NCERT Solutions For Class 9 Science Chapter 10 - Cbsestudyguru

NCERT Solutions For Class 9 Science Chapter 10

NCERT Solutions For Class 9 Science Chapter 10 Gravitation

1. Exercise Questions
2. Intext Questions

NCERT Solutions For Class 9 Science Chapter 10 Gravitation on this step-by-step Gravitation answer guide . In some of State Boards and CBSE schools, students are taught thru NCERT books. As the chapter comes to an end, students are requested few questions in an exercising to evaluate their knowledge of the chapter. Students regularly want guidance dealing with those NCERT Solutions. It’s most effective natural to get stuck withinside the exercises while solving them so that you could assist students rating better marks, we have provided grade by grade NCERT answers for all exercises of Class 9 Science Gravitation so you can be searching for assist from them. Students have to solve those exercises carefully as questions withinside the very last exams are requested from those so these exercises at once have an impact on students’ final score. Find all NCERT Solutions for Class nine Science Gravitation below and prepare for your tests easily.

NCERT Class 9 Science

Chapter 10 – Gravitation

Exercise Questions

 

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

Ans . 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 , that is ,

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

Q. 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 ?

Ans . All objects fall on ground with constant acceleration , called acceleration due to gravity ( in the absence of air resistance ) . It is constant ( g = 9.8 m / s² ) and it does not depend upon the mass of the object . Hence , heavy objects do not fall faster than light objects .

Let F be the gravitational force acting on a body of mass m , then it can be given by ,

From the above equations , we can see that

F α m

Although ‘ g ( acceleration due to gravity ) does not depend on mass ‘ m ‘ . Hence , all bodies fall with the same acceleration provided there is no air or other resistance .

Q. 3. What is the magnitude of the gravitational force between the Earth and a 1 kg object on its surface ? ( Mass of the Earth = 6 x 1024 kg and radius of the Earth is 6.4 x 106 m )

Ans . Given that ,

Mass of the body ( m ) = 1 kg

Mass of the Earth ( M ) = 6 x 10²4 kg

Radius of the Earth ( R ) = 6.4 × 106 m

where ,

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So , the magnitude of the gravitational force between the Earth and a 1 kg object on its surface is 9.8 N.

Q. 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 ?

Ans . 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 .

The universal law of gravitation , the force acting between Earth and moon will be :

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So , the magnitude of force ( F ) is same for Earth and Moon .

Q. 5. If the Moon attracts the Earth , why does the Earth not move towards the Moon ?

Ans . According to the Newton’s third law of motion , the Moon also attracts the Earth with a force equal to that with which the Earth attracts the moon . However , the mass of the Earth is much larger than the mass of the moon .

Hence , it accelerates at a rate less than the acceleration rate of the moon towards the Earth . For this reason , the Earth does not move towards the Moon .

Q. 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 ?

Ans . According to the universal law of gravitation , the force of gravitation between two objects , is given by ,

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( i ) When the mass of either body is doubled , the force is also doubled as per the following equation , F ‘ being the new force :

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F is directly proportional to the masses of the objects . If the mass of one object is doubled , then the gravitational force will also get doubled .

( ii ) When the distance between the objects is doubled , the force becomes one – fourth of the original force as per the following equation :

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F is inversely proportional to the square of the distances 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 ) When the masses of both bodies are doubled , the force becomes four times the original force because F is directly proportional to the product of mass of both objects . If the masses of both the objects are doubled , then the gravitational force becomes four times the original value .

Q. 7. What is the importance of universal law of gravitation ?

Ans . Universal law of gravitation is important because with the help of this law we can explain natural phenomena such as ,

1. the force of attraction that binds us to the Earth .

2 . the motion of planets moving around the Sun.

3. the motion of Moon around the Earth .

4. the occurring of tides due to the gravitational forces of the Sun and Moon .

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

Ans . When objects fall towards the Earth under the effect of gravitational force alone , they are said to be in free fall . The acceleration of free fall is produced when a body falls under the influence of the force of gravitation of the Earth alone .

Acceleration of free fall is 9.8 m s – 2 , which is constant for all objects ( irrespective of their masses ) .

Q. 9. What do we call the gravitational force between the Earth and an object ?

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

The weight of an object is defined as the force of gravity on the object and it is equal to the product of mass and the acceleration due to gravity .

w = mg

Since the weight is a force , so its S.I. unit is Newton .

Q. 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 ] .

Ans . 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 .

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

Ans . When a sheet paper is crumpled into a ball , then its surface area in contact with the air decreases .

