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NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power

Edited By Vishal kumar | Updated on Aug 28, 2023 11:59 AM IST

NCERT Solutions for Class 11 Physics Chapter 6 – Download Free PDF

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power play a crucial role in the mechanics section of the 11th grade. Moreover, this chapter of NCERT Solution serves as the cornerstone for several subsequent chapters. To achieve high scores in your Class 11 exams or competitive tests like JEE and NEET, it's essential to grasp work energy and power Class 11 numericals with solutions thoroughly and efficiently tackle its questions.

On this page of Careers360, you will find comprehensive class 11 physics chapter 6 exercise solutions from Question 6.1 to 6.30. along with, additional exercise questions designed by subject experts. These solutions are presented in a clear, step-by-step manner that is easy to understand.

Consider that you are carrying a 10 Kg stationary in your hand and you are not moving. In this situation, you are doing work physically according to you. But according to physics, you are doing zero work. Can you find out why work is zero here? Work Energy and Power ch 6 Physics class 11 has questions based on the concepts of work and types of work, the relation between work and energy and the concept of power. In the CBSE NCERT solutions for Class 11 Physics Chapter 6 Work Energy and Power, you will study problems on the law of conservation of energy (energy can be neither created nor destroyed).

NCERT solutions for Class 11 Physics Chapter 6 Work Energy and Power also give answers to the questions based on the concepts of collision, which are important for the exam. Solutions of NCERT are helpful in scoring well in class and board exams. Solutions for both exercise and additional exercise in the NCERT are given. Students can also download Work Energy and Power Class 11 NCERT solutions pdf by clicking on the link download PDF for free.

NCERT Class 11 Physics Chapter 6 Exercise Solutions pdf download for free.

**According to the CBSE Syllabus for the academic year 2023-24, the chapter you previously referred to as Chapter 6, "Work, Energy and Power" has been renumbered as Chapter 5.

NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

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NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power - Exercise Questions

Q1(b) The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(b) work done by gravitational force in the above case

Answer:

In this case, the direction of displacement is upward and the direction of force is downward. Thus work done is negative in nature.

Q1 (c) The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: work done by friction on a body sliding down an inclined plane,

Answer:

We know that friction acts in the direction opposite to the direction of motion. Hence work done by the frictional force is negative.

Q2 (a) A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the work done by the applied force in 10 s,

Answer:

Using Newton's law we can write :

a\ =\ \frac{F}{m}

=\ \frac{7}{2}\ =\ 3.5\ m/s^2

The frictional force is given by :

f\ =\ \mu mg

=\ 0.1\times 2\times 9.8\ =\ 1.96\ N

Its direction will be opposite of the direction of the motion. Thus acceleration produced will be negative.

a\ =\ \frac{-1.96}{2}\ =\ -0.98\ m/s^2

Thus the net acceleration is = 3.5 - 0.98 = 2.52 m/s 2 .

The total distance travelled is given by :

s\ =\ ut\ +\ \frac{1}{2}at^2

or =\ 0\ +\ \frac{1}{2}(2.52)10^2\ =\ 126\ m

Hence the work done by applied force is given by :

W\ =\ F.s\ =\ 7\times 126\ =\ 882\ J

Q2 (d) A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the change in kinetic energy of the body in 10 s

Answer:

It is given that initial velocity is zero. The final velocity can be calculated by the equation of motion :

v\ =\ u\ +\ at

or =\ 0\ +\ (2.52)10

or =\ 25.2\ m/s

Thus change is kinetic energy is :

\Delta K\ =\ \frac{1}{2}mv^2\ -\ \frac{1}{2}mu^2

or =\ \frac{1}{2}\times 2(25.2)^2\ -\ 0

or =\ 635\ J

Q3 Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
1645180951423

1645180952211

Answer:

Total energy = kinetic energy (KE) + potential energy(PE)

KE > 0 since m and v 2 is positive. If KE <0 particles cannot be find. If PE>TE, then KE<0 (now in all graph check for this condition)

In case 1 kinetic energy is negative for x<a. So at x<a particle cannot be found.

In case 2 for x<a and for x> b kinetic energy is negative. So the particle cannot be found in these regions.

