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NCERT Exemplar Class 11 Physics solutions Chapter 6 Work Energy and Power

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NCERT Exemplar Class 11 Physics solutions Chapter 6 Work Energy and Power

Edited By Safeer PP | Updated on Aug 08, 2022 03:40 PM IST

Using NCERT Exemplar Class 11 Physics solutions chapter 6, the students will very quickly learn about the topic. This solution helps them gain knowledge and solve their related doubts with easy explanation. The chapter will tell us about the concepts of work, energy, and power. It shows the inter-relation between all three of them. Not only theoretically, but the topic gives exposure to practical knowledge. NCERT Exemplar Class 11 Physics chapter 6 solutions have been prepared by the experts according to the understanding capacity of the students.

This Story also Contains
  1. NCERT Exemplar Class 11 Physics Solutions Chapter 6 MCQI
  2. NCERT Exemplar Class 11 Physics Solutions Chapter 6 MCQII
  3. NCERT Exemplar Class 11 Physics Solutions Chapter 6 Very Short Answer
  4. NCERT Exemplar Class 11 Physics Solutions Chapter 6 Short Answer
  5. NCERT Exemplar Class 11 Physics Solutions Chapter 6 Long Answer
  6. Main Subtopics in NCERT Exemplar Class 11 Physics Solutions Chapter 6 Work, Energy, and Power
  7. What Will The Students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 6?
  8. NCERT Exemplar Class 11 Solutions
  9. NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
  10. Important Topics To Cover For Exams From NCERT Exemplar Class 11 Physics Solutions Chapter 6 Work, Energy, and Power

NCERT Exemplar Class 11 Physics Solutions Chapter 6 MCQI

Question:3.1

An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another.
This is because,
(a) the two magnetic forces are equal and opposite, so they produce no net effect.
(b) the magnetic forces do no work on each particle.
(c) the magnetic forces do equal and opposite (but non-zero) work on each particle.
(d) the magenetic forces are necessarily negligible.

Answer:

The answer is the option (b) the magnetic forces do not work on each particle.
The concept used in this question is of the work-energy theorem. According to which, net work done = (change in kinetic energy). Final kinetic energy – Initial kinetic energy of the object.
\sum W = K_{f} - K_{i}
The direction of the magnetic forces will be at an angle of 90 degrees to the motion of electron and proton under mutual forces. As a result, the magnetic force acts like a centripetal force which leads the particles to undergo uniform circular motion. Hence, their speed remains constant throughout the motion. Since the speed is constant, the kinetic energy remains the same, and hence the net work done due to the change in kinetic energy by the forces amounts to zero. By the formula we know that: \overrightarrow{Fm }= q. (\overrightarrow{v} \times \overrightarrow{ B}). , B is the external magnetic field, and v is the velocity of the particle. Hence, we can ignore the magnetic force of one particle on another.

Question:6.2

A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particles is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is
a) same as the same force law is involved in the two experiments
b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens
c) more for the case of a positron, as the positron moves away a larger distance
d) same as the work done by charged particle on the stationary proton

Answer:

The answer is the option (c) more for the case of a positron, as the positron moves away a larger distance
The charges on a proton and a positron are same, but the mass of a positron is lesser than that of a proton. Hence, the force between a proton and a positron is the same as that between two protons. But, as the positron has a lower mass, it moves away through a larger amount of distance. The change in momentum remain equal for both cases, but the positron being of lighter weight moves a larger distance than a proton. Now, work done = force x distance. In our case, the forces are the same, but the positron travels a greater distance, and hence the work done in that case will be more.

Question:6.3

A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is
a) constant and equal to mg in magnitude
b) constant and greater than mg in magnitude
c) variable but always greater than mg
d) at first greater than mg, and later becomes equal to mg

Answer:

The answer is the option (d) at first greater than mg, and later becomes equal to mg.

Question:6.4

A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicyclist due to the road is 200 N and is directly opposed to the motion. The work done by the cycle on the road is
a) +2000 J
b) -200 J
c) zero
d) -20,000 J

Answer:

The answer is the option (c) Zero
Mechanical energy is not conserved here due to the presence of frictional forces. These lead to dissipation of energy. Work done by frictional force= 200 \times -10 = -2000 J. Work done by the cycle on the road is zero as the road does not move and cover the distance.

Question:6.5

A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?
a) kinetic energy
b) potential energy
c) total mechanical energy
d) total linear momentum

Answer:

The answer is the option (c) total linear momentum
Under the effect of the conservative forces, for a free-falling body, the potential energy decreases and the kinetic energy increases continuously. This leads to the total energy (KE +PE) remaining constant.
E = \frac{1}{2} mv^{2} + mgh
Velocity is zero at point A, so KE is also zero and Ea = mgH
At point B, Vb =\sqrt{ 2gh}
Eb = \frac{1}{2} m(2gh) + mg(H-h)
Eb = mgH
At point C,Vc = \sqrt{2gh}
Ec = \frac{1}{2} m(2gH) + mg (0)
Ec = mgH
Hence the total mechanical energy remains the same.

Question:6.6

During inelastic collision between two bodies, which of the following quantities always remain conserved?
a) total kinetic energy
b) total mechanical energy
c) total linear momentum
d) speed of each body

Answer:

The answer is the option (c) total linear momentum
A collision is in-elastic if the kinetic energy pre and post-collision are not equal.

