**Lakhmir Singh Manjit Kaur Class 9 Physics 3rd Chapter “Gravitation” solution**

Lakhmir Singh and Manjit Kaur Physics solution: “Gravitation” Chapter 3. Here you get easy solutions of Lakhmir Singh and Manjit Kaur Physics solution Chapter 3 . Here we have given Chapter 3 all solution of Class 9. Its help you to complete your homework.

**Board –**CBSE**Text Book****–**Physics**Class –**9**Chapter –**03

**Lakhmir Singh Manjit Kaur Class 9 Physics 3rd Chapter Solution (First Part)**

**Very short answer type Questions**

**Very short answer type Questions **

**1.) What is the value of gravitational constant G (I) on the earth, and on the moon ?**

Answer: The value of gravitational constant G on earth = 6.7 × 10^{-11} N m^{2} kg^{-2}.

**2.) Which force is responsible for the moon revolving round the earth ?**

Answer: Gravitational force is responsible for the moon revolving round the earth.

**3.) Does the acceleration produced in a freely falling body depend on the mass of the body ?**

Answer: No, acceleration produced in a freely falling body does not depend on the mass of the body.

**4.) Name the scientist who gave the three laws of planetary motion.**

Answer: Johannes Kepler gave the three laws of planetary motion.

**5.) Name the scientist who explained the motion of planets on the basis of gravitational force between the sun and planets.**

Answer: Johannes Keplerexplained the motion of planets on the basis of gravitational force between the sun and planets.

**6.) State the Kepler’s law which is represented by the relationr ^{3}**

**∝**

**T**

^{2}.Answer: Kepler’s third law states that, the cube of distance from sun is directly proportional to square of distance.

Mathematically it is expressed as,

r^{3} ∝ T^{2}.

**7.) Which of the Kepler’s laws of planetary motion led Newton to establish the inverse-square rule for gravitational force between two bodies?**

Answer: Sir Isaac Newton is famous for universal law of gravitation. He derived this law using Kepler’s third law of planetary motion. Kepler’s third law states that the ratio of cube of distance to square of time is always constant. Newton also gave a relationship between time and distance in his law of gravitation.

**8.) Name the property of earth which is responsible for extremely small acceleration being produced in it as a result of attraction by other small objects.**

Answer: Earth had extremely large mass. As we know that second law of motion.,

Force = mass × acceleration

Acceleration = force/ mass

The acceleration is inversely proportional to mass. And earth has extremely large mass so extremely small acceleration being produced in it.

**9.) What is the acceleration produced in a freely falling body of mass 10 kg ? (Neglect air resistance)**

Answer: If any objects falling freely in presence of earth’s gravitational force only then that motion is called as free fall. Free falling objects moves with constant acceleration at the rate of 9.8 m/s^{2 }.

**10.) When an object is dropped from a height, it accelerates and falls down, Name the force which accelerates the object.**

Answer: When an object is dropped from a height, it accelerates and falls down due to gravitational force exerted on that object by earth.

**11.) Give the formula for the gravitational force F between two bodies of masses M and m kept at a distance d from each other.**

Answer: Newton’s law of gravitation states that every body in the universe attract another body with a Force which is directly proportional to product of their masses and universally proportional square of distance between them.

If m and M masses separated by distance d then gravitational force is

F = G mM/d^{2}

Where G is constant.

**12.) What force is responsible for the earth revolving round the sun ?**

Answer: Gravitational force is responsible for the earth revolving round the sun.

**13.) What name has been given to the force with which two objects lying apart attract each other?**

Answer: Gravitational force is the name has been given to the force with which two objects lying apart attract each other.

**14.) What type of force is involved in the formation of tides in the sea ?**

Answer: Moon’s gravitational force is involved in the formation of tides in the sea.

**15.) Which force is responsible for holding the solar system together ?**

Answer: Gravitational force is responsible for holding the solar system together.

**16.) What is the weight of a 1 kilogram mass on the earth ? ( g = 9.8 m/s ^{2}).**

Answer: Given, Mass = 1 kg, g = 9.8 m/s^{2}

We can calculate weight using the formula

Weight = gravitational force exerted on body by earth

Weight = m × g

Weight = 1 × 9.8

Weight = 9.8 N

9.8 N is the weight of a 1 kilogram mass on the earth.

**17.) On what factor/factors does the weight of a body depend ?**

Answer: Weight is depend mass and value of gravitational force exerted by that planet.

**18.) As the altitude of a body increases, do the weight and mass both vary ?**

Answer: quantity weight is varies with altitudes. But mass is always constant.

**19.) If the same body is taken to places having different gravitational field strength, then what will vary : its weight or mass ?**

Answer: If the same body is taken to places having different gravitational field strength, then weight will vary. But mass is always constant.

**20.) If the mass of an object be 10 kg, what is its weight ? (g = 9.8 m/s ^{2}).**

Answer: Given, Mass = 10 kg, g = 9.8 m/s^{2}

We can calculate weight using the formula

Weight = gravitational force exerted on body by earth

Weight = m × g

Weight = 10× 9.8

Weight = 98 N

98 N is the weight of a 10 kg mass.

**21.) The weight of a body is 50 N. What is its mass ? (g = 9.8 m/s ^{2}).**

Answer: Given, Weight = 50 kg, g = 9.8 m/s^{2}

We can calculate mass using the formula

Weight = gravitational force exerted on body by earth

Weight = m × g

50= m × 9.8

50/9.8= mass

Mass = 5.102

5.102 kgis the mass of a 50 NWeight.

**22.) A body has a weight of 10 kg on the surface of earth. What will be its weight when taken to the centre of the earth ?**

Answer: Given, Mass = 10 kg, value of g at centre of the earth is zero.

Weight = m×g

Weight = 10×0

Weight = 0 N

The weight of that body will 0 N at the centre of earth.