In the third case , the minimum potential energy is when a<x<b . At this position, the potential energy is negative (- V 1 ).

The kinetic energy in this case is given by :

K.E.\ =\ E\ -\ (-V_1)\ =\ E\ +\ V_1

And the minimum energy of particle is - V 1 .

In the fourth case , the particle will not exist in the states which will have potential energy greater than the total energy.

Thus particle will not exist in \frac{-b}{2}<x<\frac{b}{2} and \frac{-a}{2}<x<\frac{a}{2} .

The minimum energy of particle will be - V 1 as it is the minimum potential energy.

Q4 The potential energy function for a particle executing linear simple harmonic motion is given by V ( x ) = k x^2 /2 , where k is the force constant of the oscillator. For k = 0.5 Nm^{-1} the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it
reaches x = \pm 2

1645180974254

Answer:

The total energy of the particle is given by :

E\ =\ K.E\ +\ P.E

or =\ \frac{1}{2}mv^2\ +\ \frac{1}{2} kx^2

At the extreme position, the velocity of the object is zero thus its kinetic energy at that point is zero.

E\ =\ \frac{1}{2} kx^2

or 1\ =\ \frac{1}{2} (0.5)x^2

or x^2\ =\ 4

or x\ =\ \pm \ 2

Hence the extreme position are \pm \ 2\ m .

Q5 (a) Answer the following :
The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?

Answer:

The total energy is given by :

E\ =\ K.E\ +\ P.E

or =\ \frac{1}{2}mv^2\ +\ mgh

The burning of casing results in a reduction in the mass of the rocket. This leads to a lowering in the total energy.

Thus heat required for burning is obtained from the expenses of the rocket .

Q5 (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

Answer:

This is because the gravitational force is a conservative force. And we know that the work done by a conservative force in a closed path is always zero. That's why the work done by the gravitational force is zero in a complete orbit revolution.

Q5 (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?

Answer:

The total energy of artificial satellite remains constant. Thus when it approaches towards the earth the distance between them decreases. This results in a decrease in the potential energy of the satellite. By energy conservation, the kinetic energy of satellite increases and so does the velocity.

Q6 (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.

Answer:

It is given that work done by the conservative force is positive, thus the force acts in the direction of the motion. This results in a decrease in distance between the bodies. Thus it's potential energy decreases .

Q6 (b) Work done by a body against friction always results in a loss of its kinetic/potential energy.

Answer:

Work done by the body against friction results in a decrease in the velocity of the body. Thus the kinetic energy of the body decreases.

Q6 (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.

Answer:

The internal force cannot produce a change in the total momentum as no external force is acting. Thus the change in total momentum is proportional to the external forces acting on the body.

Q6(d) In an inelastic collision of two bodies, the quantities which do not change after
the collision are the total kinetic energy/total linear momentum/total energy of
the system of two bodies.

Answer:

The conservation of total linear momentum doesn't depend upon the fact whether it is an elastic collision or an inelastic collision.

Q7(a) State if each of the following statements is true or false. Give reasons for your answer.

In an elastic collision of two bodies, the momentum and energy of each body is conserved.

Answer:

False: - The linear momentum and energy will be conserved if both are considered in a system. But for individual bodies, this conservation of momentum and energy doesn't hold. This is because the impact during the collision may transfer energy/momentum of one ball to the other ball.

Q7 (b) State if each of the following statements is true or false. Give reasons for your answer. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.

Answer:

False: - Internal forces will not change the energy of the system but external forces can change the total energy by changing their magnitude or direction.

Q7 (c) State if each of the following statements is true or false. Give reasons for your answer. Work done in the motion of a body over a closed loop is zero for every force in nature.

Answer:

False:- This is true only for conservative forces e.g. gravitational force. For e.g in case of frictional force (non-conservative force), the work done in a closed-loop cannot be zero as energy is wasted throughout.