Question:6.7

Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in Figure.

Which of the following statement is correct?
(a) Both the stones reach the bottom at the same time but not with the same speed.
(b) Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.
(c) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.
(d) Both the stones reach the bottom at different times and with different speeds.

Answer:

The answer is the option (c) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.
AB and AC are plains which inclined at angle \theta _{1} and \theta _{2}. Friction is absent as the plains are assumed to be smooth, which leads to the conservation of mechanical energy.


For both AB and AC,
\frac{1}{2} mv^{2} = mgh and
v= \sqrt{2gh}
Speed is the same for both stones.
Stone I
Acceleration
a_{1} = g \sin\theta_{1}
Stone II
Acceleration
a_{2} = g \sin\theta_{2}
Now, \theta_{2} >\theta_{1} and hence a_{2} >a_{1}. So, stone II reaches faster than stone I.

Question:6.8

The potential energy function for a particle executing linear SHM is given by \frac{1}{2}kx^{2} where k is the force constant of the oscillator (Figure). For k = 0.5N/m, the graph of V(x) versus x is shown in the figure. A particle of total energy E turns back when it reaches x = \pm xm . If V and K indicate the P.E. and K.E., respectively of the particle at x = \pm xm, then which of the following is correct?

(a) V = O, K = E
(b) V = E, K = O
(c) V < E, K = O
(d) V = O, K < E.

Answer:

The answer is the option (b) V = E, K = O
Potential energy at any position,
U = \frac{1}{2} kx^{2}
Total energy of the mass,
E = \frac{1}{2} ka^{2}

Kinetic energy at any position,
K=E-U = \frac{1}{2} ka^{2}+\frac{1}{2} kx^{2}
K= \frac{1}{2} k\left [ a^{2}-x^{2} \right ]
Now,
U_{max}= \frac{1}{2} ka^{2} at extremes.
U min = 0 {x=0 at mean}
K_{max}= \frac{1}{2} ka^{2} {x=0 at mean}
K min = 0 {extremes}
Hence, E= \frac{1}{2}ka^{2} is constant at all times
When particle is at x = xm (mean), KE =0
So, total energy is E = PE + 0 = \frac{1}{2}kx^{2}
So, option b is correct.

Question:6.9

Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V as shown in Figure.

If the collision is elastic, which of the following (Figure) is a possible result after collision?

Answer:

The answer is the option (b)
In an elastic collision, the velocities of particles are interchanged while the kinetic energy and total linear momentum are conserved.
Kinetic energy before collision:
\frac{1}{2} mv^{2} + 0 = \frac{1}{2} mv^{2}
Option a) Kinetic energy after the collision:
Ka = \frac{1}{2} (2m) \left ( \frac{v}{2} \right )^{2} = \frac{1}{4} mv^{2}
Option b) Kinetic energy after the collision:
Ka = \frac{1}{2} (m) \left ( v\right )^{2} = \frac{1}{2} mv^{2}
Option c) Kinetic energy after the collision:
Ka = \frac{1}{2} (3m) \left ( \frac{v}{3}\right )^{2} = \frac{1}{6} mv^{2}
Option d) Kinetic energy after the collision:
\\K_{a}=\frac{1}{2}mv^2+\frac{1}{2}m(\frac{v}{2})^2+\frac{1}{2}m(\frac{v}{3})^2\\\\=\frac{49}{72}mv^2
Hence, option b is correct.

Question:6.10

A body of mass 0.5 kg travels in a straight line with velocity v = a x^{\frac{3}{2}} where a = 5 m^{\frac{-1}{2}}s^{-1}. The work done by the net force during its displacement from x = 0 to x = 2 m is
(a) 1.5 J
(b) 50 J
(c) 10 J
(d) 100 J

Answer:

The answer is the option (b) 50 J
m = 0.5 kg
v = a x^{\frac{3}{2}}
a = 5 m^{\frac{-1}{2}}s^{-1}
now, acceleration= a = \frac{dv}{dt }= v \frac{dv}{dx}
= ax^{\frac{3}{2}} \frac{d}{dx} ax^{\frac{3}{2}} = ax^{\frac{3}{2}} \times \frac{3}{2} \times ax^{\frac{1}{2}} = \frac{3}{2} a^{2}x^{2}
Net force = ma = m \left (\frac{3}{2} a^{2}x^{2} \right )
Work done under the variable force
= \int_{x=0}^{x=2}F.dx = \int_{0}^{2}\frac{3}{2} ma^{2}x^{2} dx
= \frac{1}{2}ma^{2} \times 8 = \frac{1}{2} \times (0.5) \times (25) \times 8 = 50 J

Question:6.11

A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in Figure correctly shows the displacement-time curve for its motion?


Answer:

The answer is the option (b)
P = F.\frac{ds}{dt} = F. v= constant
Carrying out dimensional analysis we get,
[F][v] = constant
[MLT^{-2}] [LT^{-1}]= constant
L^{2}T^{-3}= constant
L\propto T^{\frac{3}{2}}

Question:6.12

Which of the diagrams shown in Figure most closely shows the variation in kinetic energy of the earth as it moves once around the sun in its elliptical orbit?

Answer:

The answer is the option (d)
On the path where Earth orbits around the sun, the speed of Earth is maximum when its position is closest to the sun, and the speed of Earth is minimum when it is farther away from the sun. Hence, we can conclude that kinetic energy can never be negative or zero. Hence, the graph in option d is correct.

