float.

TopperLearning provides step-by-step solutions for each question in each chapter in the NCERT textbook. Access Chapter 10 - Gravitation here for free.

Our NCERT Solutions for Class 9 Physics are by our subject matter experts. These NCERT Textbook Solutions will help you to revise the whole chapter, and you can increase your knowledge of Physics. If you would like to know more, please get in touch with our counsellor today!

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