Q7 (d) State if each of the following statements is true or false. Give reasons for your answer. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Answer:

True but not always:- In the case of inelastic collisions, few amounts of energy is converted into other forms of energy such as sound or in deformation. Thus final kinetic energy is always less as compared to initial kinetic energy. But in case of the explosion of a bomb final kinetic energy is greater than the initial kinetic energy

Q8 (a) Answer carefully, with reasons: In an elastic collision of two billiard balls, is the total kinetic energy conserved
during the short time of collision of the balls (i.e. when they are in contact)?

Answer:

No, because at the time of the collision, the kinetic energy is converted to the potential energy. Thus total kinetic energy is not constant at the collision.

Q8 (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?

Answer:

Yes, in case of elastic collision the total linear momentum of the system remains conserved as no external force is acting on the system of balls.

Q8 (c) What are the answers to (a) and (b) for an inelastic collision?

Answer:

The total kinetic energy of the system cannot be conserved in case of inelastic collision as there is loss of energy in the form of deformation. But the total linear momentum of the system remains constant even in the case of inelastic collision as no external force is acting.

Q8 (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational
potential energy).

Answer:

Since the potential energy of the system depends upon the separation between the bodies thus the forces acting on the body are conservative in nature. We know that conservative forces produce elastic collisions.

Q 9) A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(i) t^{1/2}\: \: (ii) t \: \: (iii) t^{3/2}\: \: (iv) t^2

Answer:

It is given that acceleration is constant thus force will also be constant (by Newton's law of motion F = ma).

Also,

a\ =\ \frac{dv}{dt}\ =\ constant

or dv\ =\ C\ dt

Thus v\ \propto \ t

Now, the work done by the force is given by :

P\ =\ F.v

Hence power is directly proportional to the time.

Q10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to

(i) t^{1/2}\: \: (ii) t \: \: (iii) t^{3/2}\: \: (iv) t^2

Answer:

We know that the power is given by :

P\ =\ F.v

or =\ ma.v

or =\ m\frac{dv}{dt}v

It is given that power is constant, thus :

mv\frac{dv}{dt}\ =\ constant

or vdv\ =\ \frac{C}{m}dt

By integrating both sides, we get

v\ =\ \left ( \sqrt{\frac{2Ct}{m}} \right )

Also, we can write :

v\ =\ \frac{dx}{dt}

or \frac{dx}{dt}\ =\ \sqrt{\frac{2C}{m}}t^\frac{1}{2}

By integrating we get the relation :

x\ \propto \ \ t^\frac{3}{2}

Q12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass = 9.11 \times 10 ^{-31} Kg , proton mass
1.67 \times 10 ^{-27} Kg , 1 e V = 1.60 \times 10 ^{-19} J )

Answer:

The kinetic energy of the electron is given by :

K_e\ =\ \frac{1}{2}mv_e^2

or 1.6\times 10^{-15}\ J\ =\ \frac{1}{2}\times 9.11\times 10^{-31}\times v_e^2

Thus velocity is obtained as :

v_e\ =\ \sqrt{\frac{2\times 1.6\times 10^{-15}}{9.11\times 10^{-31}}}

or =\ 5.93\times 10^7\ m/s

Similarly, we can find the velocity of the proton :

K_p\ =\ \frac{1}{2}mv_p^2

1.6\times 10^{-14}\ J\ =\ \frac{1}{2}\times 1.67\times 10^{-27}\times v_p^2

Thus velocity is obtained as :

v_p\ =\ \sqrt{\frac{2\times 1.6\times 10^{-14}}{1.67\times 10^{-27}}}

or =\ 4.38\times 10^6\ m/s

Thus the ratio of their velocities is :

\frac{v_e}{v_p}\ =\ \frac{5.93\times 10^{7}}{4.38\times 10^6}\ =\ 13.54

Q14 A molecule in a gas container hits a horizontal wall with speed 200 ms^{-1} and angle 30 o with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Answer:

The momentum is conserved in the collision as no external force is acting on the system. In the given case the rebound velocity is the same as the initial velocity thus the kinetic energy of the molecule initially and finally are same. Hence this is an elastic collision.

Q15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m 3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Answer:

Mass of the water is :

m\ =\ \rho v

or =\ 30\times10^3 \ Kg

Thus the output power is given by :

Power\ =\ \frac{Work\ done}{Time}

or =\ \frac{mgh}{t}

or =\ \frac{30\times 10^3\times 9.8\times 40}{900}

or =\ 13.067\times 10^3\ W

Also, we are given that efficiency is 30 per cent.