Question:6.13

Which of the diagrams shown in Figure represents variation of total mechanical energy of a pendulum oscillating in air as function of time?

Answer:

The answer is the option (c)
The total mechanical energy of a pendulum decreases in a continuous manner when it undergoes oscillatory motion due to the resistance offered by the air. This leads to an exponential decrease in the total mechanical energy of the pendulum. Hence, the graph c is correct, which shows an exponential curve between energy and time.

Question:6.14

A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be
(a) 250 \pi ^{2}
(b) 100 \pi ^{2}
(c) 5 \pi ^{2}
(d) 0

Answer:

The answer is the option (a)
m = 5kg, radius = 1m, f = frequency of revolution = 300 rev/min = 300/60 rev/second
angular velocity = 2\pi f = \frac{(300\times 2\times \pi )}{60} = 10\pirad/second
now v = \omega R=10\pi\times1= 10\pi m/s
kinetic energy
= \frac{1}{2}mv^{2}
= \frac{1}{2}\times 5 \times 10\pi \times 10\pi = 250 \pi^{2} J

Question:6.15

A raindrop falling from a height h above ground, attains a near terminal velocity when it has fallen through a height \left (\frac{3}{4} \right )h. Which of the diagrams shown in Figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground?

Answer:

The answer is the option (b)
At the height h, the raindrop will have zero kinetic energy and maximum potential energy. As it starts falling down, the PE starts decreasing, and the KE starts increasing proportionately. When we consider zero air resistance, the total mechanical energy remains conserved. However, if we consider air resistance, then there is an up-thrust which opposes the velocity of the droplet falling down, which makes the velocity constant as the time passes, which is known as terminal velocity. This relation is aptly depicted in graph b, and hence it is the correct option.

Question:6.16

In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1m s-1 at 45^{\circ} from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s-2, the kinetic energy of the shotput when it just reaches the ground will be
(a) 2.5 J
(b) 5.0 J
(c) 52.5 J
(d) 155.0 J

Answer:

The answer is the option (d) 155.0 J
h = 1.5 m, v = 1 m/s, m = 10 kg, g = 10 ms-2
the initial energy of shot put can be calculated as:
PE + KE = mgh + \frac{1}{2} mv^{2}
= 10 \times 10 \times 1.5 + \frac{1}{2} \times 10 \times 1 = 155 J
From the law of conservation of energy, we can write that,
(PE + KE) initial = (PE + KE) final
Hence, the final kinetic energy of the shotput: KE final + 0 = 155J
Hence the answer is 155 J

Question:6.17

Which of the diagrams in Figure correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity?


Answer:

The answer is the option (b)
The velocity of the iron sphere when free-falling in the lake increases as acceleration due to gravity increases. The up-thrust of the water, however, opposes the motion of the sphere which makes the velocity constant after some time. Hence, in this case, the velocity increases initially and then becomes constant after some time. So, kinetic energy becomes constant, eventually, which is shown in graph b.

Question:6.18

A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat.
Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be
(a) 10.5 N
(b) 21 N
(c) 1.05 ×104 N
(d) 2.1 × 104 N

Answer:

The answer is the option (c)
m = 150 g = \frac{150}{1000 }= \frac{3}{20} kg, t= 0.001 s
u =126km/h = 35 m/s
v = -126km/h = -35 m/s
change in momentum = m(v-u) = \frac{3}{20}(-70) = \frac{-21}{2}
Force
= \frac{\frac{-21}{2}}{0.001 } = -1.05 \times 10^{4} N
The negative sign indicates the opposite direction of the force. Hence the force exerted by the batman should be 1.05 x 104 N.

NCERT Exemplar Class 11 Physics Solutions Chapter 6 MCQII

Question:6.19

A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.
(a) Work done by all forces on man is equal to the rise in potential energy mgL.
(b) Work done by all forces on man is zero.
(c) Work done by the gravitational force on man is mgL.
(d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.

Answer:

The answer is the option (b) and (d)
The gravitational force acts in a downward direction. The displacement is labelled L, which is in the upward direction. So, the work done on man due to gravitational force amounts to -mgL. Also, the work is done to lift the man also amounts to the force in the direction of displacement. Hence the net force amounts to -mgL+ mgL = 0. Hence, option (b) is correct.
The displacement at the point where the force acts are zero. Hence, the amount of work done by the force is also zero. So, KE =0 and hence option (d) is correct.

Question:6.20

A bullet of mass m fired at 30^{\circ} to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target?
(a) The velocity of the bullet will be reduced to half its initial value.
(b) The velocity of the bullet will be more than half of its earlier velocity.
(c) The bullet will continue to move along the same parabolic path.
(d) The bullet will move in a different parabolic path.
(e) The bullet will fall vertically downward after hitting the target.
(f) The internal energy of the particles of the target will increase.

Answer:

The answer is the option (b), (d), (f)
If we assume KE1 and KE2 to be the kinetic energy of the bullet pre and post target then,
KE_{2} = \frac{1}{2} KE_{1}
\frac{1}{2} MV_{2}^{2} =\frac{1}{4} MV_{1}^{2}
V_{2}^{2} = \left (\frac{V_{1}}{\sqrt{2}} \right )^{2} = 0.707 V_{1}
b) V_{2} = 0.707 V_{1}
Hence the velocity of the bullet post it has reached the target is greater than the previous velocity. Hence, option b is the correct choice.
d) the path of the bullet will be parabolic as it has horizontal velocity, and the new parabolic path will have both the components (horizontal and vertical). So, after the target it will be a new parabolic path and hence option (d) is correct.
f) the internal energy of the particles of the target will increase, and some parts of the kinetic energy have converted to heat. Hence option f is correct.

