NCERT Exemplar Solution Class 9 Science Chapter 11
NCERT Exemplar Solution Class 9 Science Chapter 11 Work and Energy all Questions Answer Solution. NCERT Exemplar Class 9 Science Chapter 11 Solution PDF.
NCERT Exemplar Solution Class 9 Science Chapter 11: Overview
NCERT Exemplar Solution Class 9 Science Chapter 11 |
|
Board |
NCERT |
Topic |
Exemplar Problem Solution |
Class |
9 |
Subject |
Science |
Chapter |
11 |
Chapter Name |
Work and Energy |
NCERT Exemplar Solution Class 9 Science Chapter 11 Work and Energy
Multiple Choice Questions
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
Explanation: When a body falls freely towards the earth, then it’s total energy i.e. it’s potential energy and kinetic energy remains constant. As it obeys the law of conservation of energy.
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
Explanation: Potential energy of the car will not change because Height is not changed. As P.E =mgh potential energy is dependent upon the height.
3) In case of negative work the angle between the force and displacement is
a) 0°
b) 45°
c) 90°
d) 180°
Answer:- d) 180°
Explanation: Negative work done create an angle of 180° between force and displacement.
4) An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass 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
Explanation: Acceleration is dependent over the gravitational force. Thus both the spheres will be have same acceleration as gravitational force act same on both. But momentum, potential energy and kinetic energy all are dependent over the mass so it will varies with mass.
.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 m s-2)
a) 6× 103 J
b) 6 J
c) 6 J
d) Zero
Answer:- d) zero
Explanation: Work done against the gravitational force is zero because the work done in the opposite direction to the gravitation of earth.
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
Explanation: Kilowatt is not the unit of energy it is the SI unit of power.
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
Explanation: Work done is dependent over the displacement and force applied also the angle between force and displacement. But total workdone is not dependent on the initial velocity of the object.
8) Water stored in a dam possesses
a) No energy
b) Electrical energy
c) Kinetic energy
d) Potential energy
Answer:- d) potential energy
Explanation: The flowing water posses kinetic energy, but the stored water contains potential energy, As it is stored at one place, and not moving.
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
Explanation: When a body falls from height h then its potential energy decreases as it starts moving and comes to half, when it reaches to half height i.e. h/2 also the kinetic energy starts increasing from zero to half of it in the middle distance. Thus at h/2 the body have half potential and half kinetic energy.
Short Answer Questions
(10) A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?
Answer:- We have given,
Rocket is moving with V= initial velocity
Then the velocity is tripled = V2 = 3V1
Ratio of two kinetic energies?
Thus, we have
Initial kinetic energy = ½ mV1 2
And,
Final kinetic energy = ½ m v 2 2
= ½ m (3V1) 2
= 9 ( ½ m V12)
The ratio of two kinetic energies is
Initial kinetic energy : final kinetic energy
1:9 .
(11) Avinash can run with a speed of 8 m-1 against the fractional force of 10 N. And kapil can move with a speed of 3 ms-1 against the fractional force of 25 N. Who is more powerful and why?
Answer:-
Speed of avinashv = 8 ms -1
Fractional force of avinash = 10 N
Speed of kapilv = 3 ms -1
Fractional force of kapil = 25 N
More powerful person =?
Power of Avinash = P
P = F × V
P = 10 × 8
P = 80 W
Power of kapil = p
P = F × v
P = 25 × 3
P = 75 W
Kapil has less power than the Avinash. Thus Avinash is more powerful.
12) A boy is moving on a straight road against a fractional force of 5 N. After travelling a distance of 1.5 km he forget the correct path at a round about ( fig. 11.1) of radius 100 m. However, he moves on the circular path for one and half cycle and then he moves forward upto 2.0 km. Calculate the work done by him.
Answer:-
Given,
Fractional force = 5 N
Displacement =1500 + 200+2000 =3700J
Work done = ?
Work = force × displacement
Work = 5 × 3700
Work = 18500 J.
Therefore the work done by the boy is 18500 J.
13) Can any object have mechanical energy even if its momentum is zero? Explain.
Answer:-
Yes, the object can have mechanical energy even if its momentum is zero. Because when the velocity is zero then momentum becomes zero which means there is no Kinetic energy. But mechanical energy contains both kinetic as well as potential energy. If there is no Kinetic energy but there is potential energy then mechanical energy is present.
14) Can any object have momentum even if its mechanical energy is zero? Explain.