Thus the input power is :

P_i\ =\ \frac{13.067}{\frac{30}{100}}\times 10^3

or =\ 43.6\ KW

Q16 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. If the collision is elastic, which of the following (Fig. 6.14) is a possible result after collision?

1645181029146

Answer:

The initial kinetic energy of the system is given by :

=\ \frac{1}{2}mv^2\ -\ \frac{1}{2}2m(0)

or =\ \frac{1}{2}mv^2

Case (i):- The final kinetic energy is :

=\ \frac{1}{2}m.0\ -\ \frac{1}{2}2m(\frac{v}{2})^2\ =\ \frac{1}{4}mv^2

Thus the kinetic energy is not conserved in this case.

Case (ii):- The final kinetic energy is :

=\ \frac{1}{2}2m.0\ -\ \frac{1}{2}mv^2\ =\ \frac{1}{2}mv^2

Thus kinetic energy is conserved in this case.

Case (iii):- The final kinetic energy is:-

=\ \frac{1}{2}\times 3m\times \left ( \frac{v}{3} \right )^2

or =\ \frac{1}{2}m v^2

Thus the kinetic energy is not conserved in this case.

Q17 The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest
on a table as shown in Fig. 6.15. How high does the bob A rise after the collision? Neglect the size of
the bobs and assume the collision to be elastic.

1645181053219

Answer:

This is an elastic collision thus the transfer of momentum will take place. It is given that bob B is at rest and bob A has some velocity. So in momentum transfer, bob B will gain the velocity in the left direction whereas bob A will come to rest (complete momentum transfer takes place).

Hence bob A will not rise.

Q18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Answer:

Consider the extreme position (horizintal) :-

The kinetic energy at this position is zero as velocity is zero.

Thus total energy is given by : =\ mgl\ +\ 0 \ =\ mgl

Now consider the mean position (lowermost point) :

Here the potential energy of bob is zero.

Whereas kinetic energy is :

=\ \frac{1}{2}mv^2

Further, it is given that 5 per cent of energy is dissipated due to air resistance while coming down.

Thus energy equation becomes (conservation of energy):-

\frac{1}{2}mv^2\ =\ \frac{95}{100}\times mgl

or

v\ =\ \sqrt{\frac{2\times 95\times1.5\times 9.8}{100}}\ =\ 5.28\ m/s

Q19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.005 Kg s^{-1} . What is the speed of the trolley after the entire sandbag is empty?

Answer:

Since the sand is falling in the trolley thus the force generated on the system (trolley and sandbag) is an internal force. There is no external force thus momentum of the system doesn't change. Hence speed remains the same i.e., 27 Km/hr.

Q20 A body of mass 0.5 kg travels in a straight line with velocity v = ax ^{3/2} where a = 5 m ^{-1/2 }s ^{-1} . What is the work done by the net force during its displacement from x = 0 to x = 2 m ?

Answer:

The relation between work done and the kinetic energy is given by :

Work\ =\ \frac{1}{2}mv^2\ -\ \frac{1}{2}mu^2

Using the relation v = ax ^{3/2} we can write :

Initial velocity = 0 (at x = 0 )

And the final velocity = 10\sqrt{2}\ m/s (at x = 2).

Thus work done is :

Work\ =\ \frac{1}{2}m(v^2\ -\ u^2)

or =\ \frac{1}{2}\times 0.5\times (10\sqrt{2})^2

or =\ 50\ J

Q21 (a) The blades of a windmill sweep out a circle of area A. If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?

Answer:

The volume of wind = Av here A is the swept circle and v is the velocity.

Thus the mass of the wind is : - \rho Av , \rho is the density of the air.

Hence mass of wind flowing through windmill in time t is =\ \rho Avt .

Q21 (b) The blades of a windmill sweep out a circle of area A. What is the kinetic energy of the air?