**23.) Write down the weight of a 50 kg mass on the earth. (g = 9.8 m/s2).**

Answer: Given, Mass = 50 kg, g = 9.8 m/s^{2}

We can calculate weight using the formula

Weight = gravitational force exerted on body by earth

Weight = m × g

Weight = 50× 9.8

Weight = 98 N

490 N is the weight of a 50kg mass.

**24.) If the weight of a body on the earth is 6 N, what will it be on the moon ?**

Answer= The weight of object on moon is 1/6^{th} time weight on earth. So the weight of that object on moon is 1 N.

**25.) State whether the following statements are true or false :**

**(a)) A falling stone also attracts the earth.**

**(b) The force of gravitation between two objects depends on the nature of medium between them.**

**(c) The value of G on the moon is about 1/6 ^{th} of the value of G on the earth.**

**(d) The acceleration due to gravity acting on a freely falling body is directly proportional to the mass of the body.**

**(e) The weight of an Object on the earth is about one-sixth Of its weight on the moon.**

**Answer: **

(a)

True, according to 3^{rd} law of motion, a falling stone also attracts the earth.

(b)

False, the force of gravitation between two objects does not depend on the nature of medium between them.

(c)

False, the value of force on moon is 1/6^{th} of the value of force in earth.

(d)

False, the acceleration due to gravity is constant.

(e) The weight of an Object on the earth is about one-sixth Of its weight on the moon.

False, the weight of an Object on the moon is about one-sixth Of its weight on the earth.

**Short Answer type Questions**

**27.) Explain what is meant by the equation :**

g=G × M/R^{2}

where the symbols have their usual meanings.

Answer: The formula g=G × M/R^{2}

Gives the value of acceleration due to gravity.

Each term in this formula have their usual meanings,

g- acceleration due to gravity.

G – Universal gravitational constant

M- mass of the earth,

R – radius of earth.

28.) (a) What do you mean by the term ‘free fall’ ?

(b) During a free fall, will heavier objects accelerate more than lighter ones ?

Answer: (a) Free fall: If an object falling in presence of earth’s gravitational force only is called free fall.

(b) During a free fall, all objects accelerate with same rate because the force responsible for acceleration is earth’s gravitational force.

**29.) Can we apply Newton’s third law to the gravitational force ? Explain your answer.**

Answer: Yes, we can apply Newton’s third law to the earth’s gravitational force. Newton’s third law states that, with every action there is equal and opposite reaction. We can use this law to calculate the force. We understand this with a simple example.

A book of mass 2 kg is put on table. The force exerted by the book on table is calculated as

F = mg

F = 2 × 9.8

F = 19.6 N.

The force exerted by book on table is 19.6 N.

Now, we calculate the value of force exerted by table on book using third law,

The value of force exerted by book on table is equal to force exerted by table on book.

So The force exerted by table on book is 19.6 N.

**30.) Give reason for the following :**

**The force of gravitation between two cricket balls is extremely small but that between a cricket ball and the earth is extremely large.**

Answer: Universal law of gravitation states that the magnitude of attractional force between any two objects is directly proportional to product of their masses and inversely proportional to distance between them and it is mathematically written as,

F =G× m× M/r^{2}

Where G is constant and having value of the order of 10^{_10}.

So it is very weak force. If any value of mass is very very greater then gravitational force will become significant there fore The force of gravitation between two cricket balls is extremely small but that between a cricket ball and the earth is extremely large.

**31.) How the gravitational force between two objects depends on the distance between them. **

Answer: Gravitational force is inversely proportional to square of distance between two objects,

F ∝ 1/r^{2}

**32.) What happens to the gravitational force between two objects when the distance between them is :**

** (i) doubled ?**

**(ii) halved ?**

Answer: Law of gravitation states that gravitational force is inversely proportional to square of distance between two objects,

F ∝ 1/r^{2}

i) If distance doubled then force becomes ¼ times of initial value.

ii) If distance halved then force becomes four times of initial value.

**33.) State two applications of universal law of gravitation.**

Answer: Applications of universal law of gravitation

- By using this law, we can calculate the force between earth and moon.
- We use this law while launching satellites.

**34.) Explain why, if a stone held in our hand is released, it falls towards the earth.**

Answer: When we held stone in our hand that time gravitational force exerted by earth on stone is neutralize by our hand. If we released the stone then only gravitational force is on that stone so it falls towards earth.

**35.) Calculate the force of gravitation between two objects of masses 50 kg and 120 kg respectively kept at a distance of 10 m from one another. (Gravitational constant, G = 6.7 x 10 ^{-11}Nm^{2} kg^{-2}**

**)**

Answer: Given, Mass of first object = m = 50 kg.

Mass of second object = M = 120 kg.Distance =r = 10 m, and G= 6.7 x 10^{-11}Nm^{2} kg^{-2}

As we know,

F =G× m× M/r^{2}

F =6.7 x 10^{-11}× 50×120 /10^{2}

F =6.7 x 10^{-11}× 50×120 /10^{2}

F = 40200 ×10^{-11} /100

F = 4.02 ×10^{9}

The force of attraction between that two objects is 4.02 ×10^{9}N.

**36.) What is the force of gravity on a body of mass 150 kg lying on the surface of the earth ?**

(Mass of earth = 6 x 10^{24} kg; Radius of earth = 6.4 x 10^{6} m; G = 6.7 10-11 Nm^{2}/kg^{2})

Answer: Given, Mass of a body on surface of earth = m = 150 kg.

Mass of earth = M = 6 x 10^{24} kg. Distance =r = 6.4 x 10^{6} m, and G= 6.7 x 10^{-11}Nm^{2} kg^{-2}

As we know,

F =G× m× M/r^{2}

F =6.7 x 10^{-11}× 150×6 x 10^{24} /( 6.4 x 10^{6})^{2}

F =6.7 x 10^{-11}× 900 x 10^{24}/40.96× 10^{12}

F = 147.22×10^{24}×10^{-11} /10^{12}

F = 1472.2 N

The force on the man is 1472.2 N.

**37.) The mass of sun is 2 x 10 ^{30} kgand the mass of earth is 6 x 10^{24 }kg. If the average distance between the sunand the earth be 1.5 x 10^{8} km, calculate the force of gravitation between them.**

Answer: Given, Mass of the earth = m = 6 x 10^{24} kg.

Mass of the sun = M =2 x 10^{30} kg. Distance between earth and sun =r = 1.5 x 10^{8} km=1.5 x 10^{11}m,

and G= 6.7 x 10^{-11}Nm^{2} kg^{-2}

As we know,

F =G× m× M/r^{2}

F =6.7 x 10^{-11}× 6 x 10^{24}×2 x 10^{30} /( 1.5 x 10^{11})^{2}

F =80.4 x 10^{43} /2.25× 10^{22}

F =3.573 × 10^{22}

The Gravitational force between earth and sun is 3.573 × 10^{22} N.

**38.) A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the acceleration of the stone be 9.8 m/s ^{2} directed towards the ground, calculate the initial velocity Of the stone with which it is thrown upwards.**

Answer: Time required to gain maximum height by piece of stone = 3 seconds. Acceleration = 9.8 m/s^{2}

Acceleration opposes the notion so it’s value becomes 9.8 m/s^{2},

Finally it’s velocity becomes 0 m/s.

We use 1^{st}equation of motion,

V = u + a t

0 = u +(- 9.8)× 3

U = 29.4 m/s

The initial velocity Of stone is 29.4 m/s.

**39.) A stone falls from a building and reaches the ground 2.5 seconds later. How high is the building ? (S= 9.8 m/s ^{2})**

Answer: Time to reach a stone to ground is 2.5 seconds, the stone is released so initial velocity Of stone = 0 m/s. g = 9.8 m/s^{2}

We know second law of motion,

X = ut+ ½ at^{2}

x = 0× 2.5 + ½ 9.8(2.5)^{2}

x = 4.9 × 6.25

x = 30.625

The height of building is 30.625 m.

**40.) A stone is dropped from a height of 20 m.**

**(i) How long will it take to reach the ground ?**

**(ii) What will be its speed when it hits the ground? (g = 10 m/s ^{2})**

Answer: Height =x= 20 m, initialvelocity=u=0 m/s

g = 10 m/s^{2}

(i) We know 2^{rd} equator kinematics,

x = ut + ½ at^{2}

20 = 0× t + ½ × 10 × t^{2}

20×2/10= t^{2}

4 = t^{2}

Taking square root on both side,

t= 2 seconds

The stone requires 2 seconds to reach the ground.

(ii) we know 3^{rd} equation of kinematics,

v^{2} = u^{2} + 2ax

v^{2} = 0^{2} + 2× 10× 20

v^{2} = 0 + 400

Taking square root on both side,

v = 20 m/s

Final velocity Of stone when reach the ground is 20 m/s.

**41.) A stone is thrown vertically upwards with a speed of 20 m/s. How high will it go before it begins to fall (g = 9.8 m/s ^{2})**

Answer: Given, initial velocity = 20 m/s. When it gain maximum height then it’s speed becomes zero. v = 0 m/s. The velocity decreases because force opposes the motion in upward direction so we take negative acceleration, a= -9.8 m/s^{2}

we know 3^{rd} equation of kinematics,

v^{2} = u^{2} + 2ax

0^{2} = 20^{2} + 2× (-9.8)x

19.6 x = 400

x = 400/19.6

x = 20.4 m.

The maximum height gain by stone is 20.4 m

**42.) When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 metres.**

**(a) What was the initial speed of the ball ?**

**(b) How much time is taken by the ball to reach the highest point ? (g =10 m s ^{-2})**

Answer: Given, height = 5 m, g =10 m s^{-2}

When it gain maximum height then it’s speed becomes zero. v = 0 m/s. The velocity decreases because force opposes the motion in upward direction so we take negative acceleration,

a= -9.8 m/s^{2}

(a) we know 3^{rd} equation of kinematics,

v^{2} = u^{2} + 2ax

0^{2} = u^{2} + 2× (-10)× 5

u^{2} = 100

Taking square root on both sides

u = 10 m/s.

The initial speed of ball is 10 m/s.

(b) we can calculate time using 1^{st} equation of kinematics,

v = u + at

0 = 10 + (-10) t

10t = 10

t= 1 second.

Time required to reach maximum height is 1 second.

**43.) Write the differences between mass and weight of an object.**

Answer: Differences between mass and weight of an object-

I) The mass is constant quantity but weight is variable quantity.

II) Weight changes on earth with altitude but mass does not.

III) Definition

Mass- The total matter contained in the body is called mass

Weight- The gravitational force exerted on the body is called weight.

IV) Mass is a scalar quantity but weight is vector.

V) SI unit of mass is kg and weight is Newton.

**44.) Can a body have mass but no weight ? Give reasons for your answer.**

Answer: Yes a body can have mass but no weight. The body in space where zero gravity do not have weight.

**45.) A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration ? (g = 9.8 m s-2).**

Answer: Given, Force acting on the body F=20 N,

Weight = 9.8 N,g = 9.8 m s^{-2}).

As we know,

Weight = mass× acceleration due to gravity.

9.8 = mass × 9.8

Mass = 1 kg.

We cam calculate acceleration using second law of motion

Force = mass × acceleration

20 = 1 × a

a= 20 m/s^{2}

The mass of body is 1 kg and acceleration is 20 m/s^{2}

**46.) A stone resting on the ground has a gravitational force of 20 N acting on it. What is the weight of the stone ? What is its mass ? (g = 10 m/s2).**

Answer: Gravitational Force exerted by earth on a body is called weight. Gravitational force = 20 N so its weight is also 20 N.

We can calculate mass using second law of motion,

Weight = mass × acceleration due to gravity

20 = mass × 10.

Mass = 2 kg.

The mass of the body is 2 kg.