Hence , resistance to its motion through the air decreases and it falls faster than the sheet of paper .

Q. 12. 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 ?

Ans . Given ,

Weight of an object on the Moon = 1 / 6 x Weight of an object on the Earth

We know that ,

Weight = Mass x Acceleration

Acceleration due to gravity , g = 9.8 m / s²

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

And , weight of the same object on the moon = 1 / 6 x 98 = 16.3 N

Q. 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 .

Ans . ( i ) According to the equation of motion under gravity , v² – u² = 2gh

where , u = Initial velocity of the ball

v = Final velocity of the ball

h = Height achieved by the ball

g = Acceleration due to gravity

At maximum height , final velocity of the ball is zero , that is , v = 0 m / s and u = 49 m / s

During upwards motion , g = -9.8 m s ²

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But , Time of ascent = Time of descent

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

Q. 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 .

Ans . According to the equation of motion under gravity ,

v² – u² = 2gh

where ,

u = Initial velocity of the stone = 0 m / s

v = Final velocity of the stone

h = Height of the stone = 19.6 m

g = Acceleration due to gravity = 9.8 ms ²

Put the values ,

∴ v² – 0² = 2 x 9.8 x 19.6 ⇒ v² = 2 x 9.8 x 19.6 = ( 19.6 ) ²

⇒ v = 19.6 m s – 1

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

Q. 15. A stone is thrown vertically upward with an initial velocity of 40 m / s . Taking g = 10 m / s² , find the maximum height reached by the stone . What is the net displacement and the total distance covered by the stone ?

Ans . According to the equation of motion under gravity v² – u² = 2gh

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

v = Final velocity of the stone = 0 m / s

h = Height of the stone

g = Acceleration due to gravity = -10 ms – 2

If ‘ h ‘ be the maximum height attained by the stone .

Then , v² – u² = 2gh

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Therefore , total distance covered by the stone during its upwards and downwards journey

= 80 + 80 = 160 m

Net displacement during its upwards and downwards journey = 80 + ( -80 ) = 0 m

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

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

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Hence , the force of gravitation between the Earth and the Sun is 3.57 x 10²² N.

Q. 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 .

Ans . Let the two stones meet after a time t .

When the stone dropped from the tower ,

Initial velocity , u = 0 m / s

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

Acceleration due to gravity , g = 9.8 ms ²

From the equation of motion ,

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The combined displacement of both the stones at the meeting point is equal to the height of the tower 100 m .

In 4 s , the falling stone has covered a distance given by ( i ) as h = 4.9 x 4² = 78.4 m . Therefore , the stones will meet after 4 s at a height ( 100 -78.4 ) = 21.6 m from the ground .

Q. 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 .

Ans . ( a ) Time of ascent is equal to the time of descent . The ball takes a total of 6 s for its upwards and downwards journey .

Hence , it has taken 3 s to attain the maximum height . Final velocity of the ball at the maximum height , v = 0 m / s

Acceleration due to gravity , g = -9.8 m s ²

Using equation of motion , v = u + gt , [ ½ ]

Put the values

0 = u + ( -9.8 x 3 )

⇒ u = 9.8 x 3 = 29.4 m / s

Hence , the ball was thrown upwards with a velocity of 29.4 m / s .

( b ) Let the maximum height attained by the ball = h

Initial velocity during the upward journey , u = 29.4 m / s

Final velocity , v = 0 m / s

Acceleration due to gravity ,

g = -9.8 ms ²

Using the equation of motion ,

Hence , the maximum height is 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 m / s

Position of the ball after 4 s of the throw is given by the distance travelled by it during its downwards journey in 4 s – 3s = 1 s .

Using the equation of motion ,

Now , total height = 44.1 m

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

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

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

An object immersed in a liquid , whether it floats or sinks or stays suspended , always experiences buoyant force in the upward direction . This is because the pressure increases as we go deeper into a liquid .

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

Ans . Two forces act on an object immersed in water . One is the gravitational force , which pulls the object downwards , and the other is the buoyant force , which pushes the object upwards .

If the upward buoyant force is greater than the downward gravitational force , then the object comes up to the surface of the water as soon as it is released within water . Due to this reason , a block of plastic released under water comes up to the surface of the water .

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

Ans . 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 = Mass of the substance / Volume of the substance

= 50/20

= 2.5 g / cm³

The density of the substance is more than the density of water ( 1 g / cm³ ) . Hence , the substance will sink in water .

Q. 22. The volume of a 500 g sealed packet is 350 cm³ . Will the packet float or sink in water if the density of water is 1 g cm³ ? What will be the mass of the water displaced by this packet ?