Question:6.21

Two blocks M_{1} and M_{2} having equal mass are free to move on a horizontal frictionless surface. M_{2} is attached to a massless spring as shown in Figure. Initially M_{2} is at rest and M_{1} is moving toward M2 with speed v and collides head-on with M_{2}.

(a) While spring is fully compressed all the KE of M_{1} is stored as PE of spring.
(b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.
(c) If spring is massless, the final state of the M_{1} is state of rest.
(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.

Answer:

The answer is the option (c)
a) the kinetic energy of M_{1} is not fully transferred to the spring as its potential energy, and hence, the option a is incorrect
b) law of conservation of mass is valid here since the surface is frictionless; hence option b is incorrect.
c) if we consider the case, where the spring is totally massless, then all the kinetic energy of M_{1} gets transferred to M_{2}. As a result, m1 comes to rest, and M_{2} acquires a velocity v and starts moving. Hence, option c is correct.
d) even when the force of friction is not involved, a collision can be inelastic. Hence, we reject option d.

NCERT Exemplar Class 11 Physics Solutions Chapter 6 Very Short Answer

Question:6.22

A rough inclined plane is placed on a cart moving with a constant velocity u on horizontal ground. A block of mass M rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy?

Answer:

The block is in the state of rest on the inclined plane, as shown below.
We have, f = mg \sin\theta


Since there is no displacement in the block, the total work is done by f as well as the block amounts to zero. So, no dissipation of energy takes place as the work done is also zero.

Question:6.23

Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?

Answer:

When the elevator is descending, it is a condition of free fall due to the force exerted by gravity. But, in order to control the speed of the elevator and so that it moves at a uniform speed while descending, power is needed. There is a limit on the number of passengers as when the number of passengers increases, more power is required to stop the elevator from the condition of free fall. Since the power of the motor system is limited, the number of passengers which can travel at one time also has a limit.

Question:6.24

A body is being raised to a height h from the surface of earth. What is the sign of work done by
(a) applied force
(b) gravitational force?

Answer:

When a body is raised to a height h, the force is applied in this case in the direction of displacement of the object.
a) so, the sign of the work done by the force is positive here
work done = \overrightarrow{F}d\overrightarrow{s }=Fds \cos \theta
As Fds is positive, \theta =0
b) the force of gravity always acts in the downward direction. However, the force in our case is being applied in an upward manner. So, the value of \theta =180
work done = \overrightarrow{F}d\overrightarrow{s }=Fds \cos 180=-Fds. Hence, here the sign is negative.

Question:6.25

Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2m.

Answer:

As we know that work done =\overrightarrow{ F}d\overrightarrow{s} = Fds \cos \theta
The value of \theta =90 in this case, since the horizontal distance travelled is on a perpendicular situation to the force of gravity acting downward.
So, work done = \overrightarrow{ F}d\overrightarrow{s} = Fds \cos 90=0
Hence the car does not do any work against gravity

Question:6.26

A body falls towards earth in air. Will its total mechanical energy be conserved during the fall? Justify.

Answer:

The mechanical energy of a body is not conserved when the body is under a free-fall condition under the force of gravity. The force of air molecules acts against the motion of the body as in case of free fall, and hence the mechanical energy is not conserved. However, in the case of vacuum as a medium, the free-falling body has no force acting against the motion, and hence the energy is conserved.

Question:6.27

A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero?

Answer:

Only when the conservative force acts along the body when it moves in a closed loop. The work done can be zero. However, if a non-conservative force acts on the body when it moves in a loop (for example friction, electrostatic force, etc.), then the work done cannot be zero.

Question:6.28

In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact).
(a) Kinetic energy.
(b) Total linear momentum?
Give reason for your answer in each case.

Answer:

In the given case of the question, when two balls collide with each other, both; the Kinetic energy, as well as the total linear momentum, remains conserved. There is no presence of any non-conservative force like friction, electrostatic force, etc. in this case and also the balls do not get deformed upon collision; hence both of them remain conserved. If there is a deformation of the ball then the kinetic energy is not conserved but the total energy is conserved.

Question:6.29

Calculate the power of a crane in watts, which lifts a mass of 100 kg to a height of 10 m in 20s.

Answer:

We know from the formula that, the power = work done/ time = \frac{Fds \cos \theta}{ time}
F = mgh
mg = 100 \times 10 = 1000
h = 10 m
t = 20s
\theta =0, as the force and the displacement are in the same direction.
Hence, power = \frac{1000 \times 10 \cos 0}{20} = 500 watts

Question:6.30

The average work done by a human heart while it beats once is 0.5J. Calculate the power used by heart if it beats 72 times in a minute.

Answer:

we know from the formula that, the power = work done/ time
work done by heart in one beat = 0.5J
work done by heart in 72 beats =72 \times 0.5 = 36J
time = 1 minute = 60 seconds
so, power used by the heart = \frac{36}{60} = 0.6 watts

Question:6.31

Give example of a situation in which an applied force does not result in a change in kinetic energy.

Answer:

We are aware that work done = \overrightarrow{F}d\overrightarrow{s} = Fds \cos \theta
Now for the value of work done to be zero, \theta has to be equal to 90, since \cos 90 = 0
So, the direction between the force and the displacement needs to be at an angle of 90 degrees. Hence these situations can have work done to be zero. Some examples can be as follows:
i) A body moving in a horizontal direction in a uniform motion. Here, the work done due to gravity is totally zero.
ii) The case when a body is moving in a circular motion, the total work done is zero

Question:6.32

Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?

Answer:

The work-energy theorem states that the change in the kinetic energy is equal to the work done by the body. So, we can say that KE = WD.
It is given thatKE_{1} = KE_{2}
Hence, WD_{1} = WD_{2}
So, F_{1}S_{1} = F_{2}S_{2}
Since F1 and F2 are equal, S_{1} and S_{2} are also equal.
Hence, irrespective of the mass of the bodies, they travel an equal amount of distance/displacement.

Question:6.33

A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in Fig. 6.11. What will be the trajectory of the particle if the string is cut at

(a) Point B?
(b) Point C?
(c) Point X?

Answer:

The centripetal force is provided by the string which moves the object in a circular motion. The tension of the string is balanced by the tangential velocity of the particle. When the string is cut, the centripetal force becomes zero. So, the bob now moves under the influence of gravitational force in the direction of tangential velocity.
a) string cut off at B: tangential velocity in the downward direction, bob moves along the vertical path under the influence of gravity.
b) string cut off at c: bob has a horizontal component of velocity, it moves In half parabolic path.
c) string cut off at x: the component of velocity makes angle θ with the horizontal, bob reaches to a certain height and then adopts a parabolic path.

NCERT Exemplar Class 11 Physics Solutions Chapter 6 Short Answer

Question:6.34

A graph of potential energy V (x) verses x is shown in Figure. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.


Answer:


The graph above is between kinetic energy and x
As we know from the law of conservation of energy,
E = KE + PE
E = KE + V(x)
KE = E - V(x)
The potential energy is maximum at point A,
PE = E0
Hence, KE = E0 -E0 = 0
At point B,
Let PE = V_b, x = 0
KE = E _0- V_{0}
At point C, PE = 0
X = x_{1}
Hence, KE = E - V(x) =E0 - 0 =E0
At point D,
X = x_{2}, KE = E_0
At point E,
X = x_{3}, PE = E_0
Hence, KE = 0
(ii)
The graph is between velocity versus x
Here, KE = -mv
V = \sqrt{\frac{2K}{M}}
V = \sqrt{K}
Through point A and F, KE = 0
X_{a} = 0, X_{d} = x_{3}
At point C and D,
X_{c} = x_{1}, X_{d} = x_{2}
KE = E0
Hence V_{max} = +/- \sqrt{E_0} = V_{0}
At point B, x = 0 and KE = E_{1}
So, V_{b} = \sqrt{E_{1}} = +/- V

Question:6.35

A ball of mass m, moving with a speed 2v_{0} , collides inelastically (e > 0) with an identical ball at rest. Show that
(a) For head-on collision, both the balls move forward.
(b) For a general collision, the angle between the two velocities of scattered balls is less than 90^{\circ}.

Answer:

a) We assume, v1 and v2 to be the velocities of the balls post the collision takes place. as we know from the law of conservation of momentum,
MV_{0} = MV_{1} + MV_{2}
From given we can infer that,
2V_{0} = V_{1} + V_{2}----------(1)
E = \frac{V_{2} - V_{1}}{ V_{0} + V_{0}}
Hence, V_{2} = V_{1} + 2EV_{0}
V_{1} = V_{0} (1-e)
Since, e < 1
We can say that direction of V1 is same as V0, which is positive and in the forward direction.
b)

The angle between p_{1} and p_{2} is assumed to be θ
From the law of conservation of momentum, \overrightarrow{p}= \overrightarrow{p_{1}} + \overrightarrow{p_{2}}
A portion of kinetic energy is lost as heat when the inelastic collision takes place.
KE(i) > KE_{1} + KE_{2}
\frac{p^{2}}{2m } > \frac{p_{1}^{2}}{2m}+ \frac{p_{2}^{2}}{2m }
hence,\overrightarrow{ p}^{2} > \overrightarrow{p}_{1}^{2} + \overrightarrow{ p}_{2}^{2}
when the angle between p_{1} and p_{2} is lesser than 90 degree, the above equation holds true.

Question:6.36

Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C, and D in which the relation between potential energy V, kinetic energy (K) and total energy is as given below:
Region A: V > E
Region B: V < E
Region C: K > E
Region D: V > K
State with reason in each case whether a particle can be found in the given region or not.

Answer:

Total E = V + K --- (1)
K = E - V ---(2)
V = E - K -----(3)
1. Region A: V > E
As V>E
Kinetic energy will be negative according to equation (2)
So, this case is not possible

2. Region B: V < E
If V<E
Then the value of kinetic energy through equation 2 will be positive.
Hence, this case is possible

3.Region C: K > E
In this case, through equation 3, the value of V will become negative.
Hence, this case is not possible.

4.Region D: V > K
This case is possible; hence, potential energy value can be greater than the value of kinetic energy according to the situation.

Question:6.37

The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in the figure. If the length of the pendulum is 1 m, calculate
a) the height to which bob A will rise after collision
b) the speed with which bob B starts moving.
Neglect the size of the bobs and assume the collision to be elastic.

Answer:

a) when the bob which is in motion collides with the bob at rest, it transfers the momentum and comes to a state of rest. When the position of B is reached by bob A, kinetic energy is obtained by A as a result of the conversion of potential energy. This happens as the momentum is transferred due to elastic collision. So, bob A will not rise.
b)

PE at A1 is equal to kinetic energy at A2 which is equal to kinetic energy at B1
So,
\frac{1}{2}mv^{2} = mgh
V_{b} = \sqrt{2g\times 1} = 4.42 m/s

Question:6.38

A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m/s. Calculate
a) the loss of PE of the drop
b) the gain in KE of the drop
c) is the gain in KE equal to loss of PE? If not why?

Answer:

M = 0.001 kg
H = 1 km, v= 50 m/s, u =0
a) potential energy at highest point of drop
= mgh = 0.001 \times 10 \times 1000m = 10 J
b) Gain in kinetic energy
= \frac{1}{2} mv^{2} = \frac{1}{2} \times 0.001 \times 50 \times 50 = 1.25 J
c) as we can infer, the gain in KE is not equal to the loss in PE as there some energy lost due to resistance.

Question:6.39

Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position the two bobs are in common. One of the bobs is released after being displaced by 10o so that it collides elastically head-on with the other bob.
a) describe the motion of two bobs
b) draw a graph showing variation in energy of either pendulum with time for0 \leq t \leq 2Twhere T is the period of each pendulum.

Answer:

PE(A) = 0, PE(B) = E, KE(A) = KE(B) = 0
These are the initial values when the bobs are at rest. After this, the bobs are released.
t=\frac{T}{4}
when B reaches A, both the bobs collide elastically,
so, KE(A) = 0, KE(B) = E, PE(A) = PE(B)=0
t=\frac{2T}{4}
A will reach at the highest point or at maximum height. And, B is at its lowest point
KE(A) = 0, KE(B) = 0, PE(A) =E, PE(B)=0

t=\frac{3T}{4}
bob A hits bob B. Bob A comes to rest and Bob B starts moving
KE(A) = 0, KE(B) = E, PE(A) =0, PE(B)=0
Ea = 0, Eb = E

t=\frac{4T}{4}
B will reach at the highest point or at maximum height. And, A is at its lowest point
KE(A) = 0, KE(B) = 0, PE(A) =0, PE(B)=E

Question:6.40

Suppose the average mass of raindrops is 3.0 \times 10^{-5}kg and their average terminal velocity 9 m/s. Calculate the energy transferred by rain to each square meter of the surface at a place which receives 100 cm of rain in a year.

Answer:

Rain transfers its kinetic energy to the earth equal to \frac{1}{2}mv^{2}
Velocity of rain= 9m/s
Mass = volume/density
= 1\times 1\times 1000 = 1000kg
So, total energy transferred by rain fall = \frac{1}{2} \times 1000 \times 9 \times 9
= 40500 J

Question:6.41

An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 50,000 kg is moving with a speed of 36 km/h when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of the energy of the wagon is lost due to friction, calculate the spring constant.

Answer:

Kinetic energy is equal to
\frac{1}{2} mv^{2}
m = 50,000 kg, v = 10m/s
so, KE = \frac{1}{2} \times 50000 \times 10\times 10 = 2500000 J
now, as we know that 90% of the kinetic energy is lost as a result of friction brakes. Hence, only 10% of KE is transferred to spring.
So, the kinetic energy of spring would be
\frac{1}{2} kx^{2}
Here, x = 1m
Hence, K = 500000 J

Question:6.42

An adult weighing 600 N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilized by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilized for jogging.

Answer:

Potential energy required by the jogger to raise each step = mgh
mg = 600N, h = 0.25 m
Number of steps travelled in 6 km = 6000
Total Energy for 6000 steps = 6000 \times 600 \times 0.25
Jogging utilises 10% of the total energy = \frac{10}{100}\times 6000 \times 600 \times 0.25
= 90,000 J

Question:6.43

On complete combustion a litre of petrol gives off heat equivalent to 3\times 10^{7} J. In a test drive a car weighing 1200 kg, including the mass of driver, runs 15 km per litre while moving with a uniform speed on a surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were 0.5.

Answer:

Car engine’s efficiency = 0.5
So, energy given to car through petrol of 1 litre = 0.5 \times 3 \times 10^{7}
Now, the work done by car is F \times S = F \times 15000 J
As we know that the work done by car is against the force of friction, hence equating both can give us the value of friction.
F \times 15000 = 0.5 \times 3\times 10^{7}
F = 1000 N

NCERT Exemplar Class 11 Physics Solutions Chapter 6 Long Answer

Question:6.44

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of30^{\circ} by a force of 10 N parallel to the inclined surface. The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate
a) work done against gravity
b) work done against force of friction
c) increase in potential energy
d) increase in kinetic energy
e) work done by applied force

Answer:

Mass = 1 kg, \theta = 30, F = 10 N, d = 10m
(a) work done against gravity
= mgh
\sin 30 = \frac{h}{10}
h= \frac{10}{2}=5m
Work done = 1 \times 10 \times 5 = 50 J
(b)
Work done against force of friction = \mu f.s = \mu mg \cos \theta \times s
= 0.1 \times 1 \times 10 \cos 30 \times 10 = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} Joules
(c) increase in potential energy
This value is equal to the work done against gravity, which we have calculated above in part a. hence, the increase in potential energy = 50J
(d) increase in kinetic energy
\Delta KE= work done = -mgh + F_s - f_s
= - 50 - 5 \sqrt{3 }+ 10 \times 10 = 50 - 5\sqrt{ 3}
= 41.340 J
(e) work done by applied force.
= F.S = 10 \times 10 = 100 J

Question:6.45

A curved surface as shown in the figure. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from one by one from A which is at a slightly greater height than C. with the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has small friction and ball 3 has negligible friction.
a) for which balls is total mechanical energy conserved?
b) which ball can reach D?
c) for balls which do not reach D, which of the balls can reach back A?

Answer:

(a) For which ball is total mechanical energy conserved?
For balls 1 and 3, the total mechanical energy is conserved. In the case of ball 1, it rolls down without slipping and hence no force of friction acts against its motion and no energy is dissipated. In the case of ball 3, it has negligible friction, and hence, there is no loss of energy which leads to the conservation of mechanical energy.
(b) Which ball (s) can reach D?
Ball 1 slips due to its acquired rotational energy on the frictionless surface. Ball 2 loses its energy due to friction. Hence, it does not reach C.
(c) For balls which do not reach D, which of the balls can reach back A?
As we saw, balls 1 and 2 do not reach C, and hence they also don’t reach D.
Ball 3 has negligible friction and hence can reach D. for balls 1 and 2, no of them can reach back at A.

Question:6.46

A rocket accelerates straight up by ejecting gas downwards. In a small time interval \Delta t, it ejects a gas of mass \Delta m at a relative speed u. Calculate KE of the entire system at t+\Delta t and t and show that the device that ejects gas does work = \left ( \frac{1}{2} \right ) \Delta m u^{2} in this time interval.

Answer:

We assume the mass of the rocket at any time to be m.
Let velocity of rocket be v
Mass of gas ejected in time t can be Δm
(KE) _{t+ \Delta t} = \frac{1}{2} ( M- \Delta m) (v + \Delta v)^{2} + \frac{1}{2} \Delta m (v-u)^2
= \frac{1}{2} [Mv^{2} + M \Delta v^{2} + 2Mv \Delta v - \Delta mv^{2} - m \Delta v^{2} - 2v \Delta m \Delta v+ \Delta mv^{2}+ \Delta mu^{2} - 2uv \Delta m ]
When we neglect small terms we get,
(KE) = \frac{1}{2} MV^{2}
\Delta K = \frac{1}{2} \Delta mu^{2} + v (M \Delta v - u \Delta m)
In accordance with newton’s third law,
Reaction upward force on the rocket is equal to action force by burning of gases in downward direction
M \frac{dv}{dt }= \frac{dm}{dt} |u|
M \Delta v = \Delta mu
Substituting this value, we get,
K = \frac{1}{2} u^{2} \Delta m
As we know from work-energy theorem, \Delta KE = work done
We have, work done = \frac{1}{2} \Delta m u^{2}

Question:6.47

Two identical steel cubes(mass 50g, sides 1cm) collide head-on face to face with a speed of 10 cm/s each. Find the maximum compression of each. Young’s modulus for steel= Y = 2 \times 10^{11} N/m^{2}.

Answer:

The kinetic energies of the cubes when they collide, converts into potential energy.
From the hooks law we know that, stress α strain
Y = stress/strain
Y=\frac{FL}{A\Delta L}
F=\frac{ AY \Delta L}{L} = LY\Delta L
Work done = F. \Delta L = LY. \Delta L^{2}
Kinetic energy
= 2\left (\frac{1}{2} m v^{2} \right ) = 0.05 \times 0.1 \times 0.1 = 5 \times 10 ^{-4} J
As WD=KE,
LY. \Delta L^{2} = 5 \times 10 ^{-4}
\Delta L =\sqrt{ 25 \times 0.0001} = 5 \times 10 ^{-7} m

Question:6.48

A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that density of air is constant.

Answer:

The net buoyant force in this case = vpg
= vol of air displaced x net density in upward direction x g
= V (p(air) - p(He)) g

If we assume a to be acceleration in upward direction,
Ma = V (p(air) - p(He)) g --------- (1)
{p(air) - density of air
P(He) - density of helium}
m dv/dt = V (p(air) - p(He)) g
m dv = V (p(air) - p(He)) g. dt
when we integrate both sides,
mv = V (p(air) - p(He)) g. t
v = V/m (p(air) - p(He)) g.t
kinetic energy of balloon,
\frac{1}{2}mv^{2} = \frac{1}{2} \frac{V^{2}}{m} (p(air) - p(He))^{2} g^{2}t^{2}-------------(2)
Let the balloon rise to height h,
a = V/m (p(air) - p(He)) g
h = ut + \frac{1}{2} at^{2}
h= V/2m (p(air) - p(He)) gt2 ---------------- (3)
on rearranging the terms from the three equations we get,
\frac{1}{2} mv^{2} = {\frac{V}{2m} (p(air) - p(He)) gt^{2}} { V (p(air) - p(He)) g }
\frac{1}{2} mv^{2} = V p(air) gh - V p(He)gh
\frac{1}{2}mv^{2} + p(He)Vgh = p(air)Vgh
KE of balloon + PE of balloon = the change in potential energy in air
So, we can conclude that when the balloon goes up, an equal volume of air is displaced downwards. The PE and KE of the balloon keep increasing as the PE of air is changing.