**47.) An object has mass of 20 kg on earth. What will be its (i) mass, and (ii) weight, on the moon ? (g on moon = 1.6 m/s ^{2}).**

Answer: Given, Mass = 20 kg. Value of g on moon = 1.6 m/s^{2}

(i) The mass is always constant so value of mass of that object on moon is also 20 kg.

(ii) we cam calculate weight using second law

Weight = mass × acceleration due to gravity.

Weight = 20 × 1.6

Weight = 32 N.

The weight of that object on moon is 32 N.

**48.) Which is more fundamental, the mass of a body or its weight ? Why ?**

Answer: Mass is more fundamental quantity than force because

a) The mass is universally constant.

b) Weight is a combination of mass and acceleration.

**49.) How much is the weight of an object on the moon as compared to its weight on the earth ? Give reason for your answer.**

Answer: The weight of object on moon is 1/6^{th} that of mass on earth because,

a) Weight is a gravitational force acting on the body and gravitational force depend on mass of two objects.

b) As we know mass of earth is greater than mass of moon.

**Long Answer Type Questions**

**50.) (a) Define mass of a body. What is the SI unit of mass ?**

**(b) Define weight of a body. What is the SI unit of weight ?**

**(c) What is the relation between mass and weight of a body ?**

Answer:

a) Mass: the total matter contained in the body is called mass. SI unit of mass is kg.

b) Weight: Gravitational force acting on the body is called weight. SI unit of weight is Newton.

c) The relationship between mass and weight is derived from the second law of motion and is given as by,

Gravitational force = mass × acceleration due to gravity.

Weight = m ×g.

**51.) (a) State the universal law of gravitation. Name the scientist who gave this law. (b) Define gravitational constant. What are the units of gravitational constant ?**

Answer: (a) A very famous physicist Named Sir Isaac Newton gave the law of gravitation in 16^{th} century.

Statement: Every object in the universe attracts other object with a force, the magnitude of that force is directly proportional to product of their masses and inversely proportional to distance between them.

(b) Gravitational constant: From gravitational law we can understand that force is directly proportional to product of the masses and inversely proportional to square of distance, so we required a constant value to remove proportionality sign. That constant is called gravitational constant. It is denoted by ‘G’.

SI unit of gravitational constant is Nm^{2} / kg^{2}.

**52.) (a) What do you understand by the term ‘acceleration due to gravity of earth’ ?**

**(b) What is the usual value of the acceleration due to gravity of earth ? (c) State the SI unit of acceleration due to gravity.**

Answer: (a) According to law of gravitation, every object having mass is produces its own gravitational force. So earth has very greater mass so it also produces gravitational force. Answer second law states that force produces acceleration. Thus the acceleration produces in the object due the gravitational force is called acceleration due to gravity of earth.

(b) Acceleration due to gravity is acceleration produced by gravitational force of earth. The value of acceleration due to gravity is 9.8 m/s^{2} on the surface of earth. But earth is not spherical it is flatten at pole so value of ‘g’ increases upto 10 m/s^{2} .

Also for simplicity we take value of g is 10 m/s^{2}.

SI unit of acceleration due to gravity is m/s^{2}.

**53.) (a) Is the acceleration due to gravity of earth ‘g’ a constant ? Discuss.**

**(b) Calculate the acceleration due to gravity on the surface of a satellite having a mass of7.4 x 10 ^{22} kg and a radius of 1.74 x 10^{6} m (G= 6.7 x 10^{-11} Nm^{2}/kg^{2}). Which satellite do you think it could be?**

Answer:(a) we take acceleration due to gravity constant. But the value of g =9.8 is not always constant. We understand this by calculating value of g.

If an object is placed on surface of earth having mass ‘m’ and radius of earth is ‘R’. Suppose the mass of earth is M.

By the law of gravitation,

F = G ×m× M/R^{2} ———-1

We know second law of motion, the gravitational force on that object is,

F = mg——–2

Equate equation 1 and 2,

mg = G ×m× M/R^{2}

g = G × M/R^{2}——-3

Where, G = 6.7 × 10^{-11} is a constant term,

M= mass of earth is constant,

R = radius of earth is also constant,

So the value of g is constant.

But if the object is at greater height then value of decreases and it is calculated as,

g = G × M/(R+h)^{2}

The value of g at pole is 10 m/s^{2}.

(b) Given, Mass of satellite = M = 7.4 x 10^{22} kg

Radius of satellite= R = 1.74 x 10^{6} m,

G= 6.7 x 10^{-11} Nm^{2}/kg^{2}.

As we know,

g = G × M/(R)^{2}

g = 6.7 x 10^{-11} ×7.4 x 10^{22} /(1.74 x 10^{6})^{2}

g = 49.58 x 10^{11} /(3.03x 10^{12})

g = 1.636m/s^{2}

The value of g at surface of that satellite is 1.636 m/s^{2}.

The value of g us same for moon so that satellite will be moon.

**54.) State and explain Kepler’s laws of planetary motion. Draw diagrams to illustrate these laws.**

Answer: Kepler studied the motion of planets and explained it in three laws. These laws are called Kepler’s laws of planetary motion.

i) First law: All the planets revolving around the sun in elliptical orbit not in circle and sun lies either at one foci. This law is called as law of ellipse.Ellipse has two centre’s so sun is lies at one foci.

ii) Second law: While the planets revolving around the sun sweeps equal area in equal intervals of time.

The motion of planets around the sun is as

area AOB = area COD

When planets are close to sun then it moves with greater speed and if planets are far from sun then they moves with smaller speed hence they cover equal area on equal intervals of time. This law is called as law of area.

iii) Third law: Kepler’s third law states that the cube of distance from sun is directly proportional to square of time. It is mathematically expressed as,

r^{3} ∝ T^{2}.

Where

r- distance of planet from sun.

T – ,time required to complete one revolution.