Answer:

The kinetic energy is given by :

=\ \frac{1}{2}mv^2

or =\ \frac{1}{2}\ \rho Avt\ v^2

or =\ \frac{1}{2}\ \rho At\ v^3

Thus the kinetic energy of wind is \frac{1}{2}\ \rho At\ v^3 J.

Q21 (c) The blades of a windmill sweep out a circle of area A. Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m ^2 , v = 36 Km /h and the density of air is 1.2 Kg m^{-3} What is the electrical power produced?

Answer:

It is given that 25 per cent of wind energy is converted into electrical energy.

Thus electric energy produced is :

=\ \frac{25}{100} \times \frac{1}{2}\ \rho At v^3

or =\ \frac{1}{8}\ \rho At v^3

Now the electric power is given by :

Power\ =\ \frac{Energy}{Time}

or =\ \frac{\frac{1}{8}\ \rho At v^3}{t}\ =\ {\frac{1}{8}\ \rho A v^3}

or =\ {\frac{1}{8}\ \times 1.2\times 30 \times (10)^3}

or =\ 4.5\ KW

Q23 (a) A family uses 8 kW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

Answer:

It is given that the efficiency of energy conversion is 20 per cent.

According to question, we can write (equating power used by family) :

8\times 10^3\ =\ \frac{20}{100}\times A\times 200 (Here A is the area required.)

or A\ =\ \frac{8\times 10^3}{40}

or =\ 200\ m^2

Thus required area is 200 m 2 .

Q23 (b) A family uses 8 kW of power. Compare this area to that of the roof of a typical house.

Answer:

A typical has dimensions of 14 \times 14\ m^2 .

The area of the roof of the house is 225\ m^2 .

This is nearly equal to the area required for the production of the given amount of electricity.

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Additional Exercise

Q24 A bullet of mass 0.012 kg and horizontal speed 70 ms^{-1} strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer:

We are given :

Mass of the bullet m: 0.012 Kg

Mass of the block M: 0.4 Kg

The initial velocity of the bullet u: 70 m/s

The initial velocity of the block : 0

The final velocity of the system (bullet + block): v

For finding the final speed of system we will apply the law of conservation of momentum :

mu_b\ +\ M(0)\ =\ (m\ +\ M)v

or v\ =\ \frac{0.84}{0.412}\ =\ 2.04\ m/s

Now for the system, we will apply the law of conservation of energy :

The potential energy at the highest point = Kinetic energy at the lowest point

(m+M)gh\ =\ \frac{1}{2}(m+M)v^2

or h\ =\ \frac{1}{2}\times \frac{v^2}{g}

or =\ \frac{1}{2}\times \frac{2.04^2}{9.8}

or =\ 0.2123\ m

Hence heat produced is :

=\ \frac{1}{2}mu^2\ -\ \frac{1}{2}(m\ +\ M)v^2

or =\ \frac{1}{2}(0.012)(70)^2\ -\ \frac{1}{2}(0.412)(2.04)^2

or =\ 29.4\ -\ 0.857\ =\ 28.54\ J

Q25 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 (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given \theta _1 = 30 \degree , \theta = 60 \degree , and h = 10 m, what are the speeds and times taken by the two stones?

1645181110026

Answer:

The FBD of the track is shown in the figure below :

1645181125343

Using the law of conservation of energy we have :

\frac{1}{2}mv_1^2\ =\ \frac{1}{2}mv_2^2

or v_1\ =\ v_2

Hence both stones will reach the bottom with the same speed.

For stone 1 we can write :

F\ =\ mg\sin \Theta_1

or a_1\ =\ g\sin \Theta_1

For stone 2 we have :

a_2\ =\ g\sin \Theta_2

Also, using the equation of motion,

v\ =\ u\ +\ at

or t\ =\ \frac{v}{a}

It is given that \Theta _2 > \Theta_1

or a _2 > a_1

Thus t _1 > t_2

Hence, the stone travelling on the steep plane will reach before.

For finding speed and time we can use conservation of energy.

mgh\ =\ \frac{1}{2}mv^2

or v\ =\ \sqrt{2gh}

or =\ \sqrt{2\times 9.8\times 10}

or =\ 14\ m/s

And the time is given by :

t_1\ =\ \frac{v}{a_1}\ =\ \frac{14}{9.8\times \sin 30^{\circ}}\ =\ 2.86\ s

and t_2\ =\ \frac{v}{a_2}\ =\ \frac{14}{9.8\times \sin 60^{\circ}}\ =\ 1.65\ s

Q27 A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s ^{-1} . It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

Answer:

In this case, the heat produced is the loss in the potential energy.

Thus,

heat produced = mg h

or =\ 0.3\times 9.8\times 3

or =\ 8.82\ J

The heat produced (when the lift is stationary) will remain the same as the relative velocity of the bolt with respect lift still remains zero.

Q28 A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 ms ^{-1} relative to the trolley in a direction opposite to its motion, and jumps out of the trolley. What is the final speed of the trolley ? How much has the trolley moved from the time the child begins to run ?

Answer:

The initial momentum of the system (boy + trolley) is given as :

=\ (m\ +\ M)V

=\ (200\ +\ 20)10\ =\ 2200\ Kg\ m/s

Now assume v ' is the final velocity of the trolley with respect to the ground.

Then the final momentum will be :

=\ Mv'\ +\ m(v'\ -\ 4)\ =\ 220v'\ -\ 80

Conserving momentum :

220v'\ -\ 80\ =\ 2200

or v'\ =\ 10.36\ m/s

The time taken by the boy is :

=\ \frac{10}{4}\ =\ 2.5\ s

Hence the distance moved by the trolley is :

=\ 10.36\times 2.5 \ =\ 25.9\ m

Q29 Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

1645181171233 Answer:

The potential energy of the system depends inversely on the separation between the balls. Thus the potential energy will decrease as the balls will come closer and will become zero as they touch each other.

Thus elastic collision is best described only by the graph (v).

Q30 Consider the decay of a free neutron at rest: Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the of a neutron or a nucleus (Fig. 6.19).

1645181199357

[Note: The simple result of this exercise was one among the several arguments advanced by W.
Pauli to predict the existence of a third particle in the decay products of \beta -decay This
particle is known as the neutrino. We now know that it is a particle of intrinsic spin ½ (like
e—, p or n), but is neutral, and either massless or having an extremely small mass
(compared to the mass of an electron) and which interacts very weakly with matter. The
correct decay process of the neutron is : n \rightarrow p+ e ^- ]

Answer:

By Einstein’s mass-energy relation we can write :

Energy\ =\ \Delta mc^2

Here \Delta m and C are constant thus two-body decay is unable to explain (or account for) the continuous energy distribution in the β -decay of a neutron.

class 11 physics chapter 6 ncert solutions consist of a total of thirty questions. These questions cover topics such as basic work energy and Power, Enistein's mass-energy relation, friction in an inclined plane, total energy and more. Step-by-step solutions made by subject matter experts of Careers360 can provide students with a comprehensive understanding of these concepts and help them prepare for exams. By studying NCERT Solutions for Class 11 Physics Chapter 6 PDF, students can improve their problem-solving skills and gain confidence in their ability to tackle physics problems.

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Class 11 Physics chapter 6 exercise solutions: Important Formulas and Diagrams + eBook link

Knowing how important formulas and diagrams are in Class 11th Physics, Chapter 6 NCERT Solutions is key. These tools really help understand concepts and solve problems well. To help you learn better, here are the important formulas and diagrams for this chapter:

  • Work done

W= Fcos\theta \times \bigtriangleup s=\vec{F}.\Delta s=\int \vec{F} \cdot \mathrm{d} \vec{S}

[SI unit = joule (J) and dimension = ML2T2 ]

  • Kinetic energy

(KE) = 1/2mv2=P2/2m

  • Potential Energy

UG=mgh

USPRING = 1/2kx2

  • Work-Energy Theorem

W = Change in K. E. of a body =Δ KE

  • Power

P= w/t

Tips to score good marks in class 11 Physics Chapter 6 NCERT solutions

To score good marks in Physics Class 11 Chapter 6, you can follow these steps using NCERT Solutions:

  • Understand the Concepts: Read the chapter work power energy class 11 thoroughly to understand the concepts and principles involved in it. Try to understand the definitions, laws, and equations used in the chapter.