Answer:-
Mechanical energy contains both kinetic as well as potential energy , if the mechanical energy is zero then there is no Kinetic energy also no potential energy. If there is no energy then there is no velocity and hence no Momentum.
15) The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m? ( Given g= 10 m s-2)
Answer:-
16) Weight of a person on a planet A is about half that on the earth. He can jump upto 0.4 m height on the surface of the earth. How high he can jump on the planet A?
Answer:-
If the weight of the person on the planet A is half on the earth, that means there is less acceleration due to gravitationon the planet A . Thus if he can jump upto 0.4 m high on the earth then he can jump upto0.8 m i.e. double the distance with the same force on planet A. Because the gravitational force over the earth attract the body towards itself, that force is less on the planet A so he will jump high.
17) The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the Increase in the kinetic energy of the body is equal to the work done by the force on the body.
Answer:-
We have to prove here that the kinetic energy of the body is equal to the work done by force on the body.
So now,
V2 – u2 = 2as
Thus,
S = V2 – u2 /2a
And F= ma
Thus,
W = ma (v2– u2 ) /2a
W = ½ mv2 – ½ mu2
But , ½ mv2 = (K.E)f
And ½ mu2 = (K.E)t
Thus,
W= (K.E)f– (K.E)t
Hence it is proved.
18) Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain it with an example.
Answer:-
Yes, it is possible that an object in the state of accelerated motion due to external force, but no work is being done by the force. This is possible in the case where the body is in circular motion. Since, the force acts perpendicular to the direction of displacement.
19) A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much is being done by the force. Explain it with an example.
Answer:-
Given,
h = 10 m
g= 10 m s-2
Energy reduced = 40%
Now,
Energy = mgh = m× 10 × 10
Energy = 100 m joules
Then it reduces to 40 %
Thus remain energy will be 60 % = 60 m joules
Hence finding the height for energy 60 m joules.
60m = m × 10 × h
h = 6 m
Hence the height for the 40 % Reduced will be 6 m .
20) If an electric iron of 1200 W is used for 30 minutes everyday, find electric energy consumed in the month of April.
Answer:-
Given,
Power of Electric iron = 1200 W = 1.2 kW
Time = 30 minutes i.e 0.5 h
Days = 30
Electric energy = ?
Thus,
Electric energy = power × time × days
Electric energy = 1.2 × 0.5 × 30
Electric energy = 18 kW/h
Thus the total electric energy is used in the month of April is 18 kW/h .
Long Answer Questions
21) A light and a heavy object have the same momentum. Find out the ratio of their Kinetic energies. Which one has a larger kinetic energy?
Answer:-
We have to find out the ratio of kinetic energies of light and heavy objects having same momentums.
So now,
Equation for momentum is
P = mv
Thus,
P1 = m1v2
And P2 = m2v2
P1 = P2
Equation for kinetic energy is
(K.E)= ½ mv2
Thus ,
(K.E) 1 = ½ m1v12
(K.E)1 = ½ (m1v1)v1
(K.E)1 = ½ p1v1 ——–(m2v2=P2)
And (K.E)2 = ½ m2v2 2
(k.E)2 = ½ (m2v2) V2
(K.E)2 = ½ p2v2 ——(m2v2=P2)
Now comparing both the equation
Here, P1 = P2
Hence we take p2v1
By cancellation
We get
But We know that v1 is heavier than V2
i.e. V1>V2
Thus, we have proved that (k.E)1 >(K.E)2
22) An automobile engine propels a 1000 kg car (A) along a levelledroad at a speed of 36 km h-1. Find the power if the opposing frictional force is 100 N. Now, suppose after travelling a distance of 200 m, this car collides with another Stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now car (B) starts moving on the same level road without getting it’s engine started. Find the speed of the car (B) just after the Collision.
Answer:-
We have given,
Mass of car A= mass of car B = 1000 kg.
Velocity = 36 km/h = 10 m/s
Opposing Frictional force = 100 N.
Car A is moving with the constant speed, while the car B is stationery.
Thus, car A has
Power = force × distance
Power = 100 × 10
Power = 1000 W
Thus engine of car A has power of 1000 W.
Hence, after collision of moving Car A and stationery Car B the speed of the car B will be
= mA × vA + mB× vB
= 1000 × 10 + 1000 vB
- 1000vB = 1000 × 10
vB = 1000 × 10 /1000
vB = 10 m s-1
Thus, the velocity of the car B will be 10 m s-1 after collision.
