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

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.

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NCERT Solutions for Class 11 Physics Chapter 6 – Download Free PDF
NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power - Exercise Questions
NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power Additional Exercise
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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(a) 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 a man in lifting a bucket out of a well by means of a rope tied to the bucket.

Answer:

In this case, the direction of force and displacement are the same, this work done is positive.

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.

Q 1 (d) 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 an applied force on a body moving on a rough horizontal plane with uniform velocity

Answer:

In this case, the applied force supports the motion of the object (balances frictional force). Thus work done by the force is positive.

Q1 (e) 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 the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer:

Work done in this case is negative as the direction of force and motion are not identical.

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 (b) 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 friction in 10 s,

Answer:

The work done by frictional force will be negative as the force opposes the motion.

$W\ =\ f.s\ =\ -1.96\times 126\ =\ -247\ J$

Q2 (c) 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 net force on the body in 10 s,

Answer:

The net work done will be the sum of work done by applied force and work done by frictional force.

$W\ =\ 882\ -\ 247\ =\ 635\ J$

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

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.

Q5 (d) In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 6.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

1645180997126

Answer:

In the first case,

Work done is :

$W\ =\ F.s\ =\ Fs \cos \Theta$

or $=\ mgs \cos \Theta$

or $=\ 15\times 9.8\times2\times \cos 90^{\circ}$

or $=\ 0$

In the second case :

$W\ =\ Fs \cos \Theta$

or $=\ mgs \cos 0^{\circ}$

or $=\ 15\times 9.8\times 2\ =\ 294\ J$

Thus work done in the second case is greater than the first case.

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 :

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

Q11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by $F = - \hat i + 2 \hat j + 3 \hat k N$
where $\hat i , \hat j , \hat k$ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?

Answer:

Force is given to be :

$F = - \hat i + 2 \hat j + 3 \hat k N$

And the displacement is :

$s\ =\ 4 \hat k\ m$

Thus the work done is given by :

$W\ =\ F.s$

or $=\left ( - \hat i + 2 \hat j + 3 \hat k \right ). \left ( 4 \hat k \right )$

or $=0 +0 \ -\ 3(4)\ =\ \ 12\ J$

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 :

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

Q13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is $10 m s ^{-1}$ ?

Answer:

The volume of the drop is :

$V\ =\ \frac{4}{3}\pi r^3\ =\ \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3$

Thus the mass of raindrop is :

$m\ =\ \rho v$

or $=\10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3\ Kg$

Thus the work done is given by :

$W\ =\ F.s$

or $=\ mgs$

or $=\10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3 \times 9.8\ \times 250$

or $=\ 0.082\ J$

Now the total energy at the peak point is :

$E_p\ =\ mgh\ +\ 0\ =\ mgh$

or $=\10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3\ \times 9.8 \times 500$

or $=\ 0.146\ J$

And the energy at the ground is :

$E_b\ =\ 0\ +\ \frac{1}{2}mv^2\ =\ \frac{1}{2}mv^2$

or $=\ \frac{1}{2}\times 10^3\times \frac{4}{3}\times 3.14\times (2\times 10^{-3})^3\ \times (10)^2$

or $=\ 1.67\times 10^{-3} \ J$

Thus work done by the resistive force is :

$=\ 1.67\times 10^{-3} \ J\ -\ 0.164\ J\ =\ -\ 0.162\ J$

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):-

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

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

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 : - , $\rho$ is the density of the air.

Hence mass of wind flowing through windmill in time t is _.

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 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 $=\ 4.5\ KW$

Q22 (a) A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. How much work does she do against the gravitational force?

Answer:

The work done against the gravitational force is given by :

= Number of times the weight is lifted $\times$ work done in 1 time.

$=\ 1000\times mgh$

or $=\ 1000\times 10\times 9.8\times 0.5$

or $=\ 49\ KJ$

Q22 (b) A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. Fat supplies $3.8 \times 10 ^ 2 J$ of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Answer:

Efficiency is given to be 20 per cent.

Thus energy supplied by the person :

$=\ \frac{20}{100}\times 3.8\times 10^7$

Thus the amount of fat lost is :

$=\ \frac{49\times 10^3}{\frac{20}{100}\times 3.8\times 10^7}$

or $=\ 6.45\times 10^{-3}\ Kg$

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

Q26 A 1 kg block situated on a rough incline is connected to a spring of spring constant $100 N m^{-1}$ as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Answer:

Displacement (x) of the block is given as : = 0.1 m.