**55.) The mass of a planet is 6 x 10 ^{24} kg and its diameter is 12.8 x 10^{3 }km. If the value of gravitational constant be 6.7 x 10^{-11} Nm^{2}/kg^{2}, calculate the value of acceleration due to gravity on the surface of the planet. What planet could this be ?**

Answer: Given, Mass of planet = M = 6 x 10^{24} kg

Diameter of the planed = D = 12.8 x 10^{3 }km

As we know, D/2= R

R =12.8 x 10^{3 } / 2

R= 6.4 x 10^{3 }km

R= 6.4x 10^{6} m,

G= 6.7 x 10^{-11} Nm^{2}/kg^{2}.

As we know,

g = G × M/(R)^{2}

g = 6.7 x 10^{-11} ×6 x 10^{24 }/(6.4x 10^{6})^{2}

g = 40.2 x 10^{13} /(40.96x 10^{12})

g = 0.981 × 10

g = 9.81 m/s^{2}

The value of g at surface of that satellite is 9.81 m/s^{2}.

The value of g us same for earth so that planet will be earth.

**Lakhmir Singh Manjit Kaur Class 9 Physics 3rd Chapter Solution (Second Part)**

**Very Short Answer Type Questions:**

**1.) Write the common unit of density.**

Answer: Liter is a common unit of density.

**2.) What is the density of water in SI units ?**

Answer: 1000 kg/m^{3} is the density of water in SI units.

**3.) What is the value of relative density of water ?**

Answer: 1 is a relative density of water.

**4.) Name the quantity whose one of the units is Pascal (Pa).**

Answer: Pressure is the quantity whose one of the unit is pascal (Pa).

** 5.) State the units in which pressure is measured.**

Answer: Pascal(Pa) or N/m^{3} is the unit of pressure

**6.) State whether the following statements are true or false :**

(a) The buoyant force depends on the nature of object immersed in the liquid.

(b) Archimedes’ principle can also be applied to gases.

Answer: (a) False, The buoyant force does not dependon the nature of object immersed in the liquid.

(b) True, Archimedes’ principle can also be applied to gases.

**7.) In which direction does the buoyant force on an object due to a liquid act ? What is the other name of buoyant force ?**

Answer: Inupward direction buoyant force act on an object. Buoyant force is also called as upthrust.

**8.) What is the other name of buoyant force?**

Answer: Upthrust is the other name of buoyant force?

**9.) Name the force which makes heavy objects appear light when immersed in a liquid.**

Answer: buoyantforce makes heavy objects appear light when immersed in a liquid because buoyant force exerts force on the object in upwards direction.

**10.) What is upthrust ?**

Answer: The upward force exerted by liquid in upwards direction is called up trust.

**11.) Name the principle which gives the magnitude of buoyant force acting on an object immersed in a liquid.**

Answer: Archimedes principle gives the magnitude of buoyant force acting on an object immersed in a liquid.

**12.) The relative density of mercury is 13.6. What does this statement mean ?**

Answer: Relative density gives the information about how much heavier is that substance by water. The relative density of mercury is 13.6 it means that mercury is 13.6 times heavier than water.

**13.) What name is given to thrust per unit area?**

Answer: Thrust per unit area is called pressure.

**14.) What is the scientific name of the ‘upward force’ acting on an object immersed in a liquid ?**

Answer: Buoyant force is acting on an object immersed in liquid.

**15.) What is meant by the term ‘buoyancy’ ?**

Answer: When an object immersed in water then water has tendency to exerts upward force, that tendency is called buoyancy.

**16.) What Causes buoyant force (or upthrust) on a boat ? **

Answer: The water exerts force on boat in upward direction and pushing up therefore boat can float on water.

**17.) Why does ice float in water ?**

Answer: The density of ice is smaller than density if water therefore ice float in water.

**18.) What force acting on an area of 0.5m ^{2} : Will produce a pressure of 500 Pa ?**

Answer: Given, area = 0.5m^{2} , pressure = 500 Pa.

As we know,

Pressure = Force/area

500 = Force/0.5

Force = 250 N.

The force required to produce pressure of 500 Pa on area of 0.5 m^{2} is 250 N.

**19.) An object Of weight 200 N is floating in a liquid. What is the magnitude of buoyant force acting on it ?**

Area: If downward force is equal to upward buoyant force then the object is float on liquid.

In this case downward force is weight so buoyant force is also 200 N.

**20.) Name the scientist who gave the magnitude of buoyant force acting on a solid object immersed in a-liquid.**

Answer: Archimedes gave the magnitude of buoyant force acting on a solid object immersed in a-liquid.

**21.) The density Of gold is 19 g/cm ^{3}. Find the volume of 95 g of gold.**

Answer: Given, Density = 19 g/cm^{3}

Mass= 95 g.

As we know,

density = mass /volume

19 = 95/ volume

Volume = 95/19

Volume = 5 cm^{3}

The volume of the gold is 5 cm^{3}.

**22.) What is the mass of 5 m ^{3} of cement of density 3000 kg/ m^{3} ?**

Answer: Given, Density 3000 kg/m^{3}

Volume = 5 m^{3}

As we know,

density = mass /volume

3000 = mass/ 5

Mass = 3000 × 5

Mass = 15000 kg

The mass of cement is 15000 kg.

**23.) What is the density of a substance of mass 100 g and volume 10 cm ^{3} ?**

Answer: Given, Mass = 100 g

Volume = 10 cm^{3}

As we know,

density = mass /volume

Density= 100/ 10

Density = 10 g/cm^{3}

The density of that substance is 10 g/cm^{3}.

**24.) Why does a block of wood held under water rise to the surface when released ?**

Answer: The density of Block of wood is less than water. The buoyant force on block of wood is greater than mass of that block therefore block of wood held under water rise to the surface when released.

**25.) The density of a body is 800 kg/m ^{3}. Will it sink or float when dipped in a bucket of water ? (Density of water = 1000 kg/m^{3}).**

Answer: Given,

Density of a body = 800 kg/m3. Density of water = 1000 kg/m^{3}.