  • NCERT solution: Use chapter 6 physics class 11 ncert solutions provided on our careers360 website. These solutions provide detailed explanations and solutions to all the problems given in the textbook.

  • Practice Problems: Practice the problems provided in the NCERT Solutions chapter-wise and try to solve them on your own. This will help you understand the concepts better and improve your problem-solving skills.

  • Revise Regularly: Revise the concepts regularly to ensure that you have a strong foundation in the subject.

  • Solve Sample Papers and Previous Years' Question Papers: Practice sample papers and previous years' question and NCERT Exemplar Class 11 Physics Solutions.

By following these steps, you can use the class 11th physics chapter 6 ncert solutions to score good marks in your exams.

Some of The Main Points on Work Energy and Power Class 11 Topic-Collisions

  • For an elastic collision, Law of conservation of momentum and that of Kinetic Energy holds good
  • For inelastic collision Law of conservation of momentum hold good but kinetic energy is not conserved
  • Coefficient of restitution is the ratio of relative velocity after the collision to relative velocity before the collision
  • For a perfectly elastic collision coefficient of restitution is one
  • For inelastic collision coefficient of restitution is less than one
  • For a perfectly inelastic collision coefficient of restitution is zero

Key Features of NCERT Class 11 physics chapter 6 exercise solutions

  1. Expert-Crafted Solutions: The work energy and power class 11 numericals with solutions have been meticulously prepared by subject matter experts, ensuring a comprehensive understanding of each question. The use of simple language makes it easier for students to follow along.

  2. Thorough Explanation: All thirty questions in the exercise are explained in detail, breaking down each step to clarify the reasoning behind the solution approach. This helps students grasp not only the answer but also the underlying concepts.

  3. Additional Exercise Questions: In addition to the main exercise questions, there are extra practice questions included. These extra questions are designed to enhance problem-solving skills and reinforce the chapter's concepts.

  4. Comprehensive Coverage: The class 11th physics chapter 6 ncert solutions cover all aspects of the chapter, including key concepts, formulas, and diagrams. This ensures a well-rounded understanding of the topics discussed in Chapter 6.

  5. Easy Accessibility: Students can access these solutions conveniently, enabling them to study and revise the chapter effectively, ultimately leading to a stronger grasp of the subject matter.

Importance of NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power

  • Many questions were answered in the solutions of NCERT Class 11 Physics chapter 6 Work Energy and Power based on the concept of the work-energy theorem which is important throughout the physics course. The concept will be used in class 12 also.
  • NCERT solutions for class 11 will help to score well in-class exams and also competitive exams like NEET.
  • Learning NCERT solutions will help to solve problems from other reference books also.

NCERT Physics Exemplar Solutions Class 11 For All The Chapters:

Subject wise NCERT Exemplar solutions

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT solutions

Frequently Asked Questions (FAQs)

1. What is the weightage of Class 11 chapter Work Energy and Power for NEET exam

On an average 4% questions are asked for NEET from Work Energy and Power. To prepare for NEET exam NCERT book is the one of the preffered materials. Along with NCERT exercise questions students can also practice NCERT exemplar and NEET previous year papers.

2. How many questions comes from Work Energy and Power for JEE Main

One or two question can be expected from the chapter Work Energy and Power for JEE Main based on the previous year paper analysis. But the concepts studied in Work Energy and Power Class 11 chapter will be used in other chapters of Class 11 and Class 12 Physics.

3. What is the difference between Work Energy and Power studied in Class 9 and Class 11 NCERT

In Class 11 more analysis is done using Mathematics (vector and calculus)  and some more concepts are added. The concepts of collision is not explained in the Class 9 NCERT Science book.

4. How to work energy and power class 11 ncert pdf

To download the PDF version of work energy and power class 11 solutions click on NCERT Solutions for Class 11 Physics and select the chapter according to your comfort and click on the download PDF.

5. What are collisions covered in class 11 physics work energy and power?

A collision is an event that happens when two objects come into direct contact with each other and exert forces on each other for a relatively short period of time. This can involve two or more bodies interacting with one another, resulting in changes in their motion and momentum.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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