Using equilibrium conditions we can write :

$R\ =\ mg \cos 37^{\circ}$

and $\mu R\ =\ mg \sin 37^{\circ}$ ( is the frictional force).

We can write work done in terms of potential energy as :

$mg\ \left ( \sin 37^{\circ}\ -\ \mu \cos 37^{\circ} \right )x\ =\ \frac{1}{2}kx^2$

or $1\times g\ \left ( \sin 37^{\circ}\ -\ \mu \cos 37^{\circ} \right )x\ =\ \frac{1}{2}100\times (0.1)^2$

or $\mu\ =\ 0.125$ .

Thus the coefficient of friction is 0.125.

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.

NCERT solutions for class 11 physics chapter wise

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

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

U_G=mgh

U_SPRING = 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

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.
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.
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.
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.
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:

Chapter 1 Physical World

Chapter 2 Units and Measurements

Chapter 3 Motion in a Straight Line

Chapter 4 Motion in a Plane

Chapter 5 Laws of Motion

Chapter 6 Work, Energy, and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solids

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

Subject wise NCERT Exemplar solutions

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT solutions

Frequently Asked Question (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.

Also Read

NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 11 - Limits and Derivatives

NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 11 - Conic Section

NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Sequences and Series

NCERT Book for class 11 Biology 2023 - Download PDF

NCERT Books for Class 11 - Download All Subjects PDF here

NCERT Books for class 11 Hindi 2023 - Download PDF Here

NCERT Books for class 11 Chemistry 2023 - Download Pdf here

NCERT Books for class 11 Maths 2023 - Download Pdf

NCERT Books for Class 11 Physics 2023 (Topic wise) - Download PDF

NCERT Syllabus for Class 11 Maths 2023

NCERT Syllabus For Class 11 Chemistry 2023 - Download PDF here

NCERT Syllabus for Class 11 2023 (All Subjects) - Download Pdf Here

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NCERT Syllabus for Class 11 English 2023 - Download PDF here

NCERT Syllabus for Class 11 Biology 2023 - Download Unit Wise Syllabus

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NCERT Class 11th Notes Subject wise - Download Free PDF

Environmental Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 14 Notes - Download PDF

Hydrocarbons Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 13 Notes - Download PDF

Organic Chemistry- Some Basic Principles and Techniques Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 12 Notes - Download PDF

The P-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 11 Notes - Download PDF

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

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NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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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

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

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

Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

Also Read: Career as Nurse

3 Jobs Available

Maxillofacial Surgeon

A Maxillofacial Surgeon is a medical professional who performs facial surgeries that include tooth implant, neck, head or other surgeries such as removal of tumours, cosmetic surgeries and treatment of injuries on the face.

2 Jobs Available

Surgical Assistant

Surgical assistants are professionals in the service of saving others’ lives. They perform various medical procedures. In a career as a surgical assistant, one works in a team and contributes to the success of operations. Surgical assistants learn new procedures and update their knowledge of new medical technology and equipment. Surgical assistants clean and sterilize the tools used in surgery. In a career as a surgical assistant, individuals perform all the basic duties that allow surgeons to keep their focus on essential technical functions.

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

Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

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

Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

2 Jobs Available

Videographer

Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production.

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

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

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

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

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 cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

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

Fashion Journalist

Fashion journalism involves performing research and writing about the most recent fashion trends. Journalists obtain this knowledge by collaborating with stylists, conducting interviews with fashion designers, and attending fashion shows, photoshoots, and conferences. A fashion Journalist job is to write copy for trade and advertisement journals, fashion magazines, newspapers, and online fashion forums about style and fashion.

2 Jobs Available

Multimedia Specialist

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

Product Manager

3 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

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

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3 Jobs Available

QA Manager

Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.

2 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

Reliability Engineer

Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.

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

Corporate Executive

2 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

3 Jobs Available

ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

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

Big Data Analytics Engineer

Big Data Analytics Engineer Job Description: A Big Data Analytics Engineer is responsible for collecting data from various sources. He or she has to sort the organised and chaotic data to find out patterns. The role of Big Data Engineer involves converting messy information into useful data that is clean, accurate and actionable.

2 Jobs Available

Cloud Solution Developer

A Cloud Solutