CBSE Class 9

These NCERT Solutions for Class 9 Science Chapter 11 Work and Energy Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Work and Energy NCERT Solutions for Class 9 Science Chapter 11

Class 9 Science Chapter 11 Work and Energy InText Questions and Answers

Question 1.
A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 11.3). Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Answer:
When a force F acts on an object to displace it through a distance S in its direction, then the work done W on the body by the force is given by:
Work done = Force x Displacement
W = F × S
Where,
F = 7N
S = 8m
Therefore, work done, W = 7 × 8
= 56 Nm
= 56 J

Question 2.
When do we say that work is done?
Answer:
Work is done whenever the given conditions are satisfied:

• A force acts On the body.
• There is a displacement of the body caused by the applied force along the direction of the applied force.

Question 3.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:
When a force F displaces a body through a distance S in the direction of the applied force, then the work done W on the body is given by the expression:
Work done = Force × Displacement
W = F × s

Question 4.
Define 1 J of work.
Answer:
1 J is the amount of work done by a force of 1 N on an object that displaces it through a distance of 1 m in the direction of the applied force.

Question 5.
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?
Answer:
Work done by the bullocks is given by the expression:
Work done = Force × Displacement
W = F × d
Where,
Applied force, F = 140 N
Displacement, d = 15m
W = 140 × 15 = 2100 J
Hence, 2100 J of work is done in ploughing the length of the field.

Question 6.
What is the kinetic energy of an object?
Answer:
Kinetic energy is the energy possessed by a body by the virtue of its motion. Every moving object possesses kinetic energy. A body uses kinetic energy to do work. Kinetic energy of hammer is used in driving a nail into a log of wood, kinetic energy of air is used to run windmills, etc.

Question 7.
Write an expression for the kinetic energy of an object.
Answer:
If a body of mass ‘m’ is moving with a velocity v, then its kinetic energy E_k is given by the expression,
E_k = \(\frac {1}{2}\)mv²
Its SI unit is Joule (J j.

Question 7.
The kinetic energy of an object of mass, m moving with a velocity of 5 m s^-1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer:
Expression for kinetic energy is
E_K = \(\frac {1}{2}\)mv²
m = Mass of the object
v = Velocity of the object = 5ms^-1
Given that kinetic energy, E_K = 25 J
(i) If the velocity of an object is doubled, then v = 5 × 2 = 10 ms^-1.
Therefore, its kinetic energy becomes 4 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 4 = 100 J.
(ii) If velocity is increased three times, then its kinetic energy becomes 9 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 9 = 225 J.

Question 8.
What is power?
Answer:
Power is the rate of doing work or the rate of transfer of energy. If W is the amount of work done in time t, then power is given by the expression,
Power = \(\frac{\text { Work }}{\text { Time }}=\frac{\text { Energy }}{\text { Time }}\)
P = \(\frac{\mathrm{W}}{\mathrm{T}}\)
It is expressed in watt (W).

Question 9.
Define 1 watt of power:
Answer:
A body is said to have power of 1 watt if it does work at the rate of 1 joule in 1 s, i.e.,
1W = \(\frac{1 \mathrm{~J}}{1 \mathrm{~s}}\)

Question 10.
A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer:
Power is given by the expression,
Work done = \(\frac{\text { Workdone }}{\text { Time }}\)
Work done = Energy consumed by the lamp = 1000 J
Time = 10 s
Power = \(\frac{1000}{10}\) = 100J S^-1 = 100 W

Question 11.
Define average power.
Answer:
A body can do different amount of work in different time intervals. Hence, it is better to define average power. Average power is obtained by dividing the total amount of work done in the total time taken to do this work.
Average Power = \(\frac{\text { Total work done }}{\text { Total time taken }}\)

Class 9 Science Chapter 11 Work and Energy Textbook Questions and Answers

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.

• Suma is swimming in a pond.
• A donkey is carrying a load on its back.
• A windmill is lifting water from a well.
• A green plant is carrying out photosynthesis.
• An engine is pulling a train.
• Food grains are getting dried in the sun.
• A sailboat is moving due to wind energy.

Answer:
Work is done whenever the given two conditions are satisfied:
(i) A force acts on the body.
(ii) There is a displacement of the body by the application of force in or opposite to the direction of force.

(a) While swimming, Suma applies a force to push the water backwards. Therefore, Suma swims in the forward direction caused by the forward reaction of water. Here, the force causes a displacement. Hence, work is done by Seema while swimming.

(b) While carrying a load, the donkey has to apply a force in the upward direction. But, displacement of the load is in the forward direction. Since, displacement is perpendicular to force, the work done is zero.

(c) A windmill works against the gravitational force to lift water. Hence, work is done by the windmill in lifting water from the well.

(d) In this case, there is no displacement of the leaves of the plant. Therefore, the work done is zero.

(e) An engine applies force to pull the train. This allows the train to move in the direction of force. Therefore, there is a displacement in the train in the same direction. Hence, work is done by the engine on the train.

(f) Foodgrains do not move in the presence of solar energy. Hence, the work done is zero during the process of food grains getting dried in the Sun.

(g) Wind energy applies a force on the sailboat to push it in the forward direction. Therefore, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.

Question 2.
An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
Work done by the force of gravity on an object depends only on vertical displacement. Vertical displacement is given by the difference in the initial and final positions/heights of the object, which is zero.
Work done by gravity is given by the expression,
W = mgh
Where,
h = Vertical displacement = 0
W = mg × 0 = 0 J
Therefore, the work done by gravity on the given object is zero joule.

Question 3.
A battery lights a bulb. Describe the energy changes involved in the process.
Answer:
When a bulb is connected to a battery, then the chemical energy of the battery is transferred into electrical energy. When the bulb receives this electrical energy, then it converts it into light and heat energy. Hence, the transformation of energy in the given situation can be shown as:
Chemical Energy → electrical Energy → Light Energy + Heat energy

Question 4.
Certain force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms-1. Calculate the work done by the force.
Answer:
Kinetic energy is given by the expression,
(Ek)v = \(\frac {1}{2}\)mv2
Where,
Ek = Kinetic energy of the object moving with a velocity, v
m = Mass of the object
(i) Kinetic energy when the object was moving with a velocity 5 m s-1
(Ek)5 = \(\frac {1}{2}\) × 20 × (5)2 = 250J
ii. Kinetic energy when the object was moving with a velocity 2 ms-1
(Ek)2= \(\frac {1}{2}\) × 20 × (2)2 = 40J
Work done by force is equal to the change in kinetic energy.
Therefore, work done by force = (Ek)2 – (Ek)5
= 40 – 250 = -210 J
The negative sign indicates that the force is acting in the direction opposite to the motion of the object.

Question 5.
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer:
Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body. Therefore, work done by gravity is given by the expression,
W = mgh
Where,
Vertical displacement, h = 0
∴ W = mg × 0 = 0
Hence, the work done by gravity on the body is zero.

Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer:
No. The process does not violate the law of conservation of energy. This is because when the body falls from a height, then its potential energy changes into kinetic energy progressively. A decrease in the potential energy is equal to an increase in the kinetic energy of the body. During the process, total mechanical energy of the body remains conserved. Therefore, the law of conservation of energy is not violated.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle?
Answer:
While riding a bicycle, the muscular energy of the rider gets transferred into heat energy and kinetic energy of the bicycle. Heat energy heats the rider’s body. Kinetic energy provides a velocity to the bicycle. The transformation can be shown as:
Muscular Energy → Kinetic Energy + Heat Energy
During the transformation, the total energy remains conserved.

Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer:
When we push a huge rock, there is no transfer of muscular energy to the stationary rock. Also, there is no loss of energy because muscular energy is transferred into heat energy, which causes our body to become hot.

Question 9.
A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer:
1 unit of energy is equal to 1 kilowatt hour (kWh).
1 unit = 1 kWh
1 kWh = 3.6 × 10⁶ J
Therefore, 250 units of energy = 250 × 3.6 × 10⁶ = 9 × 10⁸ J

Question 10.
An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.
Answer:
Gravitational potential energy is given by the expression,
W = mgh Where,
h = Vertical displacement = 5 m
m = Mass of the object = 40 kg
g = Acceleration due to gravity = 9.8 ms^-2
∴ W = 40 × 5 × 9.8 = 1960 J.
At half-way down, the potential energy of the object will be \(\frac{1960}{2}\) = 980 J.
At this point, the object has an equal amount of potential and kinetic energy. This is due to the law of conservation of energy. Hence, half-way down, the kinetic energy of the object will be 980 J.

Question 11.
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer:
Work is done whenever the given two conditions are satisfied:

• A force acts on the body.
• There is a displacement of the body by the application of force in or opposite to the direction of force.

If the direction of force is perpendicular to displacement, then the work done is zero.
When a satellite moves around the Earth, then the direction of force of gravity on the satellite is perpendicular to its displacement. Hence, the work done on the satellite by the Earth is zero.

Question 12.
Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Answer:
Yes. For a uniformly moving object.
Suppose an object is moving with constant velocity. The net force acting on it is zero. But, there is a displacement along the motion of the object. Hence, there can be a displacement without a force.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer:
Work is done whenever the given two conditions are satisfied:

• A force acts on the body.
• There is a displacement of the body by the application of force in or opposite to the direction of force.

When a person holds a bundle of hay over his head, then there is no displacement in the bundle of hay. Although, force of gravity is acting on the bundle, the person is not applying any force on it. Hence, in the absence of force, work done by the person on the bundle is zero.

Question 14.
An electric heater is rated 1500 W. How much energy does if use in 10 hours?
Answer:
Energy consumed by an electric heater can be obtained with the help of the expression,
P = \(\frac{W}{T}\) Where,
Power rating of the heater, P = 1500 W = 1.5 kW
Time for which the heater has operated, T = 10 h
Work done = Energy consumed by the heater
Therefore, energy consumed = Power × Time
= 1.5 × 10 = 15 kWh
Hence, the energy consumed by the heater in 10 h is 15 kWh.

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer:
The law of conservation of energy states that energy can be neither created nor destroyed. It can only be converted from one form to another.

Consider the case of an oscillating pendulum.

When a pendulum moves from its mean position P to either of its extreme positions A or B, it rises through a height h above the mean level P. At this point, the kinetic energy of the bob changes completely into potential energy. The kinetic energy becomes zero, and the bob possesses only potential energy. As it moves towards point P, its potential energy decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches point P, its potential energy becomes zero and the bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates.

The bob does not oscillate forever. It comes to rest because air resistance resists its motion. The pendulum loses its kinetic energy to overcome this friction and stops after some time.

The law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surroundings. Hence, the total energy of the pendulum and the surrounding system remain conserved.

Question 16.
An object of mass, m is moving with a constant velocity, v. How much work should be done oh the object in order to bring the object to rest?
Answer:
Kinetic energy of an object of mass, m moving with a velocity, v is given by the expression,
E_k = \(\frac {1}{2}\)mv²
To bring the object to rest, \(\frac {1}{2}\)mv² amount of work is required to be done on the object.

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Answer:
Kinetic energy, Ek = \(\frac {1}{2}\)mv²
Where,
Mass of car, m = 1500 kg
Velocity of car, v = 60 km/h = 60 × \(\frac {5}{18}\)ms-1
∴ Ex = \(\frac {1}{2}\) × 1500 × \(\left(60 \times \frac{5}{18}\right)^{2}\) =20.8 × 104J
Hence, 20.8 × 104 J of work is required to stop the car.

Question 18.
In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

Answer:
Work is done whenever the given two conditions are satisfied:

• A force acts on the body.
• There is a displacement of the body by the application of force in or opposite to the direction of force.

Case I:

In this case, the direction of force acting on the block is perpendicular to the displacement.
Therefore, work done by force on the block will be zero.

Case II:
In this case, the direction of force acting on the block is in the direction of displacement. Therefore, work done by force on the block will be positive.

In this case, the direction of force acting on the block is in the direction of displacement.
Therefore, work done by force on the block will be positive.

Case III:

In this case, the direction of force acting on the block is opposite to the direction of displacement. Therefore, work done by force on the block will be negative.

Question 19.
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
Acceleration in an object could be zero even when several forces are acting on it. This happens when all the forces cancel out each other i. e., the net force acting on the object is zero. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.

Question 20.
Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.
Answer:
Energy consumed by an electric device
can be obtained with the help of the expression for power,
P = \(\frac{W}{T}\)
Where,
Power rating of the device, P = 500 W = 0.50 kW
Time for which the device runs, T = 10 h
Work done = Energy consumed by the device
Therefore, energy consumed = Power × Time
= 0.50 × 10 = 5 kWh
Hence, the energy consumed by four equal rating devices in 10 h will be 4 × 5 kWh = 20 kWh = 20 Units.

Question 21.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer:
When an object falls freely towards the ground, its potential energy decreases and kinetic energy increases. As the object touches the ground, all its potential energy gets converted into kinetic energy. As the object hits the hard ground, all its kinetic energy gets converted into heat energy and sound energy. It can also deform the ground depending upon the nature of the ground and the amount of kinetic energy possessed by the object.

Class 9 Science Chapter 11 Work and Energy Additional Important Questions and Answers

Multiple Choice Questions
Choose the correct option:

Question 1.
When a body falls freely towards the earth, then its total energy
(a) increases
(b) decreases
(c) remains constant
(d) first increases and then decreases
Answer:
(c) remains constant

Question 2.
A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car
(a) does not change
(b) becomes twice to that of initial
(c) becomes 4 times that of initial
(d) becomes 16 times that of initial
Answer:
(a) does not change

Question 3.
In case of negative work the angle between the force and displacement is
(a) 00
(b) 450
(c) 900
(d) 1800
Answer:
(d) 1800

Question 4.
An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
(a) acceleration
(b) momenta
(c) potential energy
(d) kinetic energy
Answer:
(a) acceleration

Question 5.
A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g =10 ms^-2)
(a) 6 × 103 J
(b) 6J
(c) 0.6 J
(d) zero
Answer:
(d) zero

Question 6.
Which one of the following is not the unit of energy?
(a) joule
(b) newton metre
(c) kilowatt
(d) kilowatt hour
Answer:
(c) kilowatt

Question 7.
The work done on an object does not depend upon the
(a) displacement
(b) force applied
(c) angle between force and displacement
(d) initial velocity of the object
Answer:
(d) initial velocity of the object

Question 8.
Water stored in a dam possesses
(a) no energy
(b) electrical energy
(c) kinetic energy
(d) potential energy
Answer:
(d) potential energy

Question 9.
A body is falling from a height h. After it has fallen a height h/2, it will possess
(a) only potential energy
(b) only kinetic energy
(c) half potential and half kinetic energy
(d) more kinetic and less potential energy
Answer:
(c) half potential and half kinetic energy

Very Short Answer Type Questions

Question 1.
A boy climbs up the staircase to reach to second, floor from first floor. Has he done any work?
Answer:
Yes, he has done work against the gravitational force by climbing up.

Question 2.
In the following activities, has work been done or not? Why?
(i) A girl pulls a trolly and trolley moves.
(ii) A book is lifted through a height ‘h’.
Answer:
(i) When a girl pulls a trolley, work is done because trolley experiences a displacement from its mean position.
(ii) When a book is lifted through a height ‘h’ because book has gained displacement in its height from the ground level.

Question 3.
Is work a scalar or vector quantity?
Answer:
Although work done = Force × Displacement, yet it is taken as scalar quantity.

Question 4.
What is the work done in lifting an object of mass ‘m’ through a height ‘h’ from the ground?
Answer:
Force acting on body, F = m × g
Displacement = h.
Therefore, workdone = Force × Displacement
= mg × h = mgh

Question 5.
What is the angle between the line of force and the direction of displacement for maximum work?
Answer:
With W = F.s. cos θ, θ = 0 for the maximum work.

Question 6.
Define energy. What is the S.I. unit?
Answer:
Energy is the capacity to do work. Its S.I. unit is joule.

Question 7.
Which physical quantity has the unit Js^-1 ?
Answer:
Power.

Question 8.
Which of the two out of work, power and energy have same unit?
Answer:
Work and energy.

Question 9.
What are the two forms of mechanical energy?
Answer:
The two forms of mechanical energy are kinetic energy and potential energy.

Question 10.
State the relation in kinetic energy of a body in motion with respect to its (i) mass and (ii) velocity.
Answer:

• Kinetic energy of a body is directly proportional to its mass.
• Kinetic energy of a body is directly proportional to the square of its velocity.

Question 11.
State the relation in potential energy of a body at rest with respect to its mass and height.
Answer:
Potential energy of a body is directly proportional to both the mass and height.

Question 12.
What is the relationship between Watt and Horsepower?
Answer:
1 Horsepower = 746 Watt.

Question 13.
The work done by the heart for each beat is 1J. Calculate the power of the heart if it beats 72 times in a minute.
Answer:
Total work done in minute = 1J × 72 = 72J
Time = 1 minute = 1 × 60 = 6 seconds
Power = \(\frac{\text { Work done }}{\text { Time }}\) = \(\frac {72}{60}\) =1.2 Watt

Question 14.
What type of energy transformation takes place in (i) dynamo and (ii) Electric motor?
Answer:
(i) Dymamo—Mechanical energy into electrical energy,
(ii) Electric motor—Electrical energy into mechanical energy

Question 15.
Which type of energy is possessed by stretched rubber band?
Answer:
Elastic potential energy

Question 16.
A rubber ball and a leather cricket ball are moving with same velocity. Which will have greater kinetic energy? Why ?
Answer:
The leather cricket ball will have greater kinetic energy because of its large mass as compared to rubber ball.

Question 17.
Which will require more force to be pushed forward, a truck in rest or a slowly rolling truck?
Answer:
A truck in state of rest would require more force because of greater static friction than rolling friction.

Question 18.
What happens to kinetic energy of a body in motion when it comes to rest?
Answer:
Kinetic energy of body-in motion gets transformed into its potential energy when brought to rest while some is lost in overcoming friction as heat.

Short Answer Type Questions

Question 1.
A student when preparing for examinations draws diagram, read hooks and discuss problems with his friends. In the process does he do any work?
Answer:
No work is done by the student in reading or discussing problems with his friends as long as he remains sitting at one position to avoid any displacement.

Question 2.
What is work, a scalar or vector quantity? why?
Answer:
Work is a scalar quantity because on the application of force, the work done is in the direction of force. Hence, there is no change in the direction for work to be called a vector quantity.

Question 3.
Derive an expression for the work done on an object when the force is acting on the body in a direction making an angle ‘θ’ with the direction of displacement of the object.
Answer:
Let a force F acts on a body of mass ‘m’ by making an angle ‘θ’ with the direction of the displacement of the object.

The component of the force in direction of the displacement = F cos θ
Let displacement of body = s
Work done on the object = (F cos θ) × s = F × s cos θ

Question 4.
What is the workdone against force of gravity if a person carries a 30 kg load on his head?
Answer:
Work done against force of gravity W = F. s. cos θ
With θ = 90° and cos θ = 0
W = 0 × s = 0
Hence, total work done will be zero.

Question 5.
What do you mean by negative work and positive work? Give example.
Answer:
When a force acts on a body in a direction opposite to the direction of displacement
Work done = – F cos θ × s
The negative sign indicates that work done is negative.

When a force acts on a body in a direction of the displacement of body, the work done is positive.
When a bullock pulls a cart, the work done is called positive while when a player tries to stop a moving ball, the work done is called negative.

Question 6.
If a ball tied to a string is whirled in a circular orbit with centre being the hand holding the free end of string, then what is the work done on the ball?
Answer:
The force acting on the ball is along the radius of the circular path i.e. the direction of the force is perpendicular to the direction of motion of the ball.
Therefore, work done,
W = F. cos θ × s
= O × s = 0

Hence, work done on the ball is zero.

Question 7.
Take a toy car, wind it using key. Place the car on the ground and answer the following questions:
(i) Did the car move? Why?
(ii) Does the energy acquired by the car depends upon the number of times you turn the key?
(iii) From where did it get energy?
Answer:
(i) The car would move because the potential energy acquired by the car, due to widing of the spring of car got transformed into its kinetic energy.

(ii) The potential energy gained by the car depends upon the number of turns of the widing of the key with increasing number of turns, the greater potential energy is acquired by the car. This acquired potential energy later changes into car’s kinetic energy.

(iii) Car got the energy to move freon the spring inside it.

Question 8.
Two bodies having equal masses are kept at a height of 20m and 30m respectively. What is the ratio of their potential energy?
Answer:
Ratio of potential energies = mgh₁ : mgh₂
mg × 20 : mg × 30
= 20 : 30
= 2 : 3.

Question 9.
State the type of energy changes taking place in the following:
(i) A body is thrown up in air.
(ii) In a green leaf during photosynthesis.
(iii) In an oscillation of a pendulum.
Answer:
(i) When a body is thrown in upward direction, the muscular energy which is chemical potential energy is transformed into the kinetic energy of the sail. As the ball keeps rising, kinetic energy continues to change into potential energy. When at the maximum height, the body has no kinetic energy but the maximum potential energy.

(ii) In a leaf during photosynthesis, solar energy changes into chemical potential energy.