23) A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 m s-1 by applying force. The trolley comes to rest after travellinga distance of 16 m.
a) how much work is done on the trolley?
b) how much work is done by the girl?
Answer:-
We have given,
Girl’s mass = 35 kg
Mass of trolley = 5 kg
Initial Velocity of trolley = 4 m s-1
Distance covered by the trolley after applying force = 16 m
Now we have to find
a) Work done on the trolley ?
Thus, find a
V2-u2 = 2as
a = v2-u2 /2s
a = 0 – 42 / 2 × 16
a = 0 -16 /2× 16
a = -0.5 m/ s2
Force acted on the trolley
F = ma
F = 40 × (-0.5)
F = -20 N
Now,
Work done on the trolley
= F × s
= 20 N × 16 m
Work done= 320 joules.
b) Work done by the girl will be zero because girl is being moved by the trolley and not by herself. Thus the work done by her is zero.
24) Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it. A) how much work is done by the men in lifting the box? B) how much work do they do in just holding it? C) why do they get tired while holding it? ( g= 10 m s-2)
Answer:-
Given,
Mass = 250 kg
Acceleration due to gravity = 10 m s-2
Height(s) = 1 m
A.)
Work done by men = ?
F = m× a
F = 250 g × 10 m
F = 2500 N
Thus force need by the men is 2500 N
Then,
W= F× s
W = 2500 × 1
W = 2500 j
Hence workdone by men is 2500 joules.
B)
No. Work is done while holding the box because no displacement is occurs over there. Hence work done is zero.
C)
They get tired while holding the boxes because they have to apply the muscular force to hold it, these muscular force is equal and opposite to the gravitational force.
25) What is power? How do you differentiate kilowatt from kilowatt hour? The Jog falls in Karnataka state are nearly 20 m high. 2000 tonnes of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilised?
(g=10 m s-2)
Answer:-
Rate of Doing work is known as power. Power has unit kilowatt . And kilowatt hour or kilowatt per hour is the unit of energy.
We have given
G= 10 m s-2
Time = 60 s
Height of water fall = 20 m
Mass = 2000 tonnes
Mass = 2000 × 10 3Kg
Thus, mass = 2× 10 6 kg
We have to find equivalent power
Power = mgh/t
Power = 2×106×10×20 / 60 s W
Power = 4/6 × 10 7W
Power = 2/3 × 10 7 W.
Thus, if all the energy is utilised then the equivalent power is 2/3 × 107 W.
26) How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of 1m s -1 vertically? (g= 10 m s-2)
Answer:-
We have given here,
G = 10 m s-2
Power = 100 W
Speed = 1 m s-1
We have to find out how many kg can a man lift at the 100 w power
So, here
Power = workdone/ time
Power = mgh / t
And speed = h/t
Thus,
Mass = power / g× speed
Mass = 100 / 10 × 1
Mass = 10 kg.
Hence, the man can lift upto 10 kg .
27) Define watt. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What force does it exert in moving the car at a speed of 20 m s -1 ?
Answer:-
The 1 watt is the power of the body which work on the rate of 1 J s-1 .
Here we have given,
Mass Of car engine = 150 kg
Each of 500 W.
So,
Total power of engine = 150 × 500
=7.5 × 10 4 W
Now, to find force
Given, Speed (velocity) of car = 20 m s-1
Force = power / velocity
Force = 7.5 × 10 4 / 20
Force = 3750 N.
Thus the car exerted the force 3750 N moving with the speed of 20 m s-1.
28) Compare the power at which each of the following is moving upwards against the force of gravity? ( g= 10 m s-2)
i) A butterfly of mass 1.0 that flies upward at a rest of 0.5 m s-1.
Answer:-
We have given
Mass of butterfly = 1.0
Velocity = o.5 m s-1
G= 10 m s-2
Now we have to find out the power of Butterfly
Power = mgv
Power = 1/1000× 10 × 0.5
Power = 5 × 10-3W.
Power of butterfly is 5 × 10 -3 W.
ii) a 250 g squirrel climbing up on a tree at a rest of 0.5 m s-1.
Answer:-
We have g = 10 m s-2
And mass of squirrel = 250 g = 250/1000 kg
Velocity of squirrel = 0.5 m s-1
Now,
We have to find out the power of squirrel
Power = mgv
Power = 250/1000× 10 × 0.5
Power = ¼ × 10 × 0.5
Power = 1.25 W.
Power of the squirrel is 1.25 W.
Thus the power of squirrel is higher than the butterfly.