The density of body is less than density of water so buoyant force is greater thus that body is float on water.

**Short Answer Type Questions**

27.) (a) What is the difference between the density and relative density of a substance ?

(b) If the relative density of a substance is 7.1, what will be its density in SI units ?

Answer: (a) Difference between density and relative density

I.) Density is the ratio of mass and volume but relative density is the ratio of density of substance to density of water.

II.) SI unit of density is kg/m3 but relative density has not unit.

III.) Relative density is derived from the concept of density.

IV.) Relative density gives the information about the substance is how much heavier than water of same volume.

V.) Density of water is 1000kg/m3. Relative density of water is 1.

(b) Given, relative density of substance = 7.1, we know the density of water = 1000 kg/m^{3}

As we know,

Relative density =density of substance/density of water

7.1 = density of substance/1000

Density of substance = 7.1 ×10^{3} kg/m^{3}

The density of that substance will be 7.1 ×10^{3} kg/m^{3}

**28.) Define thrust. What is its unit ?**

Answer: Thrust is nothing but force. The force acting on a body in perpendicular direction is called thrust.

SI unit of thrust is Newton.

**29.) A mug full of water appears light as long as it is under water in the bucket than when it is outside water. Why?**

Answer: When the object is immersed in water then the buoyant force is exert on the object by water in upward direction so the mug appears light as it is under water in the bucket.

**30.) What happens to the buoyant force as more and more volume of a solid object is immersed in a liquid ? When does the buoyant force become maximum ?**

Answer: If buoyant force is equal to volume of solid object immersed in liquid then liquid displaced as the amount of substance immersed in it. As the buoyant force as more and more volume of a solid object is immersed in a liquid then buoyant forces increases. But when object completely immersed in the liquid then buoyant force will become maximum.

**31.) Why do we feel light on our feet when standing in a swimming pool with water up to our armpits ?**

Answer: when we standing in a swimming pool with water up to our armpits then buoyant force exerts on our body and it push up us. So the value of weight decreases and we fell light.

**32.) Explain why, big boulders can be moved easily by flood.**

Answer: When big boulders sink into water of flood then buoyant force exerts on that boulders and its weight decreases and it can be moved easily.

**33.) An iron nail sinks in water but it floats in mercury. Why ?**

Answer: The density of nail is greater than water but smaller than mercury so iron nail sinks in water but it floats in mercury.

**34.) Explain why, a piece of glass sinks in water but it floats in mercury.**

Answer: The density of glass is greater than water but smaller than mercury so glass piece sinks in water but it floats in mercury.

**35.) Steel sinks in water but a steel boat floats. Why ?**

Answer: The density of steel is greater than water so steel rods sink in water. When air is filled in hollow steel rods it’s density decreases and boat floats on water.

**36.) Explain why, school bags are provided with wide straps to carry them..**

Answer: Gravitational force exerted on the bag in downward direction due to mass of the bag. The force exerts on the area. If bag has wide straps then the area increases hence pressure decrease. So it easy to carry the school bags at low pressure on shoulders.

**37.) Why does a sharp knife cut objects more effectively than a blunt knife ?**

Answer: Sharp knife has law area so force acting on that minimum area. And we know that pressure is inversely proportional to area. Due to low area pressure will be high. Thus the effect of force is also high. That is the reason behind sharp knife cut objects more effectively than a blunt knife.

**38.) Explain why, wooden (or concrete) sleepers are kept below the railway line.**

Answer: Railway lines has small area. Wooden or concrete sleepers increases the area of lines.The gravitational force of railway exerted in downward direction is applied on greater area so its effect is decreased. The railway lines does not sink into ground because of low pressure.

**39.) Explain why, a wide steel belt is provided over the wheels of an army tank.**

Answer: Wide steel belt has greater area. The army tank can work in desserts also. Due to large area its pressure reduces and army tank does not sink into ground.

**40.) Explain why, the tip of a sewing needle is sharp.**

Answer: By reducing the area pressure increases so needle can easily go into the cloth. Thus we can easily sew cloths by sharp needle.

**42.) Explain why, snow shoes stop you from sinking into soft snow.**

Answer: The area if show shoes is greater as compared to ordinary shoes so pressure by our body decreases by wearing show shoes. There fore we can not sink into soft snow.

**43.) Explain why, when a person stands on a cushion, the depression is much more than when he lies down on it.**

Answer: when a person lies down on cushion then it’s area increases as compared to standing. And we know that pressure is inversely proportional to area. Thus pressure reduces while lies down on the cushion.

**44.) Use your ideas about pressure to explain why it is easier to walk on soft sand if you have flat shoes rather than shoes with sharp heels.**

Answer: We know that pressure is inversely proportional to area. If shoes have sharp hills then it can sink into the sand. But the shoes having flat area then total weight is distributed hence pressure decreases so shoes do not sink into the sand.

**45.) Explain why, a nail has a pointed tip.**

Answer: Pressure is inversely proportional to area. Nails has negligible area as compared to its length. So the applied force act on that small area hence pressure increases. Thus the pin can easily go into the objects.

**46.) Explain why, buildings and dams have wide foundations.**

Answer: To distribute the total load of buildings and dams it has wide foundations. If the foundation of building and dams are not wide then it can easily bends due to weight of the building.

To reduce the pressure, buildings and dams have wide foundations.

**47.) Why does a ship made of iron and steel float in water whereas a small piece of iron sinks in it ?**

Answer: The density of steel and iron are greater than water so iron and steel piece sinks in it. But when we fill air into the hollow rods of steel and iron then its density decreases and it can be float on water. Ships can be made using this property. So ships made of iron and steel float in water.

**48.) Why do camels have large flat feet ?**

Answer: Camels are used to transport in desert area. The total weight of the Camels is distributed in the flat surface so pressure decreases as area increases. Therefore feet do not sink into the sands.

**49.) Name these forces :**

**(a) the upward push of water on a submerged object**

**(b) the force which wears away two surfaces as they move over one another (c) the force which pulled the apple off Isaac Newton’s tree. **

**(d) the force which stops you falling through the floor.**

**Answer: **

(a) the upward push of water on a submerged object- buoyant force.

(b) the force which wears away two surfaces as they move over one another- Frictional force.

(c) the force which pulled the apple off Isaac Newton’s tree- Gravitational force.

(d) the force which stops you falling through the floor- normal force exerted by floor (reaction force).

**50.) A pressure of 10 Pa acts on an area of 3.0 m ^{2}. What is the force acting on the area ? What force will be exerted by the application of same pressure if the area is made one-third?**

Answer: Given, pressure = 10 Pa, area = 3.0 m^{2}.

Ad we know,

Pressure = Force/ area

Force = pressure × area.

Force = 10 × 3

Force = 30 N.

If the area becomes 1/3^{rd}then New area = 1/3 × 3

New area = 1 m^{2} .

Force = pressure × area.

Force = 10 × 1

Force = 10 N.

**51.) A girl is wearing a pair of flat shoes. She weighs 550 N. The area of contact of one shoe with the ground is 160 cm2. What pressure will be exerted by the girl on the ground :**

**(a) if she stands on two feet ?**

**(b) if she stands on one foot ?**

Answer: Given, downward force = 550 N,

Area of one shoe = 160 cm^{2}

Area = 160 × 10^{-4}m^{2}

As we know,

Pressure = Force/ area

(a) If she stand on two feet then area becomes double,

Therefore area = 2 × 160 × 10^{-4}

Area = 320× 10^{-4}

Pressure = 550/320 ×10^{-4}

Pressure = 1.71875× 10^{4}Pa

Pressure = 17187.5 Pa.

If she stand on two feet then pressure is 17187.5 Pa

(b) If she stand in one foot then area becomes half and pressure becomes double.

Pressure = 2 × initial pressure

Pressure = 2 × 17187.5

Pressure = 34375 Pa.

If she stand on one foot then pressure is 34375 Pa.

**52.) Calculate the density of an object of volume 3 m ^{3} and mass 9 kg. State whether this object will float or sink in water. Give reason for your answer.**

Answer: Given, Mass = 9 kg

Volume = 3 m^{3}

As we know,

density = mass /volume

Density=9/3

Density = 3 kg/m^{3}

The density of that substance is 3 kg/m^{3}.

The density of this substance is smaller than water so it float on water.

**53.) An object weighs 500 grams in air. This object is then fully immersed in water. State whether it will weigh less in water or more in water. Give reason for your answer.**

Answer: The weight of an object in air 50p grams. I’d that object is fully immersed in water then water exert buoyant force on that object. Thus the weight of the object becomes less than in air.

**54.) (a) Write down an equation that defines density.**

**(b) 5 kg of material A occupy 20 cm ^{3} whereas 20 kg of material B occupy 90 cm^{3}. Which has the greater density : A or B ? Support your answer with calculations.**

Answer: (a) Density is the ratio of mass and volume and it is mathematically expressed as,

Density = mass / volume.

(b) Given mass of object A is 20 kg, and

volume = 20 cm^{3}

Volume = 20 × 10^{-4} m^{3}

As, density = mass /volume

Density = 20/ 20 × 10^{-4}

Density = 10^{4} N/ m^{3}

The density of object A = 10^{4} N/ m^{3}————-1

Now we calculate density of object B,

Mass of object B = 20 kg, and volume = 90 cm^{3}

Volume = 90 × 10^{-4} m^{3}

As, density = mass /volume

Density = 20/ 90× 10^{-4}

Density = 0.22 ×10^{4} N/ m^{3}————2

The density of object B= 0.22× 10^{4} N/ m^{3}

From equation 1 and equation 2 we conclude that

Density of object A is greater than density of object B.

**Long Answer Type Questions**

**55.) (a) Define buoyant force. Name two factors on which buoyant force depends.**

**(b) What is the cause of buoyant force ?**

**(c) When a boat is partially immersed in water, it displaces 600 kg of water. How much is the buoyant force acting on the boat in newtons ? (g = 10 m s ^{-2})**

Answer:

(a) Buoyant force: When an object is immersed in liquid then upward force acting on the object by liquid is called buoyant force.

Buoyant force is depend on volume of object which immersed in liquid and density of that liquid.

(b) We can understand the cause of buoyant force with the help of following figure.

Fill the water into bucket. Immerse a plastic ball in it as shown in figure.

The molecules of waters exerts force on ball in all directions. As we know that buoyant force is depend on the depth. In upper side of ball the depth is low as compared to lower part of the ball. So on lower side, water exert greater force on ball as compared to upper side. So there will be resultant force on the ball in upward direction. This is the cause of buoyant force.

(c) Given, Mass of water = 600 kg, g = g = 10 m s^{-2}

According to Archimedes principle, buoyant force is equal to displacement of water when object immersed in it.

In mathematical form,

Downward force = upward force.

Downward force is weight and upward force is buoyant force.

Weight = m ×g

Weight = 600 × 10

Weight 6000 N.

So the weight of the object is equal to buoyant force. Thus the value of buoyant force is 6000 N.

**56.) (a) State the principle of flotation.**

**(b) A floating boat displaces water weighing 6000 newton.**

**(i) What is the buoyant force on the boat ?**

**(ii) What is the weight of the boat ?**

Answer: (a) The principle of flotation: If the density of object which immersed in liquid is equal to or less than density of that liquid then the object will float in liquid.

(b) Given, Weight of the liquid = 6000 N.

(i) According to Archimedes principle, the buoyant force is equal to displacement of water after immersing the object in it. When a boat immersed in water then it can displaced water of 6000 N

Therefore buoyant force is also 6000 N.

(ii) According to Archimedes principle, the weight of boat also 6000 N.