(iii) An oscilating pendulum has the maximum potential energy at its peak position where it temporarily come to state of rest while at its mean position, it has maximum kinetic energy.

Question 10.
Suppose a hammer which falls purely on a nail placed on a wood has a mass of 1 kg. If it falls from a height of 1m, how much kinetic energy will it have just before hitting the nail?
Answer:
Mass of the hammer = 1 kg
Height of the hammer, h = 1m
Acceleration due to gravity, g = 10 ms^-2
Potential energy at the highest point = mgh = 1 × 10 × 1 = 10 J
When the hammer falls, its whole potential energy would be converted into kinetic energy.
Kinetic energy just before hitting the nail 10 J.

Question 11.
A man whose mass is 50 kg climbs up 30 stairs in 30 seconds. If each step is 20 cm high. Calculate the power used in climbing the stairs.
Answer:
Mass of the man, m = 50 kg
Number of the stairs = 30
Height of one stair = 20 cm = 0.20 m
Total height of stairs = 30 × 20 = 6 m
Workdone in climbing the stairs=m × g × h
=50 × 10 × 6 = 3000 J
Tune taken by man to climb up 30 stairs =30
Power used inclimbing stairs = \(\frac {3000}{30}\) = 100 w

Question 12.
A porter lifts a luggage of 15 kg from the ground and put it on his head 2 m above the ground. Calculate the work done by him on the luggage Tig = 10 ms^-2)
Answer:
Mass of luggage, m = 15 kg
Height or displacement, h = 2m
Work done by porter = m.g × h
= 15 × 10 × 2 = 300 J

Question 13.
A force of 10 N acts on an object in the direction of displacement. If the displacement of the object is 2 m. Calculate the work done by the force.
Answer:
Force applied, F = 10 N.
Displacement, s = 2 m
Work done = F × S = 10 × 2 = 20 N.

Question 14.
An object of mass 15 kg is moving with a constant velocity of 4 ms^-1. What is the kinetic energy possessed by the body?
Answer:
Mass of the object, m = 15 kg.
Velocity of the object, v = 4 ms^-1
Kinetic energy possessed by the body = ½ mv².
= ½ × 15 × (4)² = ½ × 15 × 16= 120 J

Question 15.
What is the work done to increase the velocity of a car from 30 km h^-1 to 60 km h^-1 if the mass of the car is 1500 kg ?
Answer:
Mass of the car, m = 1500 kg

Work done = change in kinetic energy
= ½ m (v² – u²)

Question 16.
Find the energy present in any object of mass 10 kg when it is at a height of 6m above the ground. (Take g = 10 ms^-2).
Answer:
Mass of the object, m = 10 kg
Height, h = 6 m
g = 10 ms^-2
Potential energy possessed by the object = mg × h
= 10 × 10 × 6 = 600 J

Question 17.
An object of mass 12 kg is at a certain height above the ground. If the gravitational potential energy of the object is 480 J. Find the height at which the object is with respect to the ground, (Give g =10 ms^-1).
Answer:
Mass of the object, m = 12 kg
Potential energy, P.E. = 480 J
Let height of the object from ground = h
Then P.E. = m × g × h
h = \(\frac{\mathrm{P.E}}{m \times g}=\frac{480}{12 \times 10}=4 m\)

Question 18.
A woman pulls a bucket of water of total mass 5 kg from a well which is 10 m deep in 10s. Calculate the power used by her. (Take g = 10 ms^-2).
Answer:
Mass of the bucket, m = 5 kg
Height, h = 10 m
g = 10 ms^-2
Energy consumed in pulling the bucket = m × g × h
= 5 × 10 × 10 = 500 J
Time taken = 10s

Question 19.
Two girls A and B each of weight 400 N climb up a rope to a height of 8m. Girl A takes 20 seconds while girl B takes 50 seconds to accomplish their task. What is the power spent by each girl?
Answer:
(i) Weight of the girl ‘A’ = 400 N.
Displacement (height) h = 8m
Work done by girl A = mg × h
=400 × 8 = 3200 J
Time taken by girl A, t = 20 seconds
Power spent by girl A,
= \(\frac{\text { Workdone }}{\text { Time taken }}=\frac{3200}{20}\) = 160 W

(ii) Weight of the girl B, = 400 N
Displacement (height), h = 8 m
= 400 × 8 = 3200 J
Time taken, t = 50 seconds
Power spent by girl B
= \(\frac{\text { Workdone }}{\text { Time taken }}=\frac{3200}{20}\) = 64 W

Question 20.
A boy of 50 kg runs up a staircase of 45 steps in 9 seconds. If the height of each step is 15 cm. Calculate the power of the boy.
Answer:
Mass of the boy, m = 50 kg
Height of each step = 15 cm = 0.15 m
Number of steps in stair case = 45
Total height of stair case = 45 × 0.15 = 6.75 m
Time taken to climb, t = 9 seconds

Question 21.
A boy pulls a toy car with a force of SON through a string which makes an angle of 30° with the horizontal so as to move the toy 2 m horizontally. Calculate the work done by the boy on the toy car.
Answer:
Force applied by the boy = F = SON
Angle of direction of force with the direction of displacement, θ = 30°
Displacement, S = 2m
Work done = F × s cos θ
Work done = 50 × 2 cos 30°
= 50 × 2 × \(\frac{\sqrt{3}}{2}\) = 50\(\sqrt{3}\)J
= 50 × 1.732 = 86.5 J

Question 22.
Calculate the velocity of an object of mass 100 g moving with a kinetic energy of 200 J.
Answer:
Mass of the object = 100g = 0.1 kg
Kinetic Energy (K.E.) = 20 J
Let velocity of the object = v
K.E. = ½ mu²
20 = ½ × 0.1 × v²
or v² = 400
v = 20 ms^-1

Question 23.
Calculate the power of a pump which can lift 200 kg of water to store it in a tank at height of 19m in 25 seconds. (Take g = 10 ms^-2).
Answer:
Mass of water, m =200 kg
Height, h = 19 m
Time taken to lift the water, t = 25 s
Power of the pump = \(\frac{\text { Workdone }}{\text { Time taken }}\)
= \(\frac{m \times g \times h}{t}=\frac{200 \times 10 \times 19}{25}\)
= 1520 W

Question 24.
Calculate the power of a pump in horse power (H.P.) which can lift 600 kg of water into the water tank at a height of 40m in 10 minutes.
Answer:
Mass of water m =600 kg
Height, h =40 m
g = 10ms^-2
Time taken by pump, t = 10 min = 600 s
Power of the pump = \(\frac{\text { Work done }}{\text { Time taken }}=\frac{m g}{600}\)
= \(\frac{600 \times 10 \times 40}{600}\) = 400 W
or = \(\frac {400}{746}\)H.P = 0.52H.P. (∵ 1H.P. = 746W)

Question 25.
An object of mass 2 kg is thrown vertically upward with an initial velocity of 20m/ s. What will-be the potential energy at the highest point (Take g= 10 ms^-2).
Answer:
Given u =20 m/s
g = -10 ms^-2
v = 0
Let maximum height of the object = h
We know that
v² – u² = 2gh
h = \(\frac{v^{2}-u^{2}}{2 g}=\frac{0-(20)^{2}}{2 \times(-10)}=\frac{-400}{-20}\) = 20 m
Potential energy of the object = mgh.
= 2 × 10 × 20 = 400 J

Long Answer Type Questions

Question 1.
Define kinetic energy and potential energy ? Give some examples in each case.
Answer:
Kinetic energy: The energy possessed by an object by virtue of its motion is known as its kinetic energy.
Kinetic energy (E_k) = ½ mv².
Kinetic energy is directly proportional to it is mass and its is directly proportional to the square of its velocity.

Example:

1. The energy possessed by a moving wind can move the blade of a wind-mill due to its kinetic energy.
2. The flowing water also have kinetic energy which can make the water-mill to work.
3. The speeding car, moving bullet, rotating wheel all have kinetic energy and can do work.

Potential Energy: The energy posessed by a an object by virute of its position or change in configuration is known as its potential energy.
Potential energy =m × g × h
where m = mass of the object, h = height, g = acceleration due to gravity.

Example:

1. When a body is lifted to a certain height against the force of gravitation, the work done on the body is stored in the body in the form of potential energy.
2. The water stored at a height in a dam possesses potential energy
3. The energy possessed b a stretched rubber or stretched spring or compressed spring is also an example of potential energy.

Question 2.
What is gravitational potential energy and elastic potential energy? Give examples.
Answer:
(i) Gravitational potential energy : When a body is lifted to a certain height then die work done in raising the object from the ground to that point against gravity is stored in the body and is called gravitational potential energy.
The graviational potential energy of the object = m.g.h

The work done by the gravity depends upon the difference in height of initial position and final position and not on the path along which object is moved.

Example: Any object lifted from the ground level or any other reference level possess potential energy.

(ii) Elastic potential energy : The energy possesed by an object due to change in its shape is known as elastic potential energy It is mainly associated with the compression or extension of an object.

Example :

1. A stretched spring or rubber possesses elastic potential energy.
2. A compressed spring also possesses elastic kinetic energy
3. Derive an expression for the kinetic energy of an object of mass ‘m’ moving with the velocity V.

Or

Derive the expression of kinetic energy (Ek = ½ mv², where m is the mass of the object and v is the constant velocity with which the body is moving.
Answer:
The kinetic energy of an object is the energy possessed by the object by virtue of its motion. The kinetic energ of a body moving with a vertain velocity is equal to the work done on it to make it acquire that velocity from rest.

Consider an object of mass ‘tri at rest. Letitbe displaced through a distance ‘S’ when a constant force ‘F acts on its in the direction of displacement. The work done is given by
W = F × S ……..(i)
The work done on the object will cause a change in its velocity. Let velocity of the object changed from zero (position of rest) to v. Let ‘a’ be the acceleration produced in the body, then

According to Newtons second law of motion F = m × a …….(ii)
But we know that
v² – u² = 2as
or s = \(\frac{v^{2}-u^{2}}{2 a}\)
Object starts moving from rest, therefore u = 0
or s = \(\frac{v^{2}}{2 a}\) …….(iii)
Now substituting the value of ‘F’ and ‘s’ from equation (ii) and (iii) in equation (i) we get
W = m × a\(\left(\frac{v^{2}}{2 a}\right)\)
= ½ mv²
Now, the work done is stored in the form of kinetic energy of the object.
Hence, kinetic energy (E_k) = ½ mv²

Question 4.
What is the law of conservation of energy ? Prove it when an object falls freely from a certain height under the force of gravity.
Answer:
Law of conservation of energy: Energy can neither be created nor destroyed but it can be changed from’one form into another form. The total energy after and before the transformation always remains constant i.e. total energy of a closed system remains constant.

Let an object of mass ‘m’ is dropped freely from ascertain height ‘h’ from the ground.
(i) The potential energy of the body at point A = mgh
Kinetic energy at A = 0 (Because velocity is zero at A).
Total energy at point A (highest point)
= mgh + 0 mgh

(ii) After a certain interval of time it reaches at B. It moves through a distance V due to force of gravitation in downward direction.
The potential energy at B = m.g. (h – x)

Let the velocity of the body at B = v
We know that v² – u² = 2 gs
v² = 2g(x)(u = 0)
Kinetic energ at B = ½ mv²
= ½ m(2gx)
= mgx

Total energy at B = Potential energ + kinetic energy
= m.g. (h – x) + mgx
= m.g.h. – mgx + mgx
= mgh.

(iii) As the fall continues, the potential energy goes on changing into kinetic energy. The potential energu would decrease while kinetic energy would increase, when the object is about the reach the ground then, h – 0.
The potential energy at C = mg × h = mg × 0 = 0.
Let v be the velocity of the object just before reaching the ground.
u = 0, s = 0
v² – u² = 2gh
or v² – 0 = 2gh
Kinetic energy of the object = ½ mv^-1
= ½ m (2gh) – mgh
Total energy = mgh + 0 = mgh.

The total energy i.e. the sum of potential energy and kinetic energy at all the three point is same. Similarly, the total energy at any point will be the same. This proves the law of conservation of energy.

These NCERT Solutions for Class 9 Science Chapter 10 Gravitation Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Gravitation NCERT Solutions for Class 9 Science Chapter 10

Class 9 Science Chapter 10 Gravitation InText Questions and Answers

Question 1.
State the universal law of gravitation.
Answer:
The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The 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 centers.

For two objects of masses ml 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:
F = \(\frac{G m_{1} m_{2}}{r^{2}}\)

Where, G is the universal gravitation constant given by:
G = 6.67 × 10^-11 Nm² kg^-2

Question 2.
Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Answer:
Let 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:
F = \(\frac{G m_{1} m_{2}}{r^{2}}\)

Question 3.
What do you mean by free fall?
Answer:
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 have free fall.

Question 4.
What do you mean by acceleration due to gravity?
Answer:
When an object falls towards the ground from a height, then 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.

Question 5.
What are the differences between the mass of an object and its weight?
Answer:
Mass:

1. It refer to quantity contained in an object
2. It is a scalar quantity.
3. It’s S.I. unit is Kg.
4. It remains constant at places.
5. It is measured by physical or beam balance

Weight:

1. It refers to force with which a body is attracted towards the earth.
2. It is a vector quantity
3. It’s S.I. unit is Newton.
4. It varies depending upon the value of g.
5. It is measured by spring balance.

Question 6.
Why is the weight of an object on the moon \(\frac {1}{6}\)th its weight on the earth?
Answer:
Let ME be the mass of the Earth and m be an object on the surface of the Earth. Let RE be the radius of the Earth. According to the universal law of gravitation, weight W_E of the object on the surface of the Earth is given by,
W_E = \(\frac{\mathrm{GM}_{\mathrm{E}} m}{\mathrm{R}_{\mathrm{E}}{ }^{2}}\)
Let M_M and R_M be the mass and radius of the moon.
Then, according to the universal law of gravitation, weight W_M of the object on the surface of the moon is given by,
W_M = \(\frac{\mathrm{GM}_{\mathrm{E}} m}{\mathrm{R}_{\mathrm{M}}{ }^{2}}\)
\(\frac{W_{M}}{W_{\mathrm{E}}}=\frac{M_{\mathrm{M}} R_{\mathrm{E}}{ }^{2}}{\mathrm{~W}_{\mathrm{E}} \mathrm{R}_{\mathrm{M}}{ }^{2}}\)
Where, ME = 5.98 × 10²⁴ kg, MM = 7.36 × 10²² kg
R_E = 6.4 × 10⁶ m, R_M = 1.74 × 10⁶ m
∴ \(\frac{W_{M}}{W_{\mathrm{E}}}=\frac{7.36 \times 10^{22} \times\left(6.37 \times 10^{6}\right)^{2}}{5.98 \times 10^{24} \times\left(1.74 \times 10^{6}\right)^{2}}=0.165 \approx \frac{1}{6}\)
Therefore, weight of an object on the moon is \(\frac {1}{6}\) of its weight on the Earth.

Question 7.
Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer:
It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.

Question 8.
What do you mean by buoyancy?
Answer:
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.

Question 9.
Why does an object float or sink when placed on the surface of water?
Answer:
If the density of an object is more than the density of the liquid, then it sinks in the liquid.

This is because the buoyant force acting on the object is less than the force of gravity. On the other hand, if the density of the object is less than the density of the liquid, then it floats on the surface of the liquid. This is because the buoyant force acting on the object is greater than the force of gravity.

Question 10.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
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.

Question 11.
You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?
Answer:
The bag of cotton is heavier than the 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 heavier 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 cotton bag is more than that of the iron bar.

Class 9 Science Chapter 10 Gravitation Textbook Questions and Answers

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
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.,
F ∝ \(\frac{1}{r^{2}}\)
If distance r becomes r/2, then the gravitational force will be proportional to
\(\frac{1}{\left(\frac{r}{2}\right)^{2}}=\frac{4}{r^{2}}\)
Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.

Question 2.
Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Answer:
All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistance). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

Question 3.
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m).
Answer:
According to the universal law of gravitation, gravitational force exerted on an object of mass m is given by:
F = \(\frac{\mathrm{GMm}}{r^{2}}\)
Where,
Mass of Earth, M = 6 × 10²⁴ kg
Mass of object, 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
Gravitational force, F = \(\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\)
= \(\frac{6.7 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^{6}\right)^{2}}\) = 9.8 N

Question 4.
The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
Answer:
According to the universal law of gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the moon with an equal force with which the moon attracts the earth.

Question 5.
If the moon attracts the earth, why does the earth not move towards the moon?
Answer:
The Earth and the moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the moon. Hence, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. For this reason, the Earth does not move towards the moon

Question 6.
What happens to the force between two objects, if
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?
Answer:
(i) Doubled
(ii) One-fourth and one-ninth
(iii) four times
According to the universal law of gravitation, the force of gravitation between two objects is given by:
F = \(\frac{\mathrm{G} m_{1} m_{2}}{r^{2}}\)
(i) 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) 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) 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.

Question 7.
What is the importance of universal law of gravitation?
Answer:
The universal law of gravitation proves that every object in the universe attracts every other object.

Question 8.
What is the acceleration of free fall?
Answer:
When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 ms^-2, which is constant for all objects (irrespective of their masses).

Question 9.
What do we call the gravitational force between the Earth and an object?
Answer:
Gravitational force between the earth and an object is known as the weight of the object.

Question 10.
Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator].
Answer:
Weight of a body on the Earth is given by:
W = mg
Where,
m = Mass of the body
g = Acceleration due to gravity

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

Question 11.
Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer:
When a sheet of paper is crumbled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper.

Question 12.
Gravitational force on the surface of the moon is only \(\frac {1}{5}\) as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth?
Answer:
Weight of an object on the moon = \(\frac {1}{6}\) × Weight of an object on the Earth
Also,
Weight = Mass × Acceleration
Acceleration due to gravity, g = 9.8 m/s²
Therefore, weight of a 10 kg object on the Earth = 10 × 9.8 = 98N
And, weight of the same object on the moon = \(\frac {1}{6}\) × 98 = 16.3 N

Question 13.
A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.
Answer:
(i) 122.5 m (ii) 10 s
According to the equation of motion under gravity: v² – u² = 2 gs
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v = 0
u = 49 m/s
During upward motion, g = – 9.8 m s^-2
Let h be the maximum height attained by the ball.
Hence,
0-(49)² = 2 × (-9.8) × h
h = \(\frac{49 \times 49}{2 \times 9.8}\) = 122.5 m
Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:
v = u + gt
We get,
0 = 49 + t × (-9.8)
9.8t = 49
t = \(\frac{49}{9.8}\) = 5s
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s

Question 14.
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answer:
According to the equation of motion under gravity:
v₂ – u₂ = 2 gs
Where,
u = Initial, velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 m s^-2
∴ v₂ – 0₂ = 2 × 9.8 × 19.6
v₂ = 2 × 9.8 × 19.6 = (19.6)²
v = 19.6 ms^-1
Hence, the velocity of the stone just before touching the ground is 19.6 m s^-1.

Question 15.
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answer:
According to the equation of motion under gravity:
v₂ -u₂ = 2 gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0
s = Height of the stone
g = Acceleration due to gravity = -10 m s^-2
Let h be the maximum height attained by the stone.

Therefore,
0 – (40)2 = 2 × h × (-10).
h = \(\frac{40 \times 40}{20}\) = 80 m
Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m
Net displacement of the stone during its upward and downward journey = 80 + (-80) = 0

Question 16.
Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.
Answer:
According to the universal law of gravitation, the force of attraction between the Earth and the Sim is given by:
F = \(\frac{\mathrm{GM}_{\text {Sun }} \mathrm{M}_{\text {Earth }}}{\mathrm{R}^{2}}\)
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
F = \(\frac{6.7 \times 10^{-11} \times 2 \times 10^{36} \times 6 \times 10^{24}}{\left(1.5 \times 10^{11}\right)^{2}}\) = 3.57 × 10²² N

Question 17.
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Answer:
Let the two stones meet after a time t.
For the stone dropped from 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 m s^-2
From the equation of motion,
S = ut + \(\frac {1}{2}\)gt²
= 0 × t + \(\frac {1}{2}\) × 9.8 × t² ………. (i)

(ii) For the stone thrown upwards:
Initial velocity, u = 25 m s^-1
Let the displacement of the stone from the ground in time t be s’.
Acceleration due to gravity, g = -9.8 ms^-2
Equation of motion,
s’ = ut + \(\frac {1}{2}\)gt²
= 25f – \(\frac {1}{2}\) × 9.8t²
∴ s’ = 25t – 9t² ………. (2)
The combined displacement of both the stones at the meeting point is equal to the height of the tower 100 m.
∴ s + s’ = 100
\(\frac {1}{2}\)gt² + 25t – \(\frac {1}{2}\)gt² = 100
∴ t = \(\frac {100}{25}\) = 4 s
In 4 s, the falling stone has covered a distance given by equation (1) as
s = \(\frac {1}{2}\) × 10 × 4² = 80m
Therefore, the stones will meet after 4 s at a height (100 – 80) = 20 m from the ground

Question 18.
A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.
Answer:
(a) 29.4m/s
(b) 44.1 m
(c) 39.2 m above the ground

(a) Time of ascent is equal to the time of descent The ball takes a total of 6 s for its upward and downward journey.
Hence, it has taken 3 s to attain the maximum height.
Final velocity of the ball at the maximum height, v = 0
Acceleration due to gravity, g = -9.8 ms^-2
Equation of motion, v = u + gt will give,
0 = u + (-9.8 × 3)
u = 9.8 × 3 = 29.4 ms^-1
Hence, the ball was thrown upwards with a velocity of 29.4 ms^-1.

(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u = 29.4 ms^-1
Final velocity, v = 0
Acceleration due to gravity, g = -9.8 ms^-2
From the equation of motion, s = ut + \(\frac {1}{2}\)at²
h = 29.4 × 3+ \(\frac {1}{2}\) × -9.8 × (3)² = 44.1m

(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, u = 0
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4s -3s = 1s.
Equation of motion, s = ut + \(\frac {1}{2}\)gt² will give,
s = 0 × t + \(\frac {1}{2}\) × 9.8 × 1² = 4.9 m
Total height = 44.1 m
This means that the ball is 39.2 m (44.1 m – 4.9 m) above the ground after 4 seconds.

Question 19.
In what direction does the buoyant force on an object immersed in a liquid act?
Answer:
An object immersed in a liquid experiences buoyant force in the upward direction.

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

Question 21.
The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm^-3, will the substance float or sink?
Answer:
If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.
Here, density of the substance = \(\frac{\text { Mass of the substance }}{\text { Volume of the substance }}=\frac{50}{20} 2.5 \mathrm{~g} \mathrm{~cm}^{-3}\)
The density of the substance is more than the density of water (1 g cm^-3). Hence, the substance will sink in water.

Question 22.
The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g cm^-3? What will be the mass of the water displaced by this packet?
Answer:
Density of the 500 g sealed packet
\(\frac{\text { Mass of the packet }}{\text { Volume of the packet }}=\frac{500}{350}=1.428 \mathrm{gcm}^{-3} \)
The density of the substance is more than the density of water (1 g cm^-3). Hence, it will sink in water.
The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 g.

Class 9 Science Chapter 10 Gravitation Additional Important Questions and Answers

Multiple Choice Questions
Choose the correct option:

Question 1.
Two objects of different masses falling freely near the surface of moon would
(a) have same velocities at any instant
(b) have different accelerations
(c) experience forces of same magnitude
(d) undergo a change in their inertia
Answer:
(a) have same velocities at any instant

Question 2.
The value of acceleration due to gravity
(a) is same on equator and poles
(b) is least on poles
(c) is least on equator
(d) increases from pole to equator
Answer:
(c) is least on equator

Question 3.
The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become
(a) F/4
(b) F/2
(c) F
(d) 2F
Answer:
(a) F/4

Question 4.
A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone
(a) will continue to move in the circular path
(b) will move along a straight line towards the centre of the circular path
(c) will move along a straight line tangential to the circular path
(d) will move along a straight line perpendicular to the circular path away from the boy
Answer:
(c) will move along a straight line tangential to the circular path

Question 5.
An object is put one by one in three liquids having different densities. The object floats with 1/9, 2/11, 3/7, and parts of their volumes outside the liquid surface in liquids of densities d₁, d₂ and d₃ respectively. Which of the following statement is correct?
(a) d₁ > d₂ > d₃
(b) d₁ > d₂ < d₃
(c) d₁ < d₂ > d₃
(d) d₁ < d₂ < d₃
Answer:
(d) d₁ < d₂ < d₃

Question 6.
In the relation F = GMm/d2, the quantity G
(a) depends on the value of g at the place of observation
(b) is used only when the earth is one of the two masses
(c) is greatest at the surface of the earth
(d) is universal constant of nature
Answer:
(d) is universal constant of nature

Question 7.
Law of gravitation gives the gravitational force between
(a) the earth and a point mass only
(b) the earth and Sim only
(c) any two bodies having some mass
(d) two charged bodies only
Answer:
(c) any two bodies having some mass

Question 8.
The value of quantity G in the law of gravitation
(a) depends on mass of earth only
(b) depends on radius of earth only
(c) depends on both mass and radius of earth
(d) is independent of mass and radius of the earth
Answer:
(d) is independent of mass and radius of the earth

Question 9.
Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be
(a) 1/4 times
(b) 4 times
(c) 1/2 times
(d) unchanged
Answer:
(b) 4 times

Question 10.
The atmosphere is held to the earth by
(a) gravity
(b) wind
(c) clouds
(d) earth’s magnetic field
Answer:
(a) gravity

Question 11.
The force of attraction between two unit point masses separated by a unit distance is called
(a) gravitational potential
(b) acceleration due to gravity
(c) gravitational field
(d) universal gravitational constant
Answer:
(d) universal gravitational constant

Question 12.
The weight of an object at the centre of the earth of radius R is
(a) zero
(b) infinite
(c) R times die weight at the surface of the earth
(d) 1/R₂ times the weight at surface of the earth
Answer:
(a) zero

Question 13.
An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be
(a) 2 N
(b) 8 N
(c) 10 N
(d) 12 N
Answer:
(a) 2 N

Question 14.
A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be
(a) maximum when length and breadth form the base
(b) maximum when breadth and width form the base
(c) maximum when width and length form the base
(d) the same in all the above three cases
Answer:
(b) maximum when breadth and width form the base

Very Short Answer Type Questions

Question 1.
State what causes the revolution of planets around the sun?
Answer:
Gravitational force of attraction between the sun and planets.

Question 2.
What is the direction of gravitation force acting between the two objects?
Answer:
The force of gravitation between two objects act along the line joining their centres of mass.

Question 3.
How does the gravitation force of attraction vary with the distance between them?
Answer:
The gravitation force of attraction is universely proportional to the square of distance between die centres of their mass.

Question 4.
State the effect on the gravitational force of attraction between two objects
(i) When distance is halved
(ii) When distance is doubled
Answer:
(i) When distance is halved, the force increases by four times.
(ii) When distance is doubled, the force decreases by four times.

Question 5.
How does the gravitation force of attraction between two objects depend upon their masses?
Answer:
The force of gravitation is directly proportional to the product of the masses of two objects.
F ∝ m₁.m₂

Question 6.
If the mass of a body is doubled and that of another body is tripled, what would be the effect of gravitational force?
Answer:
The gravitational force would increase by six times.

Question 7.
State any one physical quantity that does not affect the magnitude of acceleration due to gravity.
Answer:
MaSs of the body.

Question 8.
What is the magnitude of universal gravitational constant?
Answer:
6.67 × 10^-11 Nm²kg^-2.

Question 9.
Does the value of universal gravitational constant change in universe?
Answer:
No, its a constant value.

Question 10.
What is the Value of ‘g’ acceleration due to gravity on moon?
Answer:
The value of acceleration due to gravity on moon surface is 1/6 th of value of acceleration of gravity on earth i.e. 1.63 ms^-2.

Question 11.
If a body of mass ‘m’ revolves in a circle of radius ‘r’, then what would be the magnitude of the acceleration produced in the body?
Answer:
The magnitude of acceleration produced (a) = \(\frac{v^{2}}{r}\)

Question 12.
What is the relation between ‘g’ and ‘G’?
Answer:
g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
When M is mass of earth and R is the radius of earth.

Question 13.
Why do the nails and pins have pointed ends?
Answer:
The nails and pins have pointed ends to reduce area and increase the effect of applied force i.e. pressure.

Question 14.
What is the S.I. unit of density and relative density?
Answer:
Relative density has no unit but for density S.I. unit kg/m³.

Question 15.
State the principle that describes the working of hydrometer or lactometer.
Answer:
Archimedes principle.

Question 16.
What is lactometer?
Answer:
It is a device used to measure the purity of milk i.e. amount of water present in milk.

Question 17.
What is hydrometer?
Answer:
It is a device used to measure the density of a liquid.

Question 18.
State Archimedes principle.
Answer:
The principle states that when an object is immersed in fluid partially or completely, it experiences an upward thrust that is equal to the weight of the fluid displaced by it.

Question 19.
Where will a student find the mass of 1kg heavier, when weighted in air or in water? Why?
Answer:
In air because the buoyant force in air is less than in water.

Question 20.
If a student drops a feather and an iron block in vacuum from a particular height. Which of the two will land first? Why?
Answer:
Both feather and iron block would land at the same time because acceleration due gravity is independent of mass.

Short Answer Questions

Question 1.
What do you understand by gravitation and gravity?
Answer:
Gravitation refers to the force with which any two objects in universe attract each other irrespective of their masses and distance between them. Gravity refers to force with which earth attracts an object towards its centre.

Question 2.
Does earth attracts moon towards itself? If yes, why does the moon then not fall on the earth?
Answer:
Earth and moon have gravitational force that acts in between them. Moon revolving round the earth undergoes a change in its acceleration because of changing direction. Moreover, the motion of moon around earth is because of centripetal gravitational force which is balanced by moon’s centrifugal gravitational force to prevent it’s falling down on the earth.

Question 3.
It is seen that an apple is attracted towards the earth. Does the apple attract the earth? If yes, why does the earth not move towards the apple?
Answer:
According to universal law of gravitation apple also attracts the earth with the same force as the earth attracts the apple but in opposite direction.
Force acting on the apple due to earth’s attraction, F = mg.
Apple also attracts the earth with same force. Therefore, acceleration produced on the earth
(a) = \(\frac{\mathrm{F}}{\mathrm{M}_{\mathrm{e}}}=\frac{m g}{\mathrm{M}_{e}}\)

However, the mass of earth is 6.0 × 10²⁴ kg which is very large as compared to the mass of an apple. Therefore, the value of acceleration produced in the earth is very-very small or negligible for the earth to show any visible movement towards the apple.

Question 4.
State the factors which affect the gravitation force of attraction in between two objects?
Answer:
There are two factors which affect the force of attraction between two objects. These factors include:

1. Mass, because the force of gravitation is directly proportional to the product of the masses of two bodies.
2. Distance between the two bodies because of the force of attraction is inversely proportional to the square of the distance in between them.

Question 5.
What is gravitational constant ‘G’ ? What is its S.I. unit?
Answer:
The gravitational force of attraction between two bodies/objects is given by
F = \(\mathrm{G} \frac{m_{1} m_{2}}{d^{2}}\)
Where ‘G’ is universal gravitational constant.
G = \(\frac{\mathrm{F} d^{2}}{m_{1} m_{2}}\)
Putting the S.I. units of different physical quantities.
G = \(\frac{\mathrm{N} m^{2}}{\mathrm{~kg} . \mathrm{kg}}\) = nm²kg^-2
Hence, S.I. units of G is Mm² kg^-2.

Question 6.
What do you understand by acceleration due to gravity? Does it depend upon the mass?
Answer:
It refers to the acceleration gained by a free falling body towards the earth. It is represented by symbol ‘g’ and its value is 9.8 m/s² i.e. a free falling body gains an acceleration of 9.8 m/s² every second during tire course of its fall.
It does not depend upon the mass of the object.

Question 7.
What is the difference between ‘g’ and ‘G’ ?
Answer:

Question 8.
Calculate the value of ‘g’ on the surface of earth.
Answer:
We know that g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
M, mass of earth = 6.0 × 10²⁴ kg
R1 radius of earth = 6.4 × 10⁶ m.
G, Gravitational constant = 6.67 × 10^-11 Nm² kg^-2
g = \(\frac{6.67 \times 10^{-11} \mathrm{Nm} \mathrm{kg}^{-2} \times 6.0 \times 10^{24} \mathrm{~kg}}{\left(6.4 \times 10^{6}\right)^{2} \mathrm{~m}}\)
= \(\frac{400.38}{40.96}=9.8 \mathrm{~ms}^{-2}\)

Question 9.
What will happen to the weight of a body when it is taken down inside the earth?
Answer:
The weight of the body decreases because the effective mass of the earth with which it attracts the body towards it centre decreases. The decrease in weight due to decresing effecting mass of the earth overpowers the increase in weight due to decrease in distance.

Question 10.
Define mass and weight. What are their S.I. units?
Answer:
Mass is defined as the quantity of matter contained in a body that remains unchanged throughout the universe. It’s S.I. units is kilograms. Weight is defined as the force with which the earth attracts the object towards its centre.
Weight (W) = m × g
where m = mass of body
g = accelertion due to gravity
The S.I. unit of weight is Netwon(N).

Question 11.
Where do you expect the weight of a body to be more, at poles or at equator? Why?
Answer:
The earth is not a perfect spherical structure. It is compressed at poles therefore the radius of earth is greater at equator than at poles. With value of ‘g’ inversely proportional to square of distance from the centre of earth, the value of ‘g’ is comparitively more at poles than at equator. Therefore, the weight of an object is maximum at poles and minimum at equator.

Question 12.
Write the equation of motion for free falling bodies due to gravitational force of attraction on earth.
Answer:
When u = initial velocity, v = final velocity,
g = acceleration due to gravity, s = distance covered and time = t, then
V = u + gt
s = ut + \(\frac {1}{2}\)gt²
2gs = v² – u²

Question 13.
According to Cartesian convention what are the sign for acceleration due to gravity ‘g’. Velocity ‘v’ and distance covered ‘s’ ?
Answer:
(i) According to cartesian convention ‘g’ is always taken in negative since its acts in downward direction. It is always taken in negative whether a body goes up or comes down.

(ii) The value of velocity V is taken in positive when a body goes in upward direction and when it come down, it value is taken in negative. (Fig. 10.2)
(iii) Distance is also taken in positive when a body goes up and taken in negative when falls down on the earth.

Question 14.
What is thrust and pressure? How are they different?
Answer:

Question 15.
Why cutting tools like pair of scissors, knives or axes have sharp edges?
Answer:
The edges of pair of scissors, knives and axes are made sharp to increase the effect of the applied force on them. The pressure is universely proportional to area i.e. with the decreasing area, pressure increases.

Question 16.
What is bouyant force? How will you determine the bouyant force?
Answer:
Bouyant force is an upward force that is exerted on a body when it is partially or fully immersed in a liquid like water. However, the magnitude of bouyant force depends upon the density of liquid, with increasing density of fluid, the bouyant force also increases.

The magnitude of the force of bouyancy is determined by measuring the weight of the fluid displaced by the object when immersed in a liquid.

Question 17.
Does a body experience the same bouyant force in different fluids? If not, why?
Answer:
A body immersed in different fluids such as water, alcohol and cooking oil will experience different bouyant force because the bouyant force exerted on a body immersed in a fluid depends upon, the fluid’s density. The bouyant force increases with increasing density of the fluid.

Question 18.
What is law of floatation?
Answer:
The law of floatation describes, the conditions in which a body would float or sink. When immersed in a liquid, the law states that:

• If relative density of a body is more than one, it would sink in the water.
• If relative density is less than one, it will float on water.

Question 19.
Why a nail or a rupee coin sinks in water but a ship floats in water?
Answer:
The weight of water displaced by a nail or a rupee coin is less than the weight of the nail or rupee coin. Therefore, with less upward thrust applied on them, they sink in water.

A ship being large is so designed that the water displaced by the ship is more than the weight of the ship i.e. upward thrust is more on ship to keep it floating in water.

Question 20.
What do you understand by density and the relative density? What are their S.I. units?
Answer:
Density refers to the mass per unit volume of a substance.
Mass of substance = \(\frac{\text { Mass of substance }}{\text { Volume of substance }}\)
It’s S.I. unis is kg/m³.
Relative density refers to the ratio of the density of a substance to the density of pure water.
Relative density = \(\frac{\text { Density of substance }}{\text { Density of water }}\)
Relative density being a ratio has no unit.

Question 21.
What is the force of gravitation in between two boys having mass equal to 50 kg and 40 kg respectively and separated by a distance of lm from each other?
Answer:
Mass of one boy m₁ = 50 kg, Mass of other boy, m₂ = 40 kg Distance between them, d = 1m.
Value of G = 6.67 × 10^-1 Nm² kg^-2
According to law of gravitation, force of attraction between them is given by
F = G \(\frac{m_{1} m_{2}}{d^{2}}\)
= \(\frac{6.67 \times 10^{-11} \times 50 \times 40}{(1)^{2}}\)
= 1.334 × 10^-7 N

Question 22.
Calculate the force exerted by earth on the moon. If the mass of the earth 6.0 × 10²⁴ kg and that of the moon is 7.4 × 10²² kg. The distance between the earth and the moon is 3.84 × 10^-11. (G = 6.67 × 10^-11 Nm² kg^-2).
Answer:
Mass of the earth, M = 6.0 × 10²⁴ kg.
Mass of the moon, m = 7.4 × 10²² kg.
The distance between moon and earth, r = 3.84 × 10²⁴ m.
The force exerted by earth on moon is given by

Question 23.
A car falls off a ledge and drops in % seconds. Let g = -10 ms^-2.
(i) What will be its speed on striking the ground?
(ii) What is its average speed during the % second?
(iii) How high is the ledge from the ground.
Answer:
Given: Initial velocity of car = 0 ms^-1
Acceleration due to gravity, g = -10 ms^-2
lime taken, t = 1/2 second.
(i) Let speed on striking the groun d,v = ?
v = u + gt
= 0 – 10 × \(\frac {1}{2}\) = -5 ms^-1
Negative sign indicates that car is moving downward.

(ii) Average speed of the car
\(\frac{u+v}{2}=\frac{0+(-5)}{2}=2.5 \mathrm{~ms}^{-1}\)
Average speed is 2.5 ms^-1 downward.

(iii) Let the height of ledge from the ground
h = ut + \(\frac {1}{2}\)gt²
= 0 × \(\frac {1}{2}\) + \(\frac {1}{2}\) (-10)(\(\frac {1}{2}\))2
= \(-\frac{5}{4}\) = -1.25 m.
The distance travelled has a negative sign this means that the car is moving in downward direction.

Question 24.
A stone is dropped from 180 m high tower. How long does the stone tak to reach the ground ? What is the velocity when it touches the ground ? (Take g = -10 ms^-2)
Answer:
Height of the tower, h = – 180m
Initial velocity, u = 0
Let time taken to reach the ground, t = ?
Let velocity with which it touches the ground, v = ?
We know that
h = ut + \(\frac {1}{2}\)gt²
-180 = 0 × t + \(\frac {1}{2}\) (-10)t²
180 = 5t²
t2 = \(\frac {180}{5}\) = 36 or t = 6
(ii) u = u+gt = 0 – 15 × 6
= -60 m/s
Negative sign indicates that the body is moving in downward direction.

Question 25.
A body is thrown vertically upward rises to a height of 10m. Calculate.
(i) The velocity with which body was thrown upward.
(ii) The time taken by the body to reach the highest point.
Answer:
(i) Maximum height reached by the body, h = 10m.
Let initial velocity =?
Final velocity, v =0
Acceleration due to gravity g = – 9.8 ms^-2
We know that v² – u² = 2 (-9.8) × 10
u² = 196
u = \(\sqrt{196}\) = 14ms^-1

(ii) Initial velocity of the body = 14 ms^-1
v = u + gt
0 = 14 – 9.8 × t
t = \(\frac {14}{9.8}\) = 1.43 second
Time taken by the body to reach maximum height = 1.43 s.

Question 26.
The mass of an object is 50 kg on earth. What is its weight on the earth and what will be its weight on the moon?
Answer:
Mass of the object = 50 kg.
Acceleration due to gravity at earth, g = 9.8 ms^-2
Weight on earth = m × g = 50 × 9.8 = 490 N
Acceleration due to gravity, g on moon = \(\frac{9.8}{6} m \mathrm{~s}^{-2}\)
Weight on moon = m × g
= \(\frac{50 \times 9.8}{6}\) = 81.67 N

Question 27.
Find the mass of an object whose weight on the earth is 49 N. What is its mass on the surface of moon?
Answer:
Weight on the earth, W = 49 N.
Acceleration due to gravity on earth, g = 9.8 ms^-2.
We know that W = m.g.
m = \(\frac{W}{g}=\frac{49}{9.8}\) = 5 kg.
Mass of the body on earth = 5 kg.
Mass of the body on moon = 5 kg
(Because the mass remains constant at every place).

Question 28.
Calculate the mass of the moon. If the radius of the moon ‘R_m‘ is 1.60 × 10⁶ m. The acceleration due to gravity at moon is 1/6th of the gravity at earth.
Answer:
Let mass of the moon = M_m
Mass of the earth = m_e
We know that acceleration due to gravity ‘g’ at each is given by g = \(\frac{\mathrm{GM}_{e}}{\mathrm{R}_{\mathrm{e}}{ }^{2}}\)
Similarly, gravity at moon, g will be given by

Question 29.
A block of wood is a kept on tabletop. The mass of the wooden block is 5kg and its dimension are 40 cm × 20 cm × 10cm. Find the pressure exerted by the wooden block on the tabletop if it is made to lie on the tabletop with its side& of dimension (i) 20m × 10cm. (ii) 40 cm × 20 cm.
Answer:
The mass of the wooden block = 5 kg.
The force exerted by wooden block = mg = 5 × 9.8 = 49 N
(i) When wooden block lie with its 20 cm × 10 cm side on the tabletop.
Area of block in contact with table top = 20 × 10 = 200 cm² = 0.02 m²
Force = \(\frac{\text { Force }}{\text { Area }}=\frac{49}{0.02}\) = 2450 Nm²

(ii) When wooden block lie with its side 40 × 20 cm on the table top:
Area in contact with table = 40 × 20 = 800 cm² = 0.08 m².
Pressure = \(\frac{\text { Force }}{\text { Area }}=\frac{49}{0.08}\) = 612.5 Nm^-2.

Question 31.
The relative density of silver is 10.8. The density of water is 1 kg m^-3. What is the density of silver?
Answer:
Relative density of silver = 10.8 kgm^-3
Density of water = 10³kg m^-3
Relative density = \(\frac{\text { Density of silver }}{\text { Density of water }}\)
Density of silver = Relative density × Density of water
= 10.8 × 10³ kgm^-3

Question 32.
A cubical block of mass 5 kg with each side 5 cm is lying on a table top. Calculate the pressure exerted by the block.
Answer:
Mass of the block = 5 kg
Thrust exerted by the block =m × g
= 5 × 9.8 = 49.0 N
Side of cubical block = 5 cm = 0.05 m
Area of the surface in contact with table top
= (0.05)² = 0.0025 m² = 2.5 × 10^-3 m²
Pressure exerted = \(\frac{\text { Thrust }}{\text { Area }}\)
= \(\frac{49.0}{2.5 \times 10^{-3}}\) = 1.86 × 10⁴ Nm^-2 or Pa.

Long Answer Questions

Question 1.
State the Kepler’s laws that describe the motion of the planets around the earth.
Answer:
The motion of the planets are governed by the following Kepler’s laws.
(i) The orbit of a planet is elliptical with the Sun at one of the focus, as shown in the figure 10.3.
(ii) The line joining the planet and the sun sweeps equal area in equal intervals of time. Thus, in the given figure if the time of interval of moving from A to B is the same as that from C to D, then the area OAB and OCD are equal.

(iii) The cube of the mean distance of a planet from the Sun is proportional to the square of its orbital period, i.e. time taken by the planet to complete its orbit T:
r³ ∝ T²
or \(\frac{r^{3}}{\mathrm{~T}^{2}}\) = constant
where r = Mean distance of the planet from sun.
T = Orbital period.

Question 2.
What is inverse square rule? How did Newton guess or arrived the inverse square rule?
Answer:
Inverse square rule: When a planet revolves around the sun, the force of attraciton between sun and planet is inversely proportional to the square of radius of the orbit.
F ∝ \(\frac{1}{r^{2}}\)
F = Force of attraction between sun and planet.
r = Radius of the orbit of a planet.

Newton showed that the cause of planetary motion is the gravitational force of attraction that sun exert on planets. Newton used the third law of Kepler to guess the inverse square rule.

Suppose a planet is moving in an orbit. We can assume Planetary orbits as circular orbit. Let the radius of the circular orbit is ‘r’ and planet is moving with a linear velocity ‘v’.
Then, the force ‘F’ acting on the planet of mass ‘m’ is given by
F = \(\frac{m v^{2}}{r}\)
or F ∝ \(\frac{v 2}{\bar{r}}\) ………(i)
(m = mass of planet is contant)
If let T is the taken by the planet to complete one orbit, then,

Therefore, we have, v² ∝ \(\frac{1}{r}\)
combining (i) and (ii) we have
F ∝ \(\frac{1}{r^{2}}\)

Question 3.
What do you mean by acceleration due to gravity? How will you prove that acceleration due to gravity is independent of the mass and nature of the body?
Answer:
Acceleration due to gravity: When body falls freely due to the gravitational force of attraction of the earth, there is no change in its direction of motion but its velocity changes by equal amounts in each second. So the acceleration produced in the body due to earth’s gravitation force of attraction is known as the acceleration due to gravity. It is denoted by.

Let a body of mass ‘m’ is falling downward due to force of gravity and acceleration produced in the body is ‘g’. Then according the Newton’s second law of motion, gravitation force acting on the body is given by
F = mg ……..(i)
But the gravitation force of attraction between the body of mass ‘m’ and earth is given by
F = \(\frac{\mathrm{GM}_{m}}{d^{2}}\)
where M = mass of the earth
d = distance between the body and the centre of the earth from (i) and (ii)
mg = \(\frac{\mathrm{GM}_{m}}{d^{2}}\)
g = \(\frac{\mathrm{GM}}{d^{2}}\)
The expression of ‘g’ is independent of the mass of the object. Hence, acceleration due to gravity of the earth does not depend upon the mass and nature of the object.

Question 4.
Show that the weight of an object on the moon’s surface is 1/6th that of the earth.
Answer:
Let the mass of an object on moon’s surface is ‘m’.
Let weight of the object on moon’s surface = W
Let, Mass of moon =M
Radius of moon = R
Now the force of attraction by which moon attracts an object is equal to the weight of the object.
Therefore, Wm = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\) ………(1)
Let the weight of the same object on earth = We
The force with which earth attracts the object is equal to the weight of the object on earth.
Therefore We = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{e}^{2}} \) ……..(2)
Where Me = mass of the earth and Re = Radius of earth.

But Mass of the earth is 100 times that of the moon and radius of the earth is 4 times the radius of moon.
Therefore, M_e = 100 M
and R_e = 4R
Putting these values in equation (2) we have
We = \(\frac{G(100 M) \times m}{(4 R)^{2}}\)
= \(\mathrm{G} \frac{100 \mathrm{Mm}}{16 \mathrm{R}^{2}}\) ……..(3)
Dividing equation (1) by (3)

or Weight on the moon = \(\frac {1}{6}\) × weight on the earth.

Question 5.
Show with the help of an activity that whenever a body is immersed in a liquid, it loses same weight due to force acting on the body in the upward direction (bouyant force).
Or
Show with the help of an activity that whenever an object is immersed in a fluid, a force acts on the object in upward direction.
Answer:
When an object is immersed in liquid, a force acts on the body in upward direction. ‘It can be shown by the following activity. (Fig. 10.4)

Activity: Take a piece of stone and tie it to one end of a spring balance. Suspend the stone by holding the string as shown in the figure. Note the extension produced in the reading in the spring balance due to the weight of the stone. Now, slowly dip the stone in water in a container.

As the stone is dipped in water, it loses the reading on spring balance scale. It continues declining till it gets completely immersed in water.

The decrease in extension of the spring shows that a force is acting on the stone in the upward direction. This upward thrust causes or brings a loss in the weight.

(a) Observe the elongation of the rubber string due to the weight of a piece of stone suspended form it in air.
(b) The elongation decreases as the stone, is immersed in water.

Question 6.
What is Archimedes principle? State some of its applications.
Answer:
Archimedes principle states that when an object is immersed in a fluid partially or fully, it experiences an upward thrust or force that is equal to the weight of the fluid displaced. The principle has helped.

• In designing ships and submarines.
• In designing lactometer to assess the purity of milk.
• In desgining hydrometer to measure the density of liquids.
• In assessing the purity of a given substance.

These NCERT Solutions for Class 9 Science Chapter 7 Diversity in Living Organisms Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Diversity in Living Organisms NCERT Solutions for Class 9 Science Chapter 7

Class 9 Science Chapter 7 Diversity in Living Organisms InText Questions and Answers

Question 1.
Why do we classify organisms?
Answer:
There are a wide range of life forms (about 10 million – 13 million species) around us. These life forms have existed and evolved on the Earth over millions of years ago. The huge range of these life forms makes it very difficult to study them one by one. Therefore, we look for similarities among them and classify them into different classes to study these different classes as a whole. Thus, classification makes our study easier.

Question 2.
Give three examples of the range of variations that you see in life-forms around you.
Answer:
Examples of range of variations observed in daily life are:

• Variety of living organisms in terms of size ranges from microscopic bacteria to tall trees of 100 metres.
• The colour, shape, and size of snakes are. completely different from those of lizards.
• The life span of different organisms is also quite varied. For example, a crow lives for only 15 years, whereas a parrot lives for about 140 years.

Question 3.
Which do you think is a more basic characteristic for classifying organisms?
(a) The place where they live.
(b) The kind of cells they are made of. Why?
Answer:
The kind of cells that living organisms
are made up of is a more basic characteristic for classifying organisms, than on the basis of their habitat. This is because on the basis of the kind of cells, we can classify all living organisms into eukaryotes and prokaryotes. On the other hand, a habitat or the place where an organism lives is a very broad characteristic to be used as the basis for classifying organisms. For example, animals that live on land include earthworms, mosquitoes, butterfly, rats, elephants, tigers, etc.

These animals do not resemble each other except for the fact that they share a common habitat. Therefore, the nature or kind of a cell is considered to be a fundamental characteristic for the classification of living organisms.

Question 4.
What is the primary characteristic on which the first division of organisms is made?
Answer:
The primary characteristic on which the first division of organisms is made is the nature of the cell. It is considered to be the fundamental characteristic for classifying all living organisms. Nature of the cell includes the presence or absence of membrane-bound organelles.

Therefore, on the basis of this fundamental characteristic, we can classify all living organisms into two broad categories of eukaryotes and prokaryotes. Then, further classification is made on the basis of cellularity or modes of nutrition.

Question 5.
On what basis are plants and animals put into different categories?
Answer:
Plants and animals differ in many features such as the absence of chloroplasts, presence of cell wall, etc. But, locomotion is considered as the characteristic feature that separates animals from plants.

This is because the absence of locomotion in plants gave rise to many structural changes such as the presence of a cell wall (for protection), the presence of chloroplasts (for photosynthesis) etc. Hence, locomotion is considered to be the basic characteristic as further differences arose because of this characteristic feature.

Question 6.
Which organisms are called primitive and how are they different from the so-called advanced organisms?
Answer:
A primitive organism or lower organism is the one which has a simple body structure and ancient body design or features that have not changed much over a period of time. An advanced organism or higher organism has a complex body structure and organization. For example, an Amoeba is more primitive as compared to a starfish. Amoeba has a simple body structure and primitive features as compared to a starfish. Hence, an Amoeba is considered more primitive than a starfish.

Question 7.
Will advanced organisms be the same as complex organisms? Why?
Answer:
It is not always true that an advanced organism will have a complex body structure. But, there is a possibility that over the evolutionary time, complexity in body design will increase. Therefore, at times, advanced organisms can be the same as complex organisms.

Question 8.
What is the criterion for classification of organisms as belonging to kingdom Monera or Protista?
Answer:
The criterion for the classification of organisms belonging to kingdom Monera or Protista is the presence or absence of a well-defined nucleus or membrane-bound organelles. Kingdom Monera includes organisms that do not have a well-defined nucleus or membrane-bound organelles and these are known as prokaryotes.

Kingdom Protista, on the other hand, includes organisms with a well-defined nucleus and membrane-bound organelles and these organisms are called eukaryotes.

Question 9.
In which kingdom will you place an organism which is single-celled, eukaryotic and photosynthetic?
Answer:
Kingdom Protista includes single celled, eukaryotic, and photosynthetic organisms.

Question 10.
In the hierarchy of classification, which grouping will have the smallest number of organisms with a maximum of characteristics in common and which will have the largest number of organisms?
Answer:
In the hierarchy of classification a species will have the smallest number of organisms with a maximum of characteristics in common, whereas the kingdom will have the largest number of organisms.

Question 11.
Which division among plants has the simplest organisms?
Answer:
Thallophyta is the division of plants that has the simplest organisms. This group includes plants, which do not contain a well differentiated plant body. Their body is not differentiated into roots, stems, and leaves. They are commonly known as algae.

Question 12.
How are pteridophytes different from the phanerogams?
Answer:

Question 13.
How do gymnosperms and angiosperms differ from each other?
Answer:

Question 14.
How do poriferan animals differ from coelenterate animals?
Answer:

Question 15.
How do annelid animals differ from arthropods?
Answer:

Question 16.
What are the differences between amphibians and reptiles?
Answer:

Question 17.
What are the differences between animals belonging to the Aves group and those in the mammalia group?
Answer:

Class 9 Science Chapter 7 Diversity in Living Organisms Textbook Questions and Answers

Question 1.
What are the advantages of classifying organisms?
Answer:
There are a wide range of life forms (about 10 million-13 million species) around us. These life forms have existed and evolved on the Earth over millions of years ago. The huge range of these life forms makes it very difficult to study them one by one. Therefore, we look for similarities among them and classify them into different classes so that we can study these different classes as a whole. This makes our study easier.

Therefore, classification serves the following advantages:

• It determines the methods of organising the diversity of life on Earth.
• It helps in understanding millions of life forms in detail.
• It also helps in predicting the line of evolution.

Question 2.
How would you choose between two characteristics to be used for developing a hierarchy in classification?
Answer:
For developing a hierarchy of classification, we choose the fundamental characteristic among several other characteristics. For example, plants differ from animals in the absence of locomotion, chloroplasts, cell wall, etc. But, only locomotion is considered as the basic or fundamental feature that is used to distinguish between plants and animals. This is because the absence of locomotion in plants gave rise to many structural changes such as the presence of a cell wall for protection, and the presence of chloroplast for photosynthesis (as they cannot move around in search of food like animals).

Thus, all these features are a result of locomotion. Therefore, locomotion is considered to be a fundamental characteristic. By choosing the basic or fundamental characteristic, we can make broad divisions in living organisms as the next level of characteristic is dependent on these. This goes on to form a hierarchy of characteristics.

Question 3.
Explain the basis for grouping organisms into five kingdoms.
Answer:
RE Whittaker proposed a five kingdom classification of living organisms on the basis of Linnaeus system of classification. The five kingdoms proposed by Whittaker are Monera, Protista, Fungi, Plantae, and Animalia.

The basis for grouping organisms into five kingdoms is as follows:

(i) On the basis of the presence or absence of membrane-bound organelles, all living or
ganisms are divided into two broad categories of eukaryotes and prokaryotes. This division lead to the formation of kingdom Monera, which includes all prokaryotes.

(ii) Then, eukaryotes are divided a unicellular and multicellular, on the basis of cellularity. Unicellular eukaryotes form kingdom Protista, and multicellular eukaryotes form kingdom Fungi, Plantae, and Animalia.

(iii) Animals are then separated on the basis of the absence of a cell wall.

(iv) Since fungi and plants both contain a cell wall, they are separated into different kingdoms on the basis of their modes of nutrition. Fungi have saprophytic mode of nutrition, whereas plants have autotrophic mode of nutrition. This results in the formation of the five kingdoms.

Question 4.
What are the major divisions in the Plantae? What is the basis for these divisions?
Answer:
The kingdom Plantae is divided into five main divisions: Thallophyta, Bryophyta, Pteridophyta, Gymnosperms, and Angiosperms.

The classification depends on the following criteria:

• Differentiated/ Undifferentiated plant body
• Presence /absence of vascular tissues
• With/without seeds
• Naked seeds/ seeds inside fruits

(i) The first level of classification depends on whether a plant body is well differentiated or not. A group of plants that do not have a well differentiated plant body are known as Thallophyta.

(ii) Plants that have well differentiated body parts are further divided on the basis of the presence or absence of vascular tissues. Plants without specialised vascular tissues are included in division Bryophyta, whereas plants with vascular tissues are known as Tracheophyta.

(iii) Tracheophyta is again sub-divided into division Pteridophyta, on the basis of the absence of seed formation.

(iv) The other group of plants having well developed reproductive organs that finally develop seeds are called Phanerogams. This group is further sub-divided on the basis of whether the seeds are naked or enclosed in fruits. This classifies them into gymnosperms and angiosperms. Gymnosperms are seed bearing, non-flowering plants, whereas angiosperms are flowering plants in which the seeds are enclosed inside the fruit.

Question 5.
How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?
Answer:
The criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals because

• plants are fixed at a specific position while the animals are motile.
• plants are autotrophic while animals are heterotrophic therefore, their requirements and the circulation of the necessary substances in body is different.
• plants lack nervous system but animals from the phylum cinidaria onwards posses nervous system.

Question 6.
Explain how animals in Vertebrata are classified into further subgroups.
Answer:
Animals in Vertebrata are classified into five classes:

(i) Class Pisces: This class includes fish such as Scoliodon, tuna, rohu, shark, etc. These animals mostly live in water. Hence, they have special adaptive features such as a streamlined body, presence of a tail for movement, gills, etc. to live in water.

(ii) Class Amphibia: It includes frogs, toads, and salamanders. These animals have a dual mode of life. In the larval stage, the respiratory organs are gills, but in the adult stage, respiration occurs through the lungs or skin. They lay eggs in water.

(iii) Class Reptilia: It includes reptiles such as lizards, snakes, turtles, etc. They usually creep or crawl on land. The body of a reptile is covered with dry and cornified skin to prevent water loss. They lay eggs on land.

(iv) Class Aves: It includes all birds such as sparrow, pigeon, crow, etc. Most of them have feathers. Their forelimbs are modified into wings for flight, while hind limbs are modified for walking and clasping. They lay eggs.

(v) Class Mammalia: It includes a variety of animals which have milk producing glands to nourish their young ones. Some lay eggs and some give birth to young ones. Their skin has hair as well as sweat glands to regulate their body temperature.

Class 9 Science Chapter 7 Diversity in Living Organisms Additional Important Questions and Answers

Question 1.
Find out incorrect sentence
(a) Protista includes unicellular eukaryotic organisms
(b) Whittaker considered cell structure, mode and source of nutrition for classifying the organisms in five kingdoms
(c) Both Monera and Protista may be autotrophic and heterotrophic
(d) Monerans have well defined nucleus
Answer:
(d) Monerans have well defined nucleus

Question 2.
Which among the following has specialised tissue for conduction of water?
(i) Thallophyta
(ii) Bryophyta
(iii) Pteridophyta
(iv) Gymnosperms
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(c) (iii) and (iv)

Question 3.
Which among the following produce seeds?
(a) Thallophyta
(b) Bryophyta
(c) Pteridophyta
(d) Gymnosperms
Answer:
(d) Gymnosperms

Question 4.
Which one is a true fish?
(a) Jellyfish
(b) Starfish
(c) Dogfish
(d) Silverfish
Answer:
(c) Dogfish

Question 5.
Which among the following is exclusively marine?
(a) Porifera
(b) Echinodermata
(c)Mollusca
(d) Pisces
Answer:
(b) Echinodermata

Question 6.
Which among the following have open circulatory system?
(i) Arthropoda
(ii) Mollusca
(iii) Annelida
(iv) Coelenterata
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iv)
Answer:
(a) (i) and (ii)

Question 7.
In which group of animals, coelom is filled with blood?
(a) Arthropoda
(b) Annelida
(c) Nematoda
(d) Echinodermata
Answer:
(a) Arthropoda

Question 8.
Elephantiasis is caused by
(a) Wuchereria
(b) Pinworm
(c) Planarians
(d) Liver flukes
Answer:
(a) Wuchereria

Question 9.
Which one is the most striking or (common) character of the vertebrates?
(a) Presence of notochord
(b) Presence of triploblastic condition
(c) Presence of gill pouches
(d) Presence of coelom
Answer:
(a) Presence of notochord

Question 10.
Which among the following have scales?
(i) Xmphibians
(ii) Pisces
(iii) Reptiles
(iv) Mammals
(a) (i) and (iii)
(b) (iii) and (iv)
(c) (ii) and (iii)
(d) (i) and (ii)
Answer:
(c) (ii) and (iii)

Question 11.
Find out the false statement
(a) Aves are warm blooded, egg laying and have four chambered heart
(b) Aves have feather covered body, fore limbs are modified as wing and breathe through lungs
(c) Most of the mammals are viviparous
(d) Fishes, amphibians and reptiles are oviparous
Answer:
(d) Fishes, amphibians and reptiles are oviparous

Question 12.
Pteridophyta do not have
(a) root
(b) stem
(c) flowers
(d) leaves
Answer:
(c) flowers

Question 13.
Identify a member of Porifera
(a) Spongilla
(b) Euglena
(c) Penicillium
(d) Hydra
Answer:
(a) Spongilla

Question 14.
Which is not an aquatic animal?
(a) Hydra
(b) Jellyfish
(c) Corals
(d) Filaria
Answer:
(d) Filaria

Question 15.
Amphibians do not have the following
(a) Three chambered heart
(b) Gills or lungs
(c) Scales
(d) Mucus glands
Answer:
(c) Scales

Question 16.
Organisms without nucleus and cell organelles belong to
(i) fungi
(ii) protista
(iii) cyano bacteria
(iv) archae bacteria
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iv)
(d) (ii) and (iii)
Answer:
(b) (iii) and (iv)

Question 17.
Which of the following is not a criterion for classification of living organisms?
(a) Body design of the organism
(b) Ability to produce one’s own food
(c) Membrane bound nucleus and cell organelles
(d) Height of the plant
Answer:
(d) Height of the plant

Question 18.
The feature that is not a characteristic of protochordata?
(a) Presence of notochord
(b) Bilateral symmetry and coelom
(c) Jointed legs
(d) Presence of circulatory system
Answer:
(c) Jointed legs

Question 19.
The locomotory organs of Echinoderms are
(a) tube feet
(b) muscular feet
(c) jointed legs
(d) parapodia
Answer:
(a) tube feet

Question 20.
Corals are
(a) Poriferans attached to some solid support
(b) Cnidarians, that are solitary living
(c) Poriferans present at the sea bed
(d) Cnidarians that live in colonies
Answer:
(d) Cnidarians that live in colonies

Very Short Answer Questions

Question 1.
Who is called as the father of taxonomy?
Answer:
Carious Linneaus

Question 2.
Which division of plants is also considered as amphibian of plant kingdom?
Answer:
Bryophytes having motile sperms are considered as amphibian of plant kingdom.

Question 3.
Which is the first division of plants to bear true shoot and root?
Answer:
Pteridophytes are the first to bear true root and shoot.

Question 4.
What are sporophylls in ferns?
Answer:
The leaves bearing the sporangia on their lower surface are called the sporophylls.

Question 5.
What are circinate leaves in a dryopteris?
Answer:
Circinate leaves are young leaves with the top margins folded inwardly.

Question 6.
Which part of pinus plant bears naked seeds?
Answer:
The female cone in pinus plant bears the seeds.

Question 7.
Which is the lateral appendage of shoot in angiosperms that acts as the reproductive organ?
Answer:
Flower is the lateral appendage of shoot that acts as the reproductive organ.

Question 8.
Which phase is dominant in a Funaria plant, gametophytic i sporophytic?
Answer:
In funaria, the gametophytic phase is dominant over the sporophytic phase.

Question 9.
Which type of spores are produced by the Funaria sporophyte?
Answer:
The sporophyte always produces the haploid spores that give rise gametophytic thallus.

Question 10.
In phanerogams, which plants bear the tap root system, gymnosperms or angiosperms?
Answer:
In phanerogams, the gymnosperms bear the tap roots.

Question 11.
In which plants the algae and fungi are present in the symbiotic relationship?
Answer:
Lichens consist of algae and fungi in the symbiotic relationship.

Question 12.
Which type of plants do you expect to observe in the artic regions?
Answer:
Bryophytes

Question 13.
Which division of plants consists mostly of ornamental plants?
Answer:
Pteridophytes, mostly ferns.

Question 14.
Name an aquatic pteridophyte that is capable of nitrogen fixation?
Answer:
Azolla

Question 15.
Name any two products of economical value that are obtained from algae.
Answer:
Iodine from kelps and silica from diatoms.

Question 16.
Which term is used to describe the network of hyphae in Rhizopus?
Answer:
Mycelium

Question 17.
Which nutrient medium is mostly used in laboratory for culturing microorganisms?
Answer:
Agar-agar, a product obtained from an algae is widely used as the culture medium.

Question 18.
What are the male and female reproductive organs in a moss or fern?
Answer:
In a moss or fern, the male and female reproductive organs are multicellular. They are called antheridum and archegonium respectively.

Question 19.
Name the division of plants whose members show a clearly distinct phases of alternation of generation.
Answer:
The members of bryophytes shows a clearly distinct alternation of generation with gametophytic phase being dominant over the sporophytic phase.

Question 20.
How is a protozoan different from a metazoan?
Answer:
A protozoan is unicellular but a metazoan is multicellular.

Question 21.
What is reproductive nature of an earthworm, unisexual or bisexual?
Answer:
An earthworm is bisexual, although the ovaries and testes mature at the different rimes.

Question 22.
Which cockroach has the larger antennae, male or female?
Answer:
The male cockroach has larger antennae than the female cockroach to perceive the odour in the surroundings.

Question 23.
Which phylum in animal kingdom has the largest and the second largest number of animals?
Answer:
The phylum arthropods has the largest and phylum mollusca has the second largest number of animals.

Question 24.
Which animal phylum is characterized by the presence of water canal system and water vascular system?
Answer:
Animals belonging to phylum Echinodermata have water vascular system, while the animals belonging to phylum Porifera have water canal system.

Question 25.
Which is the common physical feature in a bird and a fish?
Answer:
A bird and a fish both have streamlined body.

Question 26.
Name the animal that resembles both the reptiles and mammals.
Answer:
Duck billed platypus as it lays eggs like reptiles.

Question 27.
Name an aerial and an aquatic mammal.
Answer:
The only aerial mammal is bat and whale is an aquatic mammal.

Question 28.
Give an example of a parasitic annelida?
Answer:
Leech ( Hinidinaria) is the parasitic annelida. It is parasitic on cattle fronl whom it sucks the blood.

Question 29.
Name any three animals that show the generation of their body parts.
Answer:
Star fish regenerates it lost arm, frog regenerates its limb and a hous lizard regenerates it lost tail.

Question 30.
The animals of which phylum are characterized by the presence of protective hard shell?
Answer:
In the phylum Mollusca, all animals posses protective hard shell.

Question 31.
Sea horse is a vertebrate. In which class has it been placed li classification?
Answer:
Sea horse has been placed in the classes of Pisces under Oestiochythe

Question 32.
Name the three layers present in the embryo of a triploblastic animal.
Answer:
The three layers are ectoderm, mesoderm and endoderm.

Short Answer Questions

Question 1.
Who proposed binomial nomenclature? State its significance.
Answer:
The concept of binomial nomenclature was proposed by Carolus Linnaeus It was meant to give the scientific name to an organism for

• easy identification of an organism because the first name refers to the gen while the second name is the specific epithet.
• maintaining the uniformity of the names worldwide.

Question 2.
Draw a labelled diagram of a typical bacterium.
Answer:

Question 3.
How are prokaryotes different from eukaryotes?
Answer:
The organisms with the prokaryotic cell are characterized by the abs of well defined nucleus and membrane bound cell organelles while those with t eukaryotic cell, plant or animal cell are called eukaryotes.

In five kingdom system, prokaryotes have been placed in kingdom Me while the eukaryotic organisms have been placed in the rest of the four kingdoms.

Question 4.
Why are viruses considered to be at the border line of living and non-livings?
Answer:
Viruses are the obligate parasites, they are living only when in body of a living host otherwise, they act as non-living. They can be filtered and crystallized like chemicals.

Question 5.
What are protozoa? Give examples.
Answer:
Protozoa are the eukaryotic unicellular animals which belongs to the kingdom ‘Protista’. These animals are either free living or parasitic. The free living protozoa include amoeba and paramecium while the parasitic protozoa which are the cause of infectious disease to mankind include

• Plasmodium, the causative agent of malaria.
• Trypanosoma, the causative agent of sleeping sickness.
• Entameoba histolytica, the causative agent of amoebiasis.

Question 6.
Why are the saprophytes called the natural cleaners of the environment?
Answer:
The saprophytes are called the natural cleaners of the environment because they decompose the dead organic matter of living organisms such as the agricultural wastes and animal excreta into simple inorganic form to allow the recycling of the different nutrients.

Question 7.
Name the unicellular fungi with its importance.
Answer:
One of the unicellular fungi is yeast. It is commercially an important gradient as it lays the foundation of brewery and bakery industry because of its ability of anaerobically breaking down of sugar, producing the alcohol, ethanol and carbon dioxide. The ethanol is the product used in brewery industries and produced carbon dioxide is used in bakery industry to make the bread or cake soft and spongy.

Question 8.
Differentiate in the followings:
(a) Monera and Protista
(b) Bryophytes and pteridophytes
(c) Monocots and dicots
(d) Mollusca and echinodermata
(e) Chordata and non-chordata
(f) Cartilaginous and bony fishes
(g) Platyhelminthes and nematoda
Answer:

(b) Bryophytes and pteridophytes

(c) Monocots and dicots

(d) Mollusca and echinodermata

(e) Chordata and non-chordata

(f) Cartilaginous and bony fishes

(g) Platyhelminthes and Nematoda

Question 9.
List some of the important characteristics of protochordates.
Answer:
The protochordates are primitive to chordates.

• They are triploblastic and bilaterally symmetrical with a coelom.
• They are exclusively marine animals.
• They posses a notochord at some stage of their life that separates nervous tissue with gut.
• They have muscles attached to notochord for easy movement.
  Example: Herdmania, Amphixous and Balanglossus.

Question 10.
Write the characteristic adaptive features in the followings:
(a) Fishes for swimming in water.
(b) Birds for flying in air.
Answer:
(a) Fishes are well adapted for swimming in water because they have

• streamlined body to reduce friction when swimming in water.
• scales to prevent the rotting action of water on body.
• have fins to swim and tail fin to steer their motion in water.
• have gills for the gaseous exchange.
• have swim bladder to change their buoyancy i.e. the depth in water.

(b) Birds are well adapted for flying because of the presence of

• streamlined body with kneeled chest to reduce the air friction in air.
• forelimbs modified in form of wings to help bird fly in air.
• pneumatic bones to keep the body weight light.
• air bladder to change their height in air.

Question 11.
Do all birds fly? If not, give examples.
Answer:
No, all birds do not fly. They are few non-flying birds such as

• Ostrich of Africa,
• Kiwi of Newzealand
• Penguin of Antarctica region

Long Answer Questions

Question 1.
How have the organisms being classified on the basis of their mode of nutrition?
Answer:
The organisms on the basis of their mode of nutrition have been basically classified into two types, autotrophs and heterotrophs. Autotrophs can synthesize their own food like green plants but heterotrophs are organisms that are dependent on others for their nutrition. These organisms based on that how they obtain their food have been further categorized into different types such as:

Saprophytes: Organisms who obtain their nourishment from other living organisms called host. They cause serious damage to the host as they obtain their nourishment such as malaria parasite, leech, ticks, etc.

Symbiotic: Organisms which while living together help each other in gaining the nourishment such as lichen. They are composed of algae and fungi, the fungi absorbs water and minerals while green algae synthesize the food that is equally shared by the both algae and fungi.

Question 2.
State the different characteristic features of monera.
Answer:
All members of Monera have the following characteristics:

• They lack multicellular body design.
• The body cell lack well defined nucleus and membrane bound cell organelles.
• Some posses cell wall like bacteria and cyanobacteria while others like mycoplasm lack it.
• They can be autotrophic, heterotrophic or symbiotic in their mode of nutrition.
• Example: Bacteria, mycoplasma and cyanobacteria like nostoc, anabena and oscillatoria.

Question 3.
State the different characteristic features of protista.
Answer:
The kingdom Protista include number of eukaryotic unicellular organisms.

• They vary in their structure, some have locomotory organs like cila or flagella while others don’t like amoeba.
• They vary in their mode of nutrition, some are autotrophic while others are heterotrophic from holozoic to parasitic. Euglena is the only Protista that shows both autotrophic and heterotrophic mode of nutrition.

Question 4.
State the chief characteristics of the division Thallophyta with examples.
Answer:
The chief characteristics of thallophytes include:

• The plant body is in form of individual thallus ire. not differentiated between stem, root and leaf.
• They lack vascular system and tissues for conduction of water and food.
• They are predominantly aquatic.
• Their reproductive organs are single celled.
• They are autotrophic like algae, symbiotic like lichens or saprophytic like fungi in their mode of nutrition.

Question 5.
Which conventions are followed when writing the scientific name of organism?
Answer:
When writing the scientific name of an organism, the sign conventions followed are:

• The name of genus is written first followed by the name of species.
• The name of genus starts with capital alphabet while that of species starts with small alphabet.
• When printed, it is always written in italics and if written by hand, it is underlined.

These NCERT Solutions for Class 9 Science Chapter 6 Tissues Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Tissues NCERT Solutions for Class 9 Science Chapter 6

Class 9 Science Chapter 6 Tissues InText Questions and Answers

Question 1.
What is a tissue?
Answer:
Tissue is a group of cells that are similar in structure and are organised together to perform a specific task.

Question 2.
What is the utility of tissues in multi-cellular organisms?
Answer:
In unicellular organisms, a single cell performs all the basic functions such as respiration, movement, excretion, digestion, etc. But in multicellular organisms, cells are grouped to form tissues. These tissues are specialised to carry out a particular function at a definite place in the body. For example, the muscle cells form muscular tissues which helps in movement, nerve cells form the nervous tissue which helps in transmission of messages. This is known as division of labour in multicellular organisms. It is because of this division of labour that multicellular organisms are able to perform all functions efficiently.

Question 3.
Name types of simple tissues.
Answer:
Simple permanent tissues are of three types: Parenchyma, Collenchyma, and Sclerenchyma. Parenchyma tissue is of further two types – aerenchyma and chlorenchyma.

Question 4.
Where is apical meristem found?
Answer:
Apical meristem is present at the growing tips of stems and roots. Their main function is to initiate growth in new cells of seedlings, at the tip of roots, and shoots.

Question 5.
Which tissue makes up the husk of coconut?
Answer:
The husk of a coconut is made up of sclerenchyma tissue.

Question 6.
What are the constituents of phloem?
Answer:
Phloem is the food conducting tissue in plants. It is made up of four components:

• Sieve tubes
• Companion cells
• Phloem parenchyma
• Phloem fibres

Question 7.
Name the tissue responsible for movement in our body.
Answer:
The muscular tissue is responsible for movement in our body.

Question 8.
What does a neuron look like?
Answer:
A neuron consists of a cell body with a nucleus and cytoplasm. It has two important extensions known as the axon and dendrites. An axon is a long thread-like extension of nerve cells that transmits impulses away from the cell body. Dendrites, on the other hand, are thread-like extensions of cell body that receive nerve impulses. Thus, the axon transmits impulses away from the cell body, whereas the dendrite receives nerve impulses. This coordinated function helps in transmitting impulses very quickly.

Question 9.
What are the functions of areolar tissue?
Answer:
Functions of areolar tissue:

• It helps in supporting internal organs.
• It helps in repairing the tissues of the skin and muscles.

Class 9 Science Chapter 6 Tissues Textbook Questions and Answers

Question 1.
Define the term “tissue”.
Answer:
Tissue is a group of cells that are similar in structure and are organized together to perform a specific task.

Question 2.
How many types of elements together make up the xylem tissue? Name them.
Answer:
There are four different types of cells that make up the xylem tissue. They are:

• Tracheids
• Vessels
• Xylem parenchyma
• Xylem fibres

Question 3.
How are simple tissue? different from complex tissues in plants?
Answer:
Simple tissue:

1. It consists of the cells that similar in both their structure and functions.
2. Parenchyma, collenchyma and sclerenchyma.

Complex tissue:

1. It consists of the cells that are neither similar in structure nor in function but collectively help to perform a specific function.
2. Xylem and phloem

Question 4.
Differentiate between parenchyma, collenchyma and sclerenchyma, on the basis of their cell wall.
Answer:

Question 5.
What are the functions of the stomata?
Answer:
Functions of the stomata:

• They allow the exchange of gases (CO₂ and O₂) with the atmosphere.
• Evaporation of water from the leaf surface occurs through the stomata. Thus, the stomata help in the process of transpiration.

Question 6.
Diagrammatically show the difference between the three types of muscle fibres.
Answer:
The three types of muscle fibres are: Striated muscles, smooth muscles (unstriated muscle fibre), and cardiac muscles.

Question 7.
What is the specific function of the cardiac muscle?
Answer:
The specific function of the cardiac muscle is to control the contraction and relaxation of the heart.

Question 8.
Draw a labelled diagram of a neuron.
Answer:

Question 9.
Name the following:
(a) Tissue that forms the inner lining of our mouth.
(b) Tissue that connects muscle to bone in humans.
(c) Tissue that transports food in plants.
(d) Tissue that stores fat in our body.
(e) Connective tissue with a fluid matrix.
(f) Tissue present in the brain.
Answer:
(a) Tissue that forms the inner lining of our mouth ? Epithelial tissue
(b) Tissue that connects muscle to bone in humans? Dense regular connective tissue (tendons)
(c) Tissue that transports food in plants ? Phloem
(d) Tissue that stores fat in our body ? Adipose tissue
(e) Connective tissue with a fluid matrix ? Blood
(f) Tissue present in the brain ? Nervous tissue

Question 10.
Identify the type of tissue in the following: skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Answer:
Skin: Stratified squamous epithelial tissue
Bark of tree: Simple permanent tissue
Bone: Connective tissue
Lining of kidney tubule: Cuboidal epithelial tissue
Vascular bundle: Complex permanent tissue

Question 11.
Name the regions in which parenchyma tissue is present.
Answer:
Leaves, fruits, and flowers are the regions where the parenchyma tissue is present.

Question 12.
What is the role of epidermis in plants?
Answer:
Epidermis is present on the outer surface of the entire plant body. The cells of the epidermal tissue form a continuous layer without any intercellular space. It performs the following important functions:

• It is a protective tissue of the plant body
• It protects the plant against mechanical injury
• It allows exchange of gases through the stomata

Question 13.
How does the cork act as a protective tissue?
Answer:
The outer protective layer or bark of a tree is known as the cork. It is made up of dead cells. Therefore, it protects the plant against mechanical injury, temperature extremes, etc. It also prevents the loss of water by evaporation.

Question 14.
Complete the following chart:

Answer:

Class 9 Science Chapter 6 Tissues Additional Important Questions and Answers

Question 1.
Which of the following tissues has dead cells?
(a) Parenchyma
(b) Scierenchyma
(c) Collenchyma
(d) Epithelia tissue
Answer:
(b) Scierenchyma

Question 2.
Find out incorrect sentence
(a) Parenchymatous tissues have intercellular spaces
(b) Collenchymatous tissues are irregularly thickened at corners
(c) Apical and intercalary meristems are permanent tissues
(d) Meristeniatic tissues, in its early stage, lack vacuoles
Answer:
(c) Apical and intercalary meristems are permanent tissues

Question 3.
Girth of stem increases due to
(a) apical meristem
(b) lateral meristem
(c) intercalary meristem.
(d) vertical meristem
Answer:
(b) lateral meristem

Question 4.
Which cell does not have perforated cell wall?
(a) Tracheids
(b) Companion cells
(c) Sieve tubes
(d) Vessels
Answer:
(b) Companion cells

Question 5.
Intestine absorb the digested food materials. What type of epithelial cells are responsible for that?
(a) Stratified squamous epithelium
(b) Columnar epithelium
(c) Spindle fibres
(d) Cuboidal epithelium
Answer:
(b) Columnar epithelium

Question 6.
A person met with an accident in which two long bones of hand were dislocated. Which among the following may be the possible reason?
(a) Tendon break
(b) Break of skeletal muscle
(c) Ligament break
(d) Areolar tissue break
Answer:
(c) Ligament break

Question 7.
While doing work and running, you move your organs like hands, legs etc. Which among the following is correct?
(a) Smooth muscles contract and pull the ligament to move the bones
(b) Smooth muscles contract and pull the tendons to move the bones
(c) Skeletal muscles contract and pull the ligament to move the bones
(d) Skeletal muscles contract and pull the tendon to move the bones
Answer:
(d) Skeletal muscles contract and pull the tendon to move the bones

Question 8.
Which muscles act involuntarily?
(i) Striated muscles
(ii) Smooth muscles
(iii) Cardiac muscles
(iv) Skeletal muslces
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(b) (ii) and (iii)

Question 9.
Meristematic tissues in plants are
(a) localised and permanent
(b) not limited to certain regions
(c) localised and dividing cells
(d) growing in volume
Answer:
(c) localised and dividing cells

Question 10.
Which is not a function of epidermis?
(a) Protection from adverse condition
(b) Gaseous exchange
(c) Conduction of water
(d) Transpiration
Answer:
(c) Conduction of water

Question 11.
Select the incorrect sentence
(a) Blood has matrix containing proteins, salts and hormones
(b) Two bones are connected with ligament
(c) Tendons are non-fibrous tissue and fragile
(d) Cartilage is a form of connective tissue
Answer:
(c) Tendons are non-fibrous tissue and fragile

Question 12.
Cartilage is not found in
(a) nose
(b) ear
(c) kidney
(d) larynx
Answer:
(c) kidney

Question 13.
Fats are stored in human body as
(a) cuboidal epithelium
(b) adipose tissue
(c) bones
(d) cartilage
Answer:
(b) adipose tissue

Question 14.
Bone matrix is rich in
(a) fluoride and calcium
(b) calcium and phosphorus
(c) calcium and potassium
(d) phosphorus and potassium
Answer:
(b) calcium and phosphorus

Question 15.
Contractile proteins are found in
(a) bones
(b) blood
(c) muscles
(d) cartilage
Answer:
(c) muscles

Very Short Answer Questions

Question 1.
Which tissue in plants has cells mostly found in state of division?
Answer:
Meristematic tissue present at the root and shoot tip has cells found in the state of division.

Question 2.
Which permanent tissue in plants consists of cells with thin cell wall?
Answer:
Parenchyma tissue

Question 3.
What is chlorenchyma?
Answer:
It’s a type of parenchyma whose cells possess the chloroplast containing chlorophyll.

Question 4.
Name the three types of simple permanent tissues found in plants.
Answer:
Parenchyma, collenchyma and sclerenchyma

Question 5.
Which simple permanent plant tissue have cell walls thickened at the comers?
Answer:
Collenchyma has somewhat more elongated cells with thickened walls corner.

Question 6.
Which are the two types of complex permanent tissues found in plants.
Answer:
Xylem and phloem

Question 7.
Which plant tissue cells are found in the apical part of stem and root?
Answer:
Meristematic tissue consists of parenchymatous cells with active protoplasm.

Question 8.
Name the only living component present in xylem?
Answer:
Xylem parenchyma

Question 9.
Name the only non-living component present in phloem.
Answer:
Phloem fibres

Question 10.
Name the four types of animal tissues?
Answer:
The lour types of animal tissues are epithelial tissue, connective tissues, musclar tissues and nervous tissue.

Question 11.
Which tissue forms the outer human skin?
Answer:
Epithelial tissue forms the outer human skin consisting of dead cells.

Question 12.
Which animal tissue you expect to observe when watching a thin slice Jther?
Answer:
Epithelial tissue as it makes, the skin from which the leather is made after tanning?

Question 13.
Bone is a type of tissue. In which tissue type will you classify the bones?
Answer:
Bone is a hard skeletal tissue, a type of connective tissue.

Short Answer Questions

Question 1.
How is a simple permanent tissue in plants different from compound permanent tissue?
Answer:
A simple permanent tissue consists of cells that are similar both in their structure and functions but the compound permanent tissue consists of cells that are different in their structure but are associated in performing a specific common function.

Question 2.
How is a cell from the meristematic tissue different from the cell from a simple permanent tissue?
Answer:
The meristematic tissue cells are thin walled with active protoplasm and rrninent nucleus. They also have reduced vacuole as compared to the living is of the permanent tissues.

Question 3.
What are the characteristics of the parenchymatous tissue?
Answer:
The chief characteristics of parenchymatous tissue include:

• Cells are mostly round or polygonal but can also be elongated.
• Cells have thin cell wall with uniform thickness.
• Cells are live with nucleus and large vacuoles.
• Cells may or may not have the intracellular, spaces.

Question 4.
What is specific about the cells of collenchymatous tissue in plants?
Answer:
The cells of collenchymatous tissue have thin cells walls but these walls are thickened at the corners because of the deposition of extra cellulose and pectin.

These cells also lack intercellular spaces being closely packed to each other.

Question 5.
State the functions of collenchymatous tissues with their locations in a plant.
Answer:
Collenchymatous tissue mostly found located under the epidermis helps

• providing mechanical support and elasticity to plant parts.
• providing flexibility to plant parts like leaves and branches to prevent them from breaking or tearing in blowing wind.
• manufacture of sugar and starch if they contain chloroplasts.

Question 6.
Describe the structure and function of sclerenchyma as a simple plant tissue.
Answer:
Sclerenchyma as a tissue consists of dead cells because of the excessive deposition of cutin, pectin and lignin in their cell walls. These chemicals act as cement hence, makes the cell wall hard. The cells are long, narrow and fine

Question 7.
Differentiate in the followings:
(a) Bone and cartilage
(b) Tendon and ligament
(c) Meristematic and permanent tissue
(d) Cardiac muscles and striated muscles
(e) Blood and lymph
Answer:
(a) Bone and Cartilage

(b) Tendon and ligament

(c) Meristematic and permanent tissue

(d) Cardiac muscles and striated muscles

(e) Blood and lymph

Question 8.
What do you understand by apical dominance? Why is it exercised?
Answer:
Apical dominance refers to the dominance of apical meristem over other meristems. This in most plants delays the branching of stem. Therefore, to promote: ranching in hedge plants, a gardener cuts off the apical meristem.

Long Answer Questions

Question 1.
What are meristematic tissues? State their characteristics with locations.
Answer:
Meristematic tissues are tissues found at the meristematic regions i.e. growing regions in plants. Their cells have active protoplasm, with small or no vacuole, no chloroplasts but a large active nucleus. The cells are compactly packed with less or no intercellular spaces. The cells are mostly involved in division to form new cells which later undergo differentiation to from different cells to add to plant growth.

Meristems on basis of its location has been classified into three types:
Apical meristem, present at the growing tips of stem and root. It contributes to vertical plant growth.
Intercalary meristem, present at the base of leaves and internodes where they contribute to the branching of stem.

Lateral meristem, present on the lateral side of stem and root in form of cambium. These meristems contribute to the growth in thickness i.e. increase in thickness or girth of stem and root.

Question 2.
State the functions of the different parenchymatous tissue.
Answer:
Parenchymatous tissue performs the following functions:

• They mostly store and assimilate food material.
• When in form of chlorenchyma in leaves and other green plant parts, they help in photosynthesis.
• In aquatic plants, their cells bear large air cavities and hence, help provide the necessary buoyancy to the plant parts for floating in water.
• They provide mechanical support also particularly the xylem parenchyma and pholem parenchyma.

Question 3.
What are vascular bundles? State their functions.
Answer:
Vascular bundles are the bundles comprising of xylem and pholem.

Xylem consist of four components: tracheids, vessel cells, xylem fibres and xylem parenchyma. Of these components, only xylem parenchyma is living the rest are the dead structures. Tracheids and vessels cells are tubular structures with a lumen to allow the transport of water and minerals from root to leaves. Their thick walls have pits too, for the lateral transport of water.

Phloem consists of four components: sieve cells, companion cells, phloem parenchyma and phloem fibres. In these components, phloem fibres are dead and are meant to provide, mechanical strength to the tissue, while the rest are living structures. However, the sieve cells are live but lack nucleus and hence, they are always present along with companion cells. These cells perform cellular activities for sieve cells. The sieve cells or sieve tubes allow the bidirectional flow of food.

Question 4.
Describe briefly the different types of epithelial tissues with their functions.
Answer:
The different types of epithelial tissues include:
(i) Squamous epithelial tissue: Their cells are thin, flattened, polygonal like the tiles of floor. They form the outer layer of skin, tongue, esophagus and the inner lining of mouth.

(ii) Cuboidal epithelial tissue: Their cells are mostly cuboidal in shape. They form the lining of kidney tubules and ducts of salivary glands. These cells provide mechanical support and if ciliated, they help in pushing the content forward such as in oviduct.

(iii) Columnar epithelial tissue: Their cells are long column or pillar like. They form the inner lining of the intestine and if ciliated, they help in movement, such as in respiratory tract, the ciliated columnar epithelial cells help in pushing the mucus upward from lungs to mouth.

(iv) Glandular epithelial tissue: Their cells are also columnar or cuboidal shape and involved in secretion of mucus and digestive enzymes and therefore are found in different body glands.

Question 5.
Name the different types of connective tissues with their functions.
Answer:
Areolar tissue: ft is present in between the skin and underlying muscles contains different types of cells that are responsible for the formation of different fibre types i.e. collagen fibres and elastin fibres. Some cells present also synthesize heparin, an anticoagulant and histamine.

Tendon: These are white fibrous tissues which attaches the muscles to bones.

Ligament: These are yellow fibrous tissues which attaches the two bones together at the point of their joint.

Adipose tissue: This tissue underlying the skin stores excess of fats and provide insulation to the body.

Blood is fluid connective tissue that helps in circulation of nutrients, metabolic waste, gases, etc in body along with providing the immunity or defense against any infection.

Question 6.
What are the different components of blood? What are their functions?
Answer:
Blood is the fluid connective tissue that consists of blood cells found immersed in blood plasma. Blood plasma is the fluid part of blood. In blood three different cell types are found: RBC, WBC and blood platelets.

Red blood cells(RBC) are carrier of haemoglobin and therefore responsible for the transport of oxygen from lungs to body parts.

White blood cells(WBC) are either amoeboid shaped called phagocytes or large oval cells with large nucleus called lymphocytes. Phagocytes provide defense t engulfing the germs and lymphocytes provide defense by secreting antibodies: fight to invading germs.

Blood platelets are small fragments like cells that are responsible for blood coagulation after the injury.

These NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

The Fundamental Unit of Life NCERT Solutions for Class 9 Science Chapter 5

Class 9 Science Chapter 5 The Fundamental Unit of Life InText Questions and Answers

Question 1.
Who discovered cells and how?
Answer:
Cells were discovered in 1665 by an English Botanist, Robert Hooke. He used a primitive microscope to observe cells in a cork slice.

Question 2.
Why is the cell called the structural and functional unit of life?
Answer:
Cells constitute various components of plants and animals. A cell is the smallest unit of life and is capable of all living functions. Cells are the building blocks of life. This is the reason why cells are referred to as the basic structural and functional units of life. All cells vary in their shape, size, and activity they perform. In fact, the shape and size of the cell is related to the specific functions they perform.

Question 3.
How do substances like CO₂ and water move in and out of the cell? Discuss.
Answer:
The cell membrane is selectively permeable and regulates the movement of substances in and out of the cell.

Movement of CO₂:
CO₂ is produced during cellular respiration. Therefore, it is present in high concentrations inside the cell. This CO₂ must be excreted out of the cell. In the cell’s external environment, the concentration of CO₂ is low as compared to that inside the cell. Therefore, according to the principle of diffusion, CO₂ moves from a region of higher concentration (inside the cell) towards a region of lower concentration (outside the cell). Similarly, O₂ enters the cell by the process of diffusion when the concentration of O₂ inside the cell is low as compared to its surroundings.

Movement of water:
Water moves from a region of high concentration to a region of low concentration through the plasma membrane. The plasma membrane acts as a semi-permeable membrane, and this movement of water is known as osmosis.

However, the movement of water across the plasma membrane of the cell is affected by the amount of substance dissolved in water.

Question 4.
Why is the plasma membrane called a selectively permeable membrane?
Answer:
The cell membrane or the plasma membrane is known as a selectively permeable membrane because it regulates the movement of substances in and out of the cell. This means that the plasma membrane allows the entry of only some substances and prevents the movement of some other materials.

Question 5.
Fill in the gaps in the following table illustrating differences between prokaryotic and eukaryotic cells.

Answer:

Question 6.
Can you name the two organelles we have studied that contain their own genetic material?
Answer:
Mitochondria and plastids are the two organelles that contain their own genetic material. Both these organelles have their own DNA and ribosomes.

Question 7.
If the organisation of a cell is destroyed due to some physical or chemical influence, what will happen?
Answer:
Cell is the smallest unit of life, which is capable of all living functions. If the organisation of a cell is destroyed due to some physical or chemical influence, then the ability of the cell to perform all living functions such as respiration, nutrition, excretion, etc. would be affected.

Question 8.
Why are lysosomes known as suicide bags?
Answer:
Lysosomes are membrane-bound vesicular structures that contain powerful digestive enzymes. These enzymes are capable of breaking down any foreign food particle or microbes entering the cell. Sometimes, lysosomes can cause self-destruction of a cell by releasing these digestive enzymes within the cells. Hence, they are also known as ‘suicidal bags’.

Question 9.
Where are proteins synthesized inside the cell?
Answer:
Ribosomes are the site for protein synthesis. Ribosomes are very small structures found either in a free state, suspended in the cytoplasm, or attached to the surface of the endoplasmic reticulum. They are composed of ribonucleic acids and proteins.

Class 9 Science Chapter 5 The Fundamental Unit of Life Textbook Questions and Answers

Question 1.
Make a comparison and write down ways in which plant cells are different from animal cells.
Answer:

Question 2.
How is a prokaryotic cell different from a eukaryotic cell?
Answer:
Prokaryotic cell :

1. Most prokaryotic cells are unicellular.
2. Size of the cell is generally small (0.5-5 μm).
3. Nuclear region is poorly, defined due to the absence of a nuclear membrane or the cell lacks true nucleus.
4. It contains a single chromosome.
5. Nucleolus is absent.
6. Membrane-bound cell organelles such as plastids, mitochondria, endoplasmic reticulum, Golgi apparatus, etc. are absent.
7. Cell division occurs only by mitosis.
8. Prokaryotic cells are found in bacteria and blue-green algae.

Eukaryotic cell:

1. Most eukaryotic cells are multicellular.
2. Size of the cell is generally large (50-100 μm).
3. Nuclear region is well-defined and is surrounded by a nuclear membrane, or true nucleus bound by a nuclear membrane is present in the cell.
4. It contains more than one chromosome.
5. Nucleolus is present.
6. Cell organelles such as mitochondria, plastids, endoplasmic reticulum, Golgi apparatus, lysosomes, etc. are present.
7. Cell division occurs by mitosis and meiosis.
8. Eukaryotic cells are found in fungi, plants, and animal cells.

Question 3.
What would happen if the plasma membrane ruptures or breaks down?
Answer:
If the plasma membrane of a cell is ruptured, then the cell will die. The plasma membrane regulates the movement of substances in and out of the cell by diffusion or osmosis. Thus, if the plasma membrane is ruptured, then the cell might leak out its contents.

Question 4.
What would happen to the life of a cell if there was no Golgi apparatus?
Answer:
If there was no Golgi apparatus in the cell, then most activities performed by the Golgi apparatus will not take place.
(i) Membranes of the Golgi apparatus are often connected to ER membranes. It collects simpler molecules and combines them to make more complex molecules. These are then packaged in small vesicles and are either stored in the cell or sent out as per the requirement. Thus, if the Golgi apparatus is absent in the cell, then the above process of storage, modification, and packaging of products will not be possible.

(ii) The formation of complex sugars from simple sugars will not be possible as this takes place with the help of enzymes present in Golgi bodies.

(iii) The Golgi apparatus es involved in the formation of lysosomes or peroxisomes. Thus, if the Golgi body is absent in a cell, the synthesis of lysosomes or peroxisomes will not be possible in the cell.

Question 5.
Which organelle is known as the powerhouse of the cell? Why?
Answer:
Mitochondria are known as the powerhouse of cells. Mitochondria create energy for the cell, and this process of creating energy for the cell is known as cellular respiration. Most chemical reactions involved in cellular respiration occur in the mitochondria. The energy required for various chemical activities needed for life is released by the mitochondria in the form of ATP (Adenosine triphosphate) molecules. For this reason, mitochondria are known as the powerhouse of cells.

Question 6.
Where do the lipids and proteins constituting the cell membrane get synthesized?
Answer:
Lipids and proteins constituting the cell membrane are synthesized in the endoplasmic reticulum.
SER (Smooth endoplasmic reticulum) helps in the manufacturing of lipids.
RER (Rough endoplasmic reticulum) has particles attached to its surface, called ribosomes. These ribosomes are the site for protein synthesis.

Question 7.
How does an Amoeba obtain its food?
Answer:
Amoebaobtains its food through the process of endocytosis. The flexibility of the cell membrane enables the cell to engulf the solid particles of food and other materials from its external environment.

Question 8.
What is osmosis?
Answer:
The movement of water molecules from a region of high concentration to a region of low concentration through a selectively permeable membrane is called osmosis. It is a special case of diffusion, where the medium is water.

For example, if the medium surrounding the cell has a higher water concentration than the cell i.e., if the solution is a dilute solution, then the cell will gain water by osmosis.

Question 9.
Carry out the following osmosis experiment:
Take four peeled potato halves and scoop each one out to make potato cups. One of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water.
Now-
(a) Keep cup A empty
(b) Put one teaspoon sugar in cup B
(c) Put one teaspoon salt in cup C
(d) Put one teaspoon sugar in the boiled potato cup D.
Keep these for two hours. Then observe the four potato cups and answer the following:
(i) Explain why water gathers in the hollowed portion of Band C.
(ii) Why is potato A necessary for this experiment?
(iii) Explain why water does not gather in the hollowed out portions of A and D.
Answer:
(i) Water gathers in the hollowed portions of set-up B and C because water enters the potato as a result of osmosis. Since the medium surrounding the cell has a higher water concentration than the cell, the water moves inside by osmosis. Hence, water gathers in the hollowed portions of the potato cup.

(ii) Potato A in the experiment acts as a control set-up. No water gathers in the hollowed portions of potato A.

(iii) Water does not gather in the hollowed portions of potato A because potato cup A is empty. It is a control setup in the experiment.

Water is not able to enter potato D because the potato used here is boiled. Boiling denatures the. proteins present in the cell membrane and thus, disrupts the cell membrane. For osmosis, a semi-permeable membrane is required, which is disrupted in this case. Therefore, osmosis will not occur. Hence, water does not enter the boiled potato cup.

Class 9 Science Chapter 5 The Fundamental Unit of Life Additional Important Questions and Answers

Multiple Choice Questions
Choose the correct option:

Question 1.
Which of the following can be made into crystal?
(a) A Bacterium
(b) An Amoeba
(c) A Virus
(d) A Sperm
Answer:
(c) A Virus

Question 2.
A cell will swell up if
(a) The concentration of water molecules in the cell is higher than the concentration of water molecules in surrounding medium
(b) The concentration of water molecules in surrounding medium is higher than water molecules concentration in the cell
(c) The concentration of water molecules is same in the cell and in the surrounding medium
(d) Concentration of water molecules does not matter
Answer:
(b) The concentration of water molecules in surrounding medium is higher than water molecules concentration in the cell

Question 3.
Chromosomes are made up of
(a) DNA
(b) protein
(c) DNA and protein
(d) RNA
Answer:
(c) DNA and protein

Question 4.
Which of these options are not a function of Ribosomes?
(i) It helps in manufacture of protein molecules
(ii) It helps in manufacture of enzymes
(iii) It helps in manufacture of hormones
(iv) It helps in manufacture of starch mol¬ecules
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (iv) and (i)
Answer:
(c) (iii) and (iv)

Question 5.
Which of these is not related to endoplasmic reticulum?
(a) It behaves as transport channel for proteins between nucleus and cytoplasm
(b) It transports materials between various regions in cytoplasm
(c) It can be the site of energy generation
(d) It can be the site for some biochemical activities of the cell
Answer:
(c) It can be the site of energy generation

Question 6.
Following are a few definitions of osmosis Read carefully and select the correct definition
(a) Movement of water molecules from a region of higher concentration to a region of lower concentration through a semipermeable membrane
(b) Movement of solvent molecules from its higher concentration to lower concentration
(c) Movement of solvent molecules from higher concentration to lower concentration of solution through a permeable membrane
(d) Movement of solute molecules from lower concentration to higher concentration of solution through a semipermeable membrane
Answer:
(a) Movement of water molecules from a region of higher concentration to a region of lower concentration through a semipermeable membrane

Question 7.
Plasmolysis in a plant cell is defined as
(a) break down (lysis ) of plasma membrane in hypotonic medium
(b) shrinkage of cytoplasm in hypertonic medium
(c) shrinkage of nucleoplasm
(d) none of them
Answer:
(b) shrinkage of cytoplasm in hypertonic medium

Question 8.
Which of the following are covered by a single membrane?
(a) Mitochondria
(b) Vacuole
(c) Lysosome
(d) Plastid
Answer:
(b) Vacuole

Question 9.
Find out the false sentences
(a) Golgi apparatus is involved with the formation of lysosomes
(b) Nucleus, mitochondria and plastid have DNA; hence they are able to make their own structural proteins
(c) Mitochondria is said to be the powerhouse of the cell as ATP is generated in them.
(d) Cytoplasm is called as protoplasm
Answer:
(a) Golgi apparatus is involved with the formation of lysosomes

Question 10.
Find out the correct sentence
(a) Enzymes packed in Lysosomes are made through RER (rough endoplasmic reticulum)
(b) Rough endoplasmic reticulum and
smooth endoplasmic reticulum produce lipid and protein respectively
(c) Endoplasmic reticulum is related with the destruction of plasma membrane
(d) Nucleoid is present inside the nucleoplasm of eukaryotic nucleus
Answer:
(a) Enzymes packed in Lysosomes are made through RER (rough endoplasmic reticulum)

Question 11.
Which cell organelle plays a crucial role in detoxifying many poisons and drugs in a cell?
(a) Golgi apparatus
(b) Lysosomes
(c) Smooth endoplasmic reticulum
(d) Vacuoles
Answer:
(c) Smooth endoplasmic reticulum

Question 12.
The proteins and lipids, essential for building the cell membrane, are manufactured by
(a) rough endoplasmic reticulum
(b) golgi apparatus
(c) plasma membrane
(d) mitochondria
Answer:
(a) rough endoplasmic reticulum

Question 13.
The undefined nuclear region of prokaryotes are also known as
(a) nucleus
(b) nucleolus
(c) nucleic acid
(d) nucleoid
Answer:
(d) nucleoid

Question 14.
The cell organelle involved in forming complex sugars from simple sugars are
(a) endoplasmic reticulum
(b) ribosomes
(c) plastids
(d) golgi apparatus
Answer:
(d) golgi apparatus

Question 15.
Which out of the following is not a function of vacuole?
(a) Storage
(b) Providing turgidity and rigidity to the cell
(c) Waste excretion
(d) Locomotion
Answer:
(d) Locomotion

Question 16.
Amoeba acquires its food through a process, termed
(a) exocytosis
(b) endocytosis
(c) plasmolysis
(d) exocytosis and endocytosis both
Answer:
(b) endocytosis

Question 17.
Cell wall of which one of these is not made up of cellulose?
(a) Bacteria
(b) Hydrilla
(c) Mango tree
(d) Cactus
Answer:
(a) Bacteria

Question 18.
Silver nitrate solution is used to study
(a) endoplasmic reticulum
(b) golgi apparatus
(c) nucleus
(d) mitochondria
Answer:
(b) golgi apparatus

Question 19.
Organelle other than nucleus, containing DNA is
(a) endoplasmic reticulum
(b) golgi apparatus
(c) mitochondria
(d) lysosome
Answer:
(c) mitochondria

Question 20.
Kitchen of the cell is
(a) mitochondria
(b) endoplasmic reticulum
(c) chloroplast
(d) golgi apparatus
Answer:
(c) chloroplast

Very Short Answer Questions

Question 1.
Define a cell.
Answer:
Cell is the fundamental structural and functional unit of life.

Question 2.
Which is the outermost boundary of a plant cell?
Answer:
The cell wall is the outermost boundary of a plant cell.

Question 3.
Which is the outermost boundary of an animal cell?
Answer:
Cell (Plasma) membrane is the outermost boundary of an animal cell.

Question 4.
Why do you use glycerine when preparing the temporary mount of a cell?
Answer:
Glycerine is used to prevent the dehydration of cells.

Question 5.
Name the cell organelles found in a green plant cell but not in an animal cell.
Answer:
Chloroplast containing chlorophyll that imparts the green colour to a plant cell.

Question 6.
How is cell wall different from cell membrane?
Answer:
Cell wall found in plant cell, made up of cellulose is dead but cell membrane found in both plant and animal cell, made up of protein and phospholipids is living.

Question 7.
How will you identify the nucleus in the plant cell?
Answer:
Nucleus is the dark pink stained small round body found at the periphery or the cell in a safranin stained cells.

Question 8.
How do you identify the cytoplasm in an animal cell?
Answer:
Cytoplasm is the light blue stained surface in between nucleus and cell membrane, in a cell stained with methylene blue or light pink in a cell stained with safranin.

Question 9.
Why are cells in onion peel rectangular but check cells are irregular in shape?
Answer:
Onion peel cells have rigid cell wall that imparts them a specific shape but cell membrane being flexible and elastic fails to provide it a definite shape.

Question 10.
Why a student needs to stain the cells when preparing temporary mount?
Answer:
A student needs to stain the cells when preparing temporary mount to develop a contrast in its different parts.

Question 11.
Which adjustment will you use in microscope to focus the image of the specimen?
Answer:
To focus the image of the specimen, the coarse adjustment is used by the student.

Question 12.
Which are the two main functions of a microscope?
Answer:
A microscope both resolves and magnifies the image of the specimen.

Question 13.
When you will observe less number of magnified and resolved cells of a temporary mount underflow power or high power?
Answer:
Under the high power of 40X or 45X

Question 14.
Which is the longest body cell?
Answer:
A neuron is the longest body cell.

Question 15.
Which cell in the human body has the finest cell membrane?
Answer:
Red blood cells in. the blood have the finest cell membrane.

Question 16.
Which cell in a plant is living but non-nucleated?
Answer:
Sieve cells present in the phloem are living but non-nucleated.

Question 17.
Name the largest and the smallest living cell.
Answer:
An ostrich egg is the largest living cell while the PPLO, (Pleuro pneumonia like organism) a mycoplasm is the smallest living cell.

Question 18.
In which cells the vacuoles are small and not much prominent?
Answer:
An animal cell.

Question 19.
Which cell organelle is found both in prokaryotic and eukaryotic cells?
Answer:
Ribosome

Question 20.
What are mesosomes?
Answer:
These are imaginations of plasma membrane found in prokaryotic cells. They are secretory in function and the site of cellular respiration.

Short Answer Questions

Question 1.
State the contribution of Anton Von Leuween Hooke in developments that have taken place in field of cell biology.
Answer:
Anton vop Leeuwen Hooke has observed the first living cell of unicellular organisms in a drop of pond water under his self built simple microscope. He is also considered as the pioneer of the simple microscope.

Question 2.
Do all organisms have similar type of cells?
Answer:
No, all organisms do not have similar type of cells because the shape and size of the cell depends upon the function, a cell needs to perform. Therefore, some cells are oval, some are motile with cilia or flagella and some are long like muscle fibres or neuron.

Question 3.
What are cell organelles? Give examples.
Answer:
Cell organelles are granule like structures found immersed in cell cytoplasm. Each cell organelle has a specific function to perform such as mitochondria is the seat of respiration and chloroplasts are the seat of photosynthesis in green plant cells.

Question 4.
Define the following terms: Protoplasm, Cytoplasm and Nucleoplasm.
Answer:
Protoplasm includes all the living material found in a cell invested by thin cell membrane. It is comprised of the nucleus and cytoplasm.

Cytoplasm is the jelly like substance containing call organelles that is present in between the nucleus and cell membrane. The cytoplasm close to nucleus is in sol state while the cytoplasm close to cell membrane in gel state.

Nucleoplasm is the jelly like substance that is presentin the nucleus of a cell. The network of fibre like structure called chromatin material is found immersed r it along with different types of RNA.

Question 5.
State the major postulates of cell theory.
Answer:
The major postulates of cell theory proposed by Schwan and Schleiden are:

• Cell is the fundamental structural unit of all living organisms.
• Cell is the fundamental functional unit of all living organisms.
• Every cell arises from the pre-existing cell.
• Cell is the carrier of genetic information from one generation to the next generation.

Question 6.
Differentiate in the followings:
(i) Cell wall and cell membrane
(ii) DNA and RNA
Answer:
(i)

(ii)

Question 7.
What are genes? State their functions.
Answer:
Genes are the segments of DNA arranged in a linear order in a chromosome.

Genes are the units of hereditary characteristics as when transferred from one generation to the next generation, they carry the codes of different characters with Item.

Question 8.
Give examples of non-membranous single-membranous and double membranous cell organelles.
Answer:
Non-membranous cell organelles—Ribosome and nucleolus.
Single membranous organelles—Lysosome and vacuole
Double membranous organelles—Mitochondria and chloroplast

Question 9.
Which are the three functional regions of a cell?
Answer:
The three functional regions of a cell include:

• Nucleus, it acts as the control centre of all cellular activities.
• Cytopiasirvjelly like substance with cell organelles carrying out different cellular activities.
• Plasma membrane, active in regulation of the transport of different substances across it.

Question 10.
State the functions of the following organelles :
Mitochondria, chloroplasts, lysosomes, golgi apparatus, ribosome, S.E.R.
Answer:
Mitochondria: They are the seat of cellular respiration and generation of energy in form of ATP and hence, also called as the powerhouse of cell.

Chloroplasts: They are the seat of photosynthesis in cells of green plant parts hence, also called as the kitchen of a green plant cell.

Lysosomes: These are single membranous vacuole like structures that are filled with hydrolytic enzymes and hence, responsible for intracellular digestion.

Golgi apparatus: These are the stacks of cisternae responsible for the formation of primary lysosomes, secretion and excretion from the cell.

Ribosomes: These are non-membranous granule like structures which are the seat of protein synthesis and hence, also called as the protein factories of a cell.

S.E.R.: These are endoplasmic reticulum that are devoid of ribosome on their surface. They are known to take active part in the steroid synthesis in a cell.

Question 11.
What happens when a R.B.C and a plant cell is immersed in distilled water? Why?
Answer:
In both RBC and plant cell the endosmosis takes place. The distilled water act as hypotonic solution and the cell cytoplasm act as hypertonic solution.

The plant cell swells with incoming of water and become turgid but the RBC burst open with incoming water as cell membrane alone being flexible and elastic fail to bear the pressure exerted by the incoming water.

Question 12.
What do you expect to happen if a student immerses
(i) few grapes in concentrated sugar solution? Why?
(ii) few raisins in distilled water.
Answer:
(i) The grapes will undergo plasmolysis, shrink and take the form of raisins. The outside solution is hypertonic as compared to inside solution in grapes hence the exosmosis will occur leading to the loss of water from inside of the grapes to the outside.

(ii) Raisins immersed in distilled water would swell because of imbibitions of water and endosmosis.

Question 13.
What are centrosomes? State their fucntions.
Answer:
These are special structures found close to nucleus in an animal cell. They consist of two granules like structures, one large and other small called centriole.

They play an important role during cell division in spindle formation and lower animals in the development of the locomotory organs such as cilia aa flagella.

Question 14.
What are prokaryotic and eukaryotic cells? Give two examples.
Answer:
Procaryotic cells: These are cells without a well defined nucleus and membrane bound cell organelles e.g. bacteria, mycoplasm, blue green algae, etc.

Eucaryotic cells: These are cells with a well defined nucleus and membrane bound cell organelles e.g. cells in higher plants and animals like human cells, etc.

Question 15.
How is diffusion different from osmosis?
Answer:
Diffusion involves the free movement of substances i.e. both solute and solvent from higher concentration to the low concentration.

Osmosis involves the movement of solvents like water from region of hypotonic solution into hypertonic solution across a semi-permeable membrane.

In both diffusion and osmosis , the movement continues until a state of equilibrium is achieved in two adjoining systems.

Long Answer Questions

Question 1.
Draw the diagram of some cells in human body showing the diversity of shape of cells.
Answer:
In body different cells perform different functions, therefore they posses different shape to make their work easier.

Question 2.
Draw a labelled diagram of a compound simple microscope and state the notions of its different parts.
Answer:
A compound microscope is an optical:
device that is used to see minute things such as cells or microorganisms that are invisible to naked eyes. The main parts of a compound microscope includes the followings:

Reflector: A concave mirror that focuses the light on the specimen in slide for the vision.
Condenser: It condenses the beam of light on its way to the specimen.
Specimen stage: It’s a platform with a hole where the specimen on the slide is kept for viewing through eyepiece.
Objective piece: It consists of a biconvex lens of small focal length. It comes closer to the specimen to form its image.

Eyepiece: It also consists of a biconvex lens but of large focal length. The specimen is observed by keeping the eyes slightly above it.

Question 3.
What are the different types of plastids found in a plant cell?
Answer:

Leucoplasts are plastids without any pigment. They help in storage of food.
Amyloplasts are leucoplasts which stores starch.
Aleuroplasts are leucoplasts which stores proteins.
Elaioplasts are leucoplasts which stores fats and oils.
Chromoplasts are plastids with a pigment that impart them a specific colour.
Chloroplasts bear chlorophyll as pigment and are green in colour.
Phaeoplasts bear fucoxanthin and xanthophylls as pigment arid are brown in colour.
Rhodoplasts bear phycoerythrin as pigment and are red in colour.

Question 4.
Draw a labelled diagram of an animal cell showing different cell organelles.
Answer:

Question 5.
How will you prepare the temporary mount of onion peel for its observation in the microscope.
Answer:
To prepare the temporary mount of an onion peel, a student is expected follow the following steps:

• Take a small piece from an onion bulb.
• With the help of set of forceps peel off the epidermis (skin) from the concave side.
• Put it in a petridish or watch glass having some water.

• Transfer the peel to another watch glass containing some stain or iodine.
• Wait for 2-3 minutes to allow the peel to imbibe the stain.
• Wash the peel using brush and then transfer it in the centre of a slide on top of a drop or two of glycerine.
• Carefully plate a coverslip to avoid any air bubble.
• Put the slide on the stage and observe.
• Record your observations with a labelled diagram.

Question 6.
Draw a labelled diagram of a typical plant cell showing different cell organelles.
Answer:

Question 7.
Write the different parts of a cell observed under microscope with their functions.
Answer:
Different parts seen under the microscope include the followings:
(1) Plasma membrane: It is living and outermost covering of the animal cell. In plant cell it is covered by cell wall.
Functions: It regulates the movement of molecules in and out of the cell and protects the inner parts of the cell.

(2) Nucleus: It is surrounded by a double layered nuclear membrane, embedded in the cytoplasm. It is placed centrally in the animal cell. Nucleolus and chromation material are is the other component of nucleus.

Functions:
(i) It contains nucleic acid, RNA and DNA.
(ii) DNA stores the hereditary informations. Segments of DNA are called genes.
(iii) RNA rules over the activities, taking place in cytoplasm and in synthesis of ribosomes.

(3) Cytoplasm: It is a viscous, translucent part of cell inner to plasma membrane. It contains 90% of water, organic and inorganic molecules in it. All cell organelles are embedded in it. Nucleus and cytoplasm together consitute protoplasm.

(4) Cell organelles:
(a) Endoplasmic Reticulum (E.R.): It is an extensive network of membrane bound tubes and sheets that run through the cytoplasm. These are of two types (i) Rough Endopolasmic Reticulum (RER)—It is termed so because it has ribosomes attached on its furface which gives it a rough appearance. (2) Smooth Endoplasmic Reticulum (SER), they are devoid of ribosomes.

Functions: (i) RER—systesized proteins. SER—Secretes organic molecules called lipids.
(ii) It provides mechanical strength to the cell.

(b) Ribosomes—Grannular bodies attached to RER, help in the synthesis of proteins.

(c) Mitrochondria: They are known as the powerhouse of the cell as they store energy in the form of ATR They are double membrane structures. Their inner membrane is deeply folded. These folds create large surface area for chemical reaction. They have their own DNA and ribosomes.

(d) Lysosomes—Digestive enzymes containing sac like structure, help in digestion of un-wanted material and part of worn out cell organelles. During breakdown of cell structure i.e. when the cell gets damaged, lysosomes may burst and the enzymes eat up their own cell hence, known as suicide bags.

(e) Golgi Apparatus: Consists of a set of smooth flattened sack like structures stacked together usually placed one above the other near the nucleus.
It is termed as dictyosomes if present in plant cell.
Function: Responsible for all types of cell secretion and excretion.

(f) Plastide: Tey are found in plant cell only. They are of three types based on the pigments present in them. The most important is chloroplast which contain green pigment essential for photosynthesis. Chromoplast—Contain pigments of different colurs in different fruits and flowers. Leucoplast contain colouless pigments, responsible for storange of food.

(g) Vacuoles: These are found in plant cell as well as in animal cells. In plant cell, they are very large and occupies maximum space whereas in animal cell these are minute and large in numbers.

These NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Atoms and Molecules NCERT Solutions for Class 9 Science Chapter 3

Class 9 Science Chapter 3 Atoms and Molecules InText Questions and Answers

Question 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid ? sodium ethanoate + carbon dioxide + water
Answer:
In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water.

Mass of sodium carbonate = 5.3 g (Given)
Mass of ethanoic acid = 6 g (Given)
Mass of sodium ethanoate = 8.2 g (Given)
Mass of carbon dioxide = 2.2 g (Given)
Mass of water = 0.9 g (Given)
Now, total mass before the reaction = (5.3 + 6) g = 11.3 g
And, total mass after the reaction = (8.2 + 2.2 + 0.9) g = 11.3 g
∴ Total mass before the reaction = Total mass after the reaction
Hence, the given observations are in agreement with the law of conservation of mass.

Question 2.
Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Answer:
It is given that the ratio of hydrogen and oxygen by mass to form water is 1 : 8.
Then, the mass of oxygen gas required to react completely with 1 g of hydrogen gas is 8 g.
Therefore, the mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 × 3g = 24 g.

Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer:
The postulate of Dalton’s atomic theory which is a result of the law of conservation of mass is:
Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:
The postulate of Dalton’s atomic theory which can explain the law of definite proportion is:
The relative number and kind of atoms in a given compound remains constant.

Question 5.
Define atomic mass unit.
Answer:
Mass unit equal to exactly one-twelfth the mass of one atom of carbon-12 is called one atomic mass unit. It is written as ‘u’.

Question 6.
Why is it not possible to see an atom with naked eyes?
Answer:
The size of an atom is so small that it is not possible to see it with naked eyes. Also, the atom of an element does not exist independently.

Question 7.
Write down the formulae of the followings:
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Answer:
(i) Sodium oxide → Na₂O
(ii) Aluminium chloride → AlCl₃
(iii) Sodium sulphide → Na₂S
(iv) Magnesium hydroxide → Mg(OH)₂

Question 8.
Write down the names of compounds represented by the following formulae:
(i) Al₂(SO₄)₃
(ii) CaCl₂
(iii) K₂SO₄
(iv) KNO₃
(v) CaCO₃
Answer:
(i) Al₂(SO₄)₃ → Aluminium sulphate
(ii) CaCl₂ → Calcimn chloride
(iii) K₂SO₄ → Potassium sulphate
(iv) KNO₃ → Potassium nitrate
(v) CaCO₃ → Calcium carbonate

Question 9.
What is meant by the term chemical formula?
Answer:
The chemical formula of a compound means the symbolic representation of the composition of a compound. From the chemical formula of a compound, we can know the number and kinds of atoms of different elements that constitute the compound.

For example, from the chemical formula CO₂ of carbon dioxide, we come to know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.

Question 10.
How many atoms are present in a
(i) H₂S molecule and
(ii) PO₄^3- ion?
Answer:
(i) In an H₂S molecule, three atoms are present; two of hydrogen and one of sulphur.
(ii) In a PO₄^3- ion, five atoms are present; one of phosphorus and four of oxygen.

Question 11.
Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄; C₂H₆, C₂H₄, NH₃, CH₃OH.
Answer:
Molecular mass of H₂ = 2 × Atomic mass of H
= 2 × 1
= 2u
Molecular mass of O₂ = 2 × Atomic mass of O
= 2 × 16
= 32 u
Molecular mass of Cl₂ = 2 × Atomic mass of Cl
= 2 × 35.5
= 71 u
Molecular mass of CO₂ = Atomic mass of C + 2 × Atomic mass of O
= 12 + 2 × 16
= 44 u
Molecular mass of CH₄ = Atomic mass of C + 4 × Atomic mass of H
= 12 + 4 × 1
= 16 u
Molecular mass of C₂H₆ = 2 × Atomic mass of C+ 6 × Atomic mass of H
= 2 × 12 + 6 × 1
= 30 u
Molecular mass of C₂H₄ = 2 × Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1
= 28u
Molecular mass of NH₃ = Atomic mass of N + 3 × Atomic mass of H
= 14 + 3 × 1
= 17u
Molecular mass of CH₃OH = Atomic mass of C + 4 × Atomic mass of H + Atomic mass of O
= 12 + 4 × 1 + 16
= 32 u

Question 12.
Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Answer:
Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65 + 16
= 81 u
Formula unit mass of Na₂O = 2 × Atomic mass of Na + Atomic mass of O
= 2 × 23 + 16
= 62 u
Formula unit mass of K₂CO₃ = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O
= 2 × 39 + 12 + 3 × 16
= 138 u

Question 13.
If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?
Answer:
One mole of carbon atoms weighs 12 g
(Given)
i. e., mass of 1 mole of carbon atoms = 12 g
Then, mass of 6.022 × 10²³ number of carbon atoms = 12 g
Therefore, mass of 1 atom of carbon 12
= \(\frac{12}{6.022 \times 10^{23}} \mathrm{~g}\)
= 1.9926 × 10^-23 g

Question 14.
Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?
Answer:
Atomic mass of Na = 23 u (Given)
Then, gram atomic mass of Na = 23 g
Now, 23 g of Na contains = 6.022 × 10²³
number of atoms
Thus, 100 g of Na contains \(\frac{6.022 \times 10^{23}}{23} \times 100\)
number of atoms
= 2.6182 × 10²⁴ number of atoms
Again, atomic mass of Fe = 56 u(Given)
Then, gram atomic mass of Fe = 56 g
Now, 56 g of Fe contains = 6.022 × 10²³ number of atoms
Thus, 100 g of Fe contains = \(\frac{6.022 \times 10^{23}}{56} \times 100\)
= 1.0753 × 10²⁴ number of atoms
Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

Class 9 Science Chapter 3 Atoms and Molecules Textbook Questions and Answers

Question 1.
A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer:
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)
Mass of sample = 0.24 g (Given)
Thus, percentage or boron by weight in the compound = \(\frac{0.096}{0.25}\) × 100%
= 40%
And, percentage of oxygen by weight in the compound = \(\frac{0.144}{0.25}\) × 100%
= 60%

Question 2.
When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combinations will govern your answer?
Answer:
Carbon + Oxygen → Carbon dioxide
3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide.

If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left unreactive.

In this case also, only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.

Question 3.
What are polyatomic ions? Give examples?
Answer:
A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion (NH₄⁺), hydroxide ion (OH^–), carbonate ion (CO₃^2-), sulphate ion (SO₄^2-).