Ans . 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 of a liquid , then it floats on the surface of the liquid .

Density of the 500 g sealed packet = Mass of the packets / Volume of the packet

= 500/350 = 1.428 g / cm³

The density of the substance is more than the density of water ( 1 g / cm³ ) . Hence , it will sink in water .

The mass of water displaced by the packet is equal to the volume of the packet , that is , 350 g ,

Intext Questions

 

Page 134

Q. 1. State the universal law of gravitation . [ NCERT Q. 1 , Page 134 ] [ KVS , Patna Region , 2017-18 , SA – I , 2015-16 ]

Ans . The universal law of gravitation states that , ” Any two objects exert a gravitational force of attraction on each other . The direction of the force is along the line joining the objects . The magnitude of the force is proportional to the product of the masses of the objects , and inversely proportional to the square of the distance between them . “

For two objects of masses m₁ and m₂ and the distance between them r , the force ( F ) of attraction acting between them is given by the universal law of gravitation as :

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where , G is the universal gravitation constant and its value is 6.67 × 10 ¹¹ N m² kg ² .

Q. 2. Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the Earth . [ NCERT Q. 2 , Page 134 ]

Ans . Let ‘ ME ‘ 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 :

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Page 136

Q. 3. What do you mean by free fall ? [ NCERT Q. 1 , Page 136 ]

Ans . Gravity of the Earth attracts every object towards its centre .

When an object is released from a height , it falls towards the surface of the Earth under the influence of gravitational force . The motion of the object is said to be free fall .

Q. 4. What do you mean by acceleration due to gravity ? [ NCERT Q. 2 , Page 136 ]

Ans . When an object falls towards the ground from a height , its velocity changes during the fall .

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 .

Page 138

Q. 5. What are the differences between mass of an object and its weight ? [ NCERT Q. 1 , Page 138 ]

Ans . Differences between Mass and Weight :

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Q. 6. Why is the weight of an object on the moon 1 / 6th its weight on the Earth ? [ NCERT Q. 2 , Page 138 ]

Ans . Let ‘ ME ‘ be the mass of the Earth and ‘ m ‘ be an object on the surface of the Earth and ‘ 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 ,

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Now again , let MM and RM be the mass and radius of the moon respectively .

Then , according to the universal law of gravitation , weight WM of the object on the surface of the moon is given by ,

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Now , take the ratio of WM to WE

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Therefore , weight of an object on the Moon is 1 / 6th of its weight on the Earth .

Page 141

Q. 1. Why is it difficult to hold a school bag having a strap made of a thin and strong string ? [ NCERT Q. 1 , Page 141 ]

Ans . The force exerted by a thin and strong string is equally spread on a very less area and hence the force applied due to the bag is more .

The pressure is inversely proportional to its area . So , the pressure exerted on the body by thin straps of the school bag will be more and hence will be more painful to carry the bag .

If the area of the given bag is reduced , pressure of the bag increases .

Q. 2 . What do you mean by buoyancy ? [ NCERT Q. 2 , Page 141 ]

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

When you try to immerse an object in water , then you can feel an upward force exerted on the object , which increases as you push the object deeper into water .

Q.3 . Why does an object float or sink when placed on the surface of water ? [ NCERT Q. 3 , Page 141 ]

Ans . 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 .

That means density of an object is a key factor to decide whether it’ll float or sink in a liquid .

Page 142

Q. 4. You find your mass to be 42 kg on a weighing machine . Is your mass more or less than 42 kg ? [ NCERT Q. 1 , Page 142 ]

Ans . When you weigh your 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 .

Q.5 . 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 ? [ NCERT Q. 2 , Page 142 ]

Ans . The bag of cotton is heavier than iron bar . This is because the surface area of the cotton bag is larger than the iron bar .

Hence , more buoyant force acts on the bag than that on an iron bar . This makes the cotton bag lighter than its actual value .

For this reason , the iron bar and the bag of cotton show the same mass on the weighing machine , but actually the mass of the cotton bag is more than that of the iron bar .

 

Benefits of NCERT Solutions For Class 9 Science Chapter 10 Gravitation

NCERT Solutions For Class 9 Science Chapter 10 Gravitation contains extremely important points, and for each chapter, each concept has been simplified to make it easier to remember and increase your chances of achieving excellent exam results. Exam Preparation References Here are some tips on how these solutions can help you prepare for the exam.

  1. This helps students solve many of the problems in each chapter and encourages them to make their concepts more meaningful.
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