From the point of doubts and understanding, students can take the help of NCERT Exemplar Class 11 Physics solutions chapter 6 PDF download, prepared by experts, will help them clear their difficulties.

Also, read NCERT Class 11 Physics Syllabus

For the students to solve their queries regarding the chapter and understand concepts easily. NCERT Exemplar Class 11 Physics chapter 6 solutions PDF download, can be done by the students to easily score marks in their examinations of NEET and JEE Main.

NCERT Exemplar Class 11 Physics chapter 6 solution includes the following topics.

Main Subtopics in NCERT Exemplar Class 11 Physics Solutions Chapter 6 Work, Energy, and Power

  • 6.1 Introduction
  • 6.2 Notions of work and kinetic energy: The work-energy theorem
  • 6.3 Work
  • 6.4 Kinetic energy
  • 6.5 Work done by a variable force
  • 6.6 The work-energy theorem for a variable force
  • 6.7 The concept of potential energy
  • 6.8 The conservation of mechanical energy
  • 6.9 The potential energy of a spring
  • 6.10 Various forms of energy: the law of conservation of energy
  • 6.11 Power
  • 6.12 Collisions

Also, check NCERT Solution subject wise -

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Also, check NCERT Notes subject wise -

What Will The Students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 6?

The students, with the help of NCERT Exemplar Class 11 Physics chapter 6 solutions, will learn about an important concept of physics stating the inter-relation of energy, work, and power. Kinetic energy is said to be the motion of objects, the chapter and solution both cover the topic. Newton’s third law, i.e. law of conservation of energy, with practical explanation, is explained based on mechanical forces and its relativity to constant. Learning about collisions forces, how two or more forces exert each other, even for a short period, is taught here. The NCERT Exemplar Class 11 Physics solutions chapter 6 are said to be simpler in terms of understanding and grasping knowledge.

NCERT Exemplar Class 11 Solutions

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

Important Topics To Cover For Exams From NCERT Exemplar Class 11 Physics Solutions Chapter 6 Work, Energy, and Power

There are certain students who learn important topics whose concepts need to be cleared.

· NCERT Exemplar Class 11 Physics chapter 6 solution will help the students understand the difference between energy, work and power, along with their correlation.

· The famous topic of kinetic energy, about how this energy is the work of the net force of the body. It shows how the working of the body forms this energy.

· The chapter covers the functioning of the mechanical energy of the body, by how if the forces of the body were conservative, then only our mechanical energy would be constant- law of conservation of energy.

· The scalar quantity is also explained in NCERT Exemplar Class 11 Physics solutions chapter 6 where it clears the differentiation of positive and negative scalar quantities. It shows how doing work is a scalar quantity.

· Collisions are the two or more exerting forces, on each other for a relatively short period, is very well explained in the chapter.

Check Class 11 Physics Chapter-wise Solutions

Chapter 1Physical world
Chapter 2Units and Measurement
Chapter 3Motion in a straight line
Chapter 4Motion in a Plane
Chapter 5Laws of Motion
Chapter 6Work, Energy and Power
Chapter 7System of Particles and Rotational motion
Chapter 8Gravitation
Chapter 9Mechanical Properties of Solids
Chapter 10Mechanical Properties of Fluids
Chapter 11Thermal Properties of Matter
Chapter 12Thermodynamics
Chapter 13Kinetic Theory
Chapter 14Oscillations
Chapter 15Waves

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. How many questions have been answered in the NCERT?

 A collaboration of 44 questions has been answered in the chapter. The questions are of MCQ, short answer and long answer type.

2. Why should we be relying on NCERT solutions

Prepared by the experts over the subject, NCERT Exemplar Class 11 Physics solutions Chapter 6 will make concepts and doubts of these chapters easy to understand.

3. Does the solution comprise all the required topics?

The NCERT exemplar solutions for class 11 Physics chapter 6 will consist of all the important, required topics for better learning. Questions based on all the main topicsof chapter 6 physics are covered here.

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Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available
Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
Finance Executive
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Stock Analyst

Individuals who opt for a career as a stock analyst examine the company's investments makes decisions and keep track of financial securities. The nature of such investments will differ from one business to the next. Individuals in the stock analyst career use data mining to forecast a company's profits and revenues, advise clients on whether to buy or sell, participate in seminars, and discussing financial matters with executives and evaluate annual reports.

2 Jobs Available
Researcher

A Researcher is a professional who is responsible for collecting data and information by reviewing the literature and conducting experiments and surveys. He or she uses various methodological processes to provide accurate data and information that is utilised by academicians and other industry professionals. Here, we will discuss what is a researcher, the researcher's salary, types of researchers.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available
Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Field Surveyor

Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Veterinary Doctor
5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Anatomist

Are you searching for an ‘Anatomist job description’? An Anatomist is a research professional who applies the laws of biological science to determine the ability of bodies of various living organisms including animals and humans to regenerate the damaged or destroyed organs. If you want to know what does an anatomist do, then read the entire article, where we will answer all your questions.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

2 Jobs Available
Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
QA Manager
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
ITSM Manager
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

2 Jobs Available

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