Question 4.
Write the chemical formulae of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer:
(a) Magnesium chloride → MgCl₂
(b) Calcium oxide → CaO
(c) Copper nitrate → Cu (NO₃)₂
(d) Aluminium chloride → AlCl₃
(e) Calcium carbonate → CaCO₃

Question 5.
Give the names of the elements present in the following compounds:
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer:

Question 6.
Calculate the molar mass of the following substances:
(a) Ethyne, C₂H₂
(b) Sulphur molecule, S₈
(c) Phosphorus molecule, P₄ (atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO₃
Answer:
(a) Molar mass of ethyne, C₂H₂ = 2 × 12 + 2 × 1 = 26g
(b) Molar mass of sulphur molecule, S₈ = 8 × 32 = 256 g
(c) Molar mass of phosphorusjnolecule, P₄ = 4 × 31 = 124 g
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g
(e) Molar mass of nitric acid, HNO₃ = 1 + 14 + 3 × 16 = 63 g

Question 7.
What is the mass of
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na₂SO₃)?
Answer:
(a) The mass of 1 mole of nitrogen atoms is 14 g
(b) The mass of 4 moles of aluminium atoms is (4 × 27) g = 108 g
(c) The mass of 10 moles of sodium sulphite (Na₂SO₃) is
10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

Question 8.
Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
Answer:
(a) 32 g of oxygen gas = 1 mole
Then, 12 g of oxygen gas = \(\frac {12}{32}\) = 0.375 mole
(b) 18 g of water = 1 mole
Then, 20 g of water= \(\frac {20}{18}\) = 1.11 moles (approx)
(c) 44 g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide = \(\frac {22}{44}\) = 0.5 mole

Question 9.
What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer:
(a) Mass of one mole of oxygen atoms = 16 g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2g
(b) Mass of one mole of water molecule=18 g
Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g

Question 10.
Calculate the number of molecules of sulphur (S₈) present in 16 g of solid sulphur.
Answer:
1 mole of solid sulphur (S₈) = 8 × 32 g = 256 g
i. e., 256 g of solid sulphur contains = 6.022 × 10²³ molecules
Then, 16 g of solid sulphur contains \(\frac{6.022 \times 10^{23}}{256} \times 16\) molecules
= 3.76 × 10²² molecules (approx)

Question 10.
Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)
Answer:
1 mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102 g
i. e., 102 g of Al₂O₃ = 6.022 × 10²³ molecules of Al₂O₃
Then, 0.051 g of Al₂O₃ contains
= \(\frac{6.022 \times 10^{23}}{102} \times 0.051\)
= 3.011 × 10²⁰ molecules of Al₂O₃
The number of aluminium ions (Al³⁺) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al³⁺) present in 3.011 × 10²⁰ molecules (0.051 g) of aluminium pxide (Al₂O₃) = 2 × 3.011 × 10²⁰
= 6.022 × 10²⁰

Class 9 Science Chapter 3 Atoms and Molecules Additional Important Questions and Answers

Multiple Choice Questions
Choose the correct option:

Question 1.
Which of the following correctly represents 360 g of water?
(i) 2 moles of H₂O
(ii) 20 moles of water
(iii) 6.022 × 1023 molecules of water
(iv) 1.2044 × 1025 molecules of water
(a) (i)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(d) (ii) and (iv)

Question 2.
Which of the following statements is not true about an atom?
(a) Atoms are not able to exist independently
(b) Atoms are the basic units from which molecules and ions are formed
(c) Atoms are always neutral in nature
(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch
Answer:
(a) Atoms are not able to exist independently

Question 3.
The chemical symbol for nitrogen gas is
(a) Ni
(b) N₂
(c) N⁺
(d) N
Answer:
(b) N₂

Question 4.
The chemical symbol for sodium is
(a) So
(b) Sd
(c) NA
(d) Na
Answer:
(d) Na

Question 5.
Which of the following would weigh the highest?
(a) 0.2 mole of sucrose (C₁₂H₂₂O₁₁)
(b) 2 moles, of CO₂
(c) 2 moles of CaCO₃
(d) 10 moles of H₂O
Answer:
(c) 2 moles of CaCO₃

Question 6.
Which of the following has maximum number of atoms?
(a) 18g of H₂O
(b) 18g of O₂
(c) 18g of CO₂
(d) 18g of CH₄
Answer:
(d) 18g of CH₄

Question 7.
Which of the following contains maximum number of molecules?
(a) 1g CO₂
(b) 1g N₂
(c) 1g H₂
(d) 1g CH₄
Answer:
(c) 1g H₂

Question 8.
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
(a) 6.68 × 10²³
(b) 6.09 × 10²²
(c) 6.022 × 10²³
(d) 6.022 × 10²¹
Answer:
(a) 6.68 × 10²³

Question 9.
A change in the physical state can be brought about
(a) only when energy is given to the system
(b) only when energy is taken out from the system
(c) when energy is either given to, or taken out from the system
(d) without any energy change
Answer:
(c) when energy is either given to, or taken out from the system

Very Short Answer Type Questions

Question 1.
Define law of conservation of mass.
Answer:
In a chemical reaction mass can neither be created nor destroyed.
E.g., 2Na + Cl₂ → 2NaCl
2 × 23 + 2 × 35.5 → 2(23 + 35.5)

Question 2.
Explain law of constant proportion.
Answer:
In a chemical substance the elements are always present in definite proportions by mass.
E.g., In water, the ratio of the mass of hydrogen to the mass of oxygen H : O is always 1 : 8

Question 3.
Who coined the term atom?
Answer:
John Dalton coined the term atom.

Question 4.
Define atom.
Answer:
The smallest particle of matter, which can take part in a chemical reaction is called atom.

Question 5.
Define molecule.
Answer:
The smallest particle of an element or compound which can exist independently is called molecule.

Question 6.
Define atomicity.
Answer:
The number of atoms constituting a molecule is known as its atomicity.

Question 7.
What is atomic mass unit?
Answer:
The sum of the atomic masses of all die atoms in a molecule of the substance is expressed in atomic mass unit.
E.g., H₂O = (1 × 2) + 16 = 18 amu

Question 8.
How do atoms exist?
Answer:
Atoms exist in the form of atom, molecule or ions.

Question 9.
Give the atomicity of phosphorous and nitrogen.
Answer:
The atomicity of phosphorus is P₄ i.e., 4.
The atomicity of nitrogen is N₂ i.e., 2.

Question 10.
What is an ion?
Answer:
Charged atom is called as an ion. The ion can be positively charged called cation or negatively charged called anion.

Question 11.
Give one example of cation and anion.
Answer:
Cation ⇒ Na⁺
Anion ⇒ Cl^–

Question 12.
Give one difference between cation and anion.
Answer:
Cations are positively charged ion. Anions are negatively charged ion.

Question 13.
Give the chemical formula for ammonium sulphate.
Answer:
Ammonium sulphate NH₄⁺ SO₄^2-
Chemical formula → (NH₄)₂SO₄.

Question 14.
What is Avogadro’s constant?
Answer:
The Avogadro’s constant (6.022 × 10²³) is defined as the number of atoms that are present inexactly 12 g of carbon-12.

Question 15.
Find the molecular mass of H₂O.
Answer:
Molecular mass of H₂O=(2 × 1) + (16) = 2 + 16 = 18 u

Short Answer Type Questions

Question 1.
Give the unit to measure size of atom and give size of hydrogen atom.
Answer:
The unit to measure size of atom is nanometer, size of hydrogen atom is 10^-10 m.

Question 2.
What is IUPAC, give its one function?
Answer:
IUPAC stands for International Union of Pure and Applied Chemistry. It approves the names of elements.

Question 3.
Give the Latin name for sodium, potassium, gold and mercury.
Answer:
Sodium → Natrium, Gold → Aurum Potassium → Kalium, Mercury → Hydrargyrum

Question 4.
What is the ratio by mass of combining elements in H₂O,CO₂ and NH₃?
Answer:
H₂O ratio by mass of combining elements 2 : 16 → 1 : 8 (H : O)
CO₂ ratio by mass of combining elements 12 : 32 → 3 : 8 (C : O)
NH₃ ratio by mass of combining elements 14 : 3 → 14 : 3 (N : H)

Question 5.
Define valency and give the valency for the following elements:
Magnesium, Aluminium, Chlorine and Copper.
Answer:
Valency: The combining capacity of an element is called its valency.
Valency of the following elements:
Magnesium — 2
Aluminium — 3
Chlorine — 1
Copper — 2

Question 6.
What is polyatomic ion? Give one example.
Answer:
A group of atoms carrying a charge is known as a polyatomic ion.
E.g., Ammonium — NH₄^–
Nitrate — NO₃^–

Question 7.
Write down the formula for:
Copper nitrate, calcium sulphate and aluminium hydroxide.
Answer:
Chemical formula:
Copper nitrate → Cu(NO₃)
Calcium sulphate → CaSO₄
Aluminium hydroxide → Al(OH)₃

Question 8.
What is formula unit mass? How is it different from molecular mass?
Answer:
The formula unit mass of a substance is a sum of the atomic masses of all atoms in a formula unit of a compound. The constituent particles of formula unit mass are ions and the constituent particles of molecular mass are atoms.

Question 9.
Find the number of moles in the following:
(i) 50g of H₂O
(ii) 7g of Na
Answer:
Number of moles in
(i) Molar mass of H₂O = 18g
Given mass of H₂O = 50 g
∴ No. of moles in 50g of H₂O
= \(\frac {50}{18}\) = 2.78 moles.

(ii) Molar mass of Na = 23g
Given mass of Na = 7g
∴ No. of moles in 7g of Na = \(\frac {7}{23}\) = 0.304 moles.

Question 10.
Find the number of atoms in the following:
Answer:
(i) 0.5 mole of C atom
(ii) 2 mole of N atom
Answer:
(i) 0.5 mole of C atom:
Number of atoms in 1 mole of C atom
= 6.022 × 10²³ atoms
Number of atoms in 0.5 mole of C atom
= 6.022 × 10²³ × 0.5
= 3.011 × 10²³ atoms

(ii) 2 mole of N atom:
Number of atoms in 1 mole of N atom = 6.022 × 10²³ atoms
∴ Number of atoms in 2 mole of N atom
= 6.022 × 2 × 10²³
= 1.2044 × 10²⁴ atoms

Question 11.
Find the mass of the following:
(i) 6.022 × 10²³ number of O₂ molecules
(ii) 1.5 mole of CO₂ molecule
Answer:
(i) 6.022 × 10²³ number of O₂ molecules:
Mass of 1 mole of O₂ molecule = 6.022 × 10²³ molecules = 32 g

(ii) 1.5 mole of CO₂ molecule:
Mass of 1 mole of CO₂ molecule = 6.022 × 10²³ molecules = 44 g
Mass of 1.5 mole CO₂ molecule = 44 × 1.5 = 66 g

Question 12.
What are the rules for writing the symbol of an element?
Answer:
IUPAC → International Union of Pure and Applied Chemistry approves name of elements.

Symbols are the first one or two letters of the element’s name in English. The first letter of a symbol is always written as a capital letter (upper case) and the second letter as a small letter (lower case).
e.g., Hydrogen → H
Helium → He

Some symbols are taken from the names of elements in Latin, German or Greek.
e.g., Symbol of iron is Fe, its Latin name is Ferrum.
Symbol of sodium is Na, its Latin name is Natrium.

Question 13.
Explain relative atomic mass and relative molecular mass.
Answer:
Relative atomic mass: It can be defined as the number of times one atom of given element is heavier than \(\frac {1}{2}\) th of the mass of an atom of carbon-12.

Relative Molecular Mass: It is defined as the number of times one molecule of a substance or given element is heavier than vyth of the mass of one atom of carbon-12.

Question 14.
The formula of carbon-dioxide is CO₂. What information do you get from this formula?
Answer:

• CO₂ represents carbon-dioxide.
• CO₂ is one molecule of carbon-dioxide.
• CO₂ is one mole of carbon-dioxide i.e., it contains 6.022 × 10²³ molecules of carbon dioxide.
• CO₂ contains 1 atom of carbon and two atoms of oxygen.
• CO₂ represents 44 g of molar mass.

Question 15.
State 3 points of difference between an atom and an ion.
Answer:
Atom:

1. An atom has no change.
2. Number of electrons = number
3. Atom is reactive.

Ion:

1. Anion has either positive or negative charge.
2. Number of electrons ≠ number of protons.
3. Ion is stable.

Question 16.
Calculate the formula unit mass of NaCl and CaCl₂.
(Na = 23, Cl = 35.5, Ca = 40)
Answer:
Formula unit mass of NaCl = 23 + 35.5
= 58.5 u
Formula unit mass of CaCl₂ = 40 + (2 × 35.5)
= 40 + 71
= 111 u

Question 17.
Write down the chemical formula for the following compounds:
(a) Aluminium carbonate
(b) Calcium sulphide
(c) Zinc carbonate
(d) Copper phosphate
(e) Magnesium bicarbonate
(f) Aluminium hydroxide.
Answer:
The chemical formula are:

Question 18.
Give the atomicity of the following compounds:
(a) Ca(OH)₂
(b) Mg(HCO₃)₂
(c) Cu₂O
(d) H₂SO₄
(e) Al₂(SO₄)
(f) MgCl₂
Answer:
The atomicity of the molecules are:
(a) Ca(OH)₂ → 05
(b) Mg(HCO₃)₂ → 11
(c) Cu₂O → 03
(d) H₂SO₄ → 07
(e) Al₂(SO₄)₃ → 17
(f) MgCl₂ → 03

Question 19.
Explain the difference between 2O, O₂ and O₃.
Answer:
2O → It represents 2 atoms of oxygen (cannot exist independently).
O₂ → It represents one molecule of oxygen (made up of 2 atom) can exist freely.
O₃ → It represents one molecule of ozone (made up of 3 atoms) it can exist independently.

Question 20.
1.50 g sample of barium hydroxide was dissolved in water. The total volume of the solution was 100 cm³.
A 25.0 cm³ portion of the barium hydroxide solution was titrated against hydrochloric acid. The volume of hydrochloric acid required was 18.75 cm³.
Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O
(i) Calculate how many moles of barium hydroxide were in the 25.0 cm³ portion used in the titration.
(ii) Calculate the concentration of the hydrochloric acid used.
Answer:
(i) Moles of barium hydroxide = 0.00219 mol.
(ii) Moles of HCl = 2 × 0.00219 = 0.00438.

Question 21.
Analysis of a certain compound showed that 39.348 grams of it contained 0.883 grams of hydrogen, 10.497 grams of Carbon, and 27.968 grams of Oxygen. Calculate the empirical formula of the compound.
Answer:
First divide the amount by the atomic mass to get the number of moles of each kind of atom in the formula

Analysis of the ratio shows that the first two are identicaland that the third is twice the other two. Therefore the ratio of H to C to O is 1 to 1 to 2. The empirical formula is HCO₂.

Long Answer Type Questions

Question 1.
(a) How do atoms exist?
(b) What is atomicity?
(c) What are polytom1c Ions?
Answer:
(a) Atoms of some elements are not able to exist independently. For such elements atoms
form molecules and ions. In case of metals and inert gases atoms can exist independently.

Atoms of metals and inert gases: E.g.,
\(\frac{\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}}{\text { Metals }} \frac{\mathrm{He}, \mathrm{Ne}, \mathrm{Ar}}{\text { Inert gases }}\)
Non-metals: E.g., H₂, Cl₂, P₄, S₈
Exceptional non-metal C

(b) The number of atoms constituting a molecule is known as its atomicity.
E.g., O₃ → atomicity is 3 .
O₂ → atomicity is 2

(c) Polyatomic ions: When more than two atoms combine together and act like an atom with a charge on it is called polyatomic ion.
E.g., OH^–, NO₃^–,NH₄⁺

Question 2.
Calculate
(a) the mass of one atom 6f oxygen
(b) the mass of one molecule of oxygen
(c) the mass of one mole of oxygen gas
(d) the mass of one ion of oxygen
(e) the number of atoms in 1 mole of oxygen molecule
Answer:
(a) Mass of one atom of oxygen
1 mole of oxygen atom = 16gm = 6.022 × 10²³ atoms.
∴ Mass of one atom of oxygen
= \(\frac{16}{6.022 \times 10^{23}}\) = 2.65 × 10^-23
(b) Mass of one molecule of oxygen 1 molecule of oxygen = O₂
= 2 × 16
= 32 u

(c) Mass of one mole of oxygen gas
1 mole of oxygen gas is O₂ = 32 u or 32 g
(d) Mass of one ion of oxygen
One mole of oxygen = 6.022 × 10²³ atoms = 16 g
Mass of one ion of oxygen = \(\frac{16}{6.022 \times 10^{23}}\) = 2.65 × 10^-23

(e) Number of atoms in one mole of oxygen molecule
1 mole of oxygen molecule i.e.,
O₂ = 6.022 × 10²³ molecules.
1 molecule of O₂ = 2 atoms.
∴ Number of atoms in 1 mole of oxygen molecule = 6.022 × 10²³ × 2 atoms
= 1.2044 × 10²⁴ atoms

Question 3.
What is meant by atomic mass, gram atomic mass of an element? Why is the mass have different expressions i.e., ‘u’ and ‘g’?
Answer:
The atoms are very tiny and their individual mass cannot be calculated as it is negligible. Hence the mass of atoms is expressed in units with respect to a fixed standard. Initially hydrogen atom with mass 1 was taken as standard unit by Dalton. Later, it was replaced by oxygen atom (O = 16). But due to the isotopes the masses were found in fractions instead of whole number. Hence, carbon (C = 12) isotope was taken as standard unit and was universally accepted.

The atomic mass unit is equal to one twelfth (\(\frac {1}{12}\)) the mass of an atom of carbon-12, its unit is u.

Gram atomic mass: When the atomic mass of an element is expressed in grams, it is called the gram atomic mass of the element. The mass of atoms, molecules is expressed in ‘u’ and the mass of moles i.e., molar mass is expressed in g.

Question 4.
Define a mole. Give the significance of the mole.
Answer:
Mole-One mole of any species (atoms, molecules, ions or particles) is that quantity or number having a mass equal to its atomic or molecular mass in grams. 1 mole = 6.022 × 10²³ in number (atoms, molecules, ions or particles)

Significance of the mole:
1. A mole gives the number of entities present i.e, 6.022 × 10²³ particles of the substance.
2. Mass of 1 mole is expressed as M grams.
3. Mass of 1 mole=mass of 6.022 × 10²³ atoms of the element.
E.g., 1 mole of O₂ = 6.022 × 10²³ atoms
1 mole of O₂ = 2 × 16 = 32g
6.022 × 2 × 10²³= 1.2044 × 104 atoms
1 mole of (compound) HCl = 6.022 × 10²³ atoms of H and Cl atoms
(1 + 35.5 = 36.5 g) (6.022 × 10²³ molecules of HCl)

Question 5.
Barium carbonate decomposes when heated
BaCO₃ (s) → BaO(s) + CO₂(g)
(a) A student heated a 10.0 g sample of barium carbonate until it was fully decomposed.
(i) Calculate the number of moles of barium carbonate the student used.
(ii) Calculate the volume of carbon dioxide gas produced at room temperature and pressure. Give your answer in dm³.
(b) The student added 2.00 g of the barium oxide produced to water.
BaO + H2O → Ba(OH)₂
Calculate the mass of barium hydroxide that can be made from 2.00 g of barium oxide. The molecular mass of Ba(OH)₂ is 171.
Answer:
(a) (i) Relative formula mass BaCO₃ = 197,
moles of barium carbonate = 10.0/197 = 0.0508 mol.
(ii) Volume of carbon dioxide = 1.22 dm³
(b) Mass of barium hydroxide = 2.24 g

Question 6.
Magnesium sulphate crystals are hydrated. A student heated some hydrated magnesium sulphate crystals in a crucible and obtained the following results. Mass of hydrated magnesium sulfate crystals = 4.92 g Mass of water removed = 2.52 g
(i) Calculate the number of moles of water removed,
(ii) Calculate the number of moles of anhydrous magnesium sulfate remaining in the crucible. The molecular mass of anhydrous nlagnesium sulfate is 120.
(iii) Calculate the ratio of moles of anhydrous magnesium sulfate: moles of water.
Answer:
(i) Moles of water= 2.52/18 = 0.14. moles.
(ii) Moles of anhydrous magnesium sulfate = 0.02 moles,
(iii) Ratio = 0.02/0.02 : 0.14/0.02 = ratio in whole numbers is 1 : 7.

Question 7.
Calculate the mass of each element in potassium carbonate, K₂CO₃.
Answer:
First calculate the formula mass for K₂CO₃. Find the atomic mass of each element from the periodic table. Multiply it by the number, of times it appears in the formula and add up the total

To find the percent of each element divide the part of the formula mass that pertains to that element with the total formula mass.
Percent of Potassium
K= \(\frac{78.20}{138.21} \times 100\) = 56.58%
Percent of Carbon
C = \(\frac{12.01}{138.21} \times 100\) = 8.69%
Percent of Oxygen
O = \(\frac{48.00}{138.21} \times 100\) = 34.73%

Question 8.
For each of the following calculate the empirical formula:
(a) When 2.20 g of a hydrocarbon, D, is burnt in excess oxygen, 6.90 g of CO₂ and 2.83 g of water are produced.
(b) When 1.52 g of compound E, which contains carbon, hydrogen and oxygen only, is burnt in excess oxygen, 3.04 g CO₂ and 1.24 g H₂O are produced.
Answer:
(a) Moles of CO₂ = 6.90 ÷ 44.01 = 0.157 mol
Moles of H₂O = 2.83 ÷ 18.02 = 0.157 mol
Moles of C = 0.157 mol moles of H = 2 ÷ 0.
157 = 0.304 mol
Empirical formula = CH₂

(b) Mass of C in CO₂ = 12.01 ÷ 44.01
3.04 = 0.830 g
Mass of H in H₂O = 2.02 ÷ 18.02
1.24 = 0.139g
Mass of O = 1.52 – (0.830 + 0.139) = 0.551 g
Ratio of moles C : H : O = 0.0691 : 0.139 : 0.0344
Whole number ratio = 2.01 : 4.04 : 1
Empirical formula: C₂H₄O