**57.) (a) Define density. What is the SI unit of density ?**

**(b)Define relative density. What is the Sl unit of relative density ?**

**(c)The density of turpentine is 840 kg/m ^{3}. What will be its relative density ? (Density of water = 1000 kg/m^{3})**

Answer: (a) Density: Density is defined as the ratio of mass of the object to volume of that object.

Mathematically it is given as,

Density = mass/ volume.

SI unit of density is kg/m^{3} .

(b) Relative density: Relative density is derived form density. It is defined as the ratio of density of substance to the density of water.

Mathematically it is given as,

Relative density = density of substance/ density of water.

Relative density = (Mass of substance/ volume of substance) × (volume of water/mass of water)

Relative density = mass of the substance / mass of an equal volume of water.

Relative density is also defined as ratio ofmass of the substance to mass of equal volume of water.

It is the ratio of same quantity so it has not unit.

(c) Given, density of turpentine is 840 kg/m^{3}, Density of water = 1000 kg/m^{3},

As we know

Relative density = density of substance/ density of water.

Relative density = 840/1000

Relative density = 0.84

The relative density of turpentine is 0.84.

**58.) (a) Define pressure.**

**(b) What is the relation between pressure, force and area ?**

**(c) Calculate the pressure when a force of 200 N is exerted on an area of :**

**(i) 10 m ^{2}**

**(ii) 5 m ^{2}**

Answer: (a) Pressure is defined as force acting perpendicular to the unit area.

(b) Thepressure is the ratio of force upon area.

It is mathematically written as,

Pressure = force/ area.

This equation gives the relationship between pressure, force and area.

(c) Given, Force = 200 N,

i) Area = 10 m^{2}.

As we know,

Pressure = force/ area.

Pressure = 200/ 10

Pressure= 20 N/m^{2} .

Force acting on 10 m^{2} is 20 pa.

ii) Area = 5 m^{2}.

As we know,

Pressure = force/ area.

Pressure = 200/ 5

Pressure= 40 N/m^{2 }.

Force acting on 5 m^{2} is 40 pa.

**59.) (a) What are fluids ? Name two common fluids.**

**(b) State Archimedes’ principle.**

**(c) When does an object float or sink when placed on the surface of a liquid ?**

Answer: (a) The substances which can flows easily are called as fluids or fluid is the combination of solid and liquids.

Air and water are most common example of fluid.

(b) Archimedes principle states that when an object completely immersed in a liquid then it displaced a liquid is equal to weight of that object.

(c) If the average density of the object is less than or equal to density of liquid then that objects float on the liquid.

But If the average density of the object is greater than the density of liquid then that objects sink in the liquid.

**60.) (a) How does a boat float in water ?**

**(b) A piece of steel has a volume of 12 cm ^{3}, and a mass of 96 g. What is its density :**

(i) in g/cm^{3} ?

(ii) in kg/m^{3} ?

Answer: (a) As we know that density of iron and steel is greater than density of water hence iron and steel sink into water. But if we fill air into the pipes of steel rod then it’s average density decrease and it can float on water. This phenomenon is used for making boats. So boat floats on water

(b) Given, volume of piece of steel = 12 cm^{3}, mass = 96 g.

We have to find density of piece of steel.

i) as we know,

Density = mass / volume

Density = 96/12

Density = 8 g / cm^{3}

The density of piece of steel is 8 g / cm^{3}

ii) The density of steel is 8 g / cm^{3}

It is in g/ cm3 . We have to convert it into kg/m3 .

8 g / cm^{3}= 8 × 10^{ -3} / 10^{-6}kg/m^{3}

8 g / cm^{3}= 8 × 10^{3 }kg/m^{3}

The density of steel in kg/m^{3}is 8 × 10^{3 }kg/m^{3}

**61.) An elephant weighing 40,000 N stands on one foot of area 1000 cm ^{2} whereas a girl weighing 400 N is standing on one ‘stiletto’ heel of area 1 cm^{2}.**

**(a) Which of the two, elephant or girl, exerts a larger force on the ground and by how much ?**

**(b) What pressure is exerted on the ground by the elephant standing on one foot ?**

**(c) What pressure is exerted on the ground by the girl standing on one heel ?**

**(d) Which of the two exerts larger pressure on the ground : elephant or girl ?**

**(e) What is the ratio of pressure exerted by the girl to the pressure exerted by the elephant?**

Answer: (a)Elephant has greater mass so it exert larger force,

Force exerted by elephant = 40000 N.

Force exerted by girl = 400 N.

Difference in force = Force exerted by elephant – Force exerted by girl.

Difference in force = 40000 – 400

Difference = 39600 N.

Elephant exerts a larger force on the ground and by 39600 N.

(b) Given,

Force exerted by elephant = 40000 N.

Area on elephant exert force = 1000cm^{2}

Area on elephant exert force=1000× 10^{-4}m^{2}

Area on elephant exert force = 0.1 m^{2}

As we know, Pressure = force / area

Pressure = 40000/0.1

Pressure = 400000 N/m^{2} .

The force exerted by elephant is 400000 N/m^{2}

(c) Given,

Force exerted by girl= 400 N.

Area on girl exert force = 1 cm^{2}

Area on elephant exert force=1× 10^{-4 }m^{2}

As we know, Pressure = force / area

Pressure = 400/10^{-4}

Pressure = 4000000 N/m^{2} .

The force exerted by girl is 4000000 N/m^{2}

(d) The pressure exerted by elephant is 400000 N/m^{2}and The force exerted by girl is 4000000 N/m^{2}.

So girl exert more Pressure than elephant.

(e) The pressure exerted by elephant is 400000 N/m^{2}and The force exerted by girl is 4000000 N/m^{2}.

The ratio of pressure exerted by the girl to the pressure exerted by the elephant

= (4000000 N/m^{2})/(400000 N/m^{2} )

= 10

The ratio of pressure exerted by the girl to the pressure exerted by the elephant is 10:1.

**For more Chapter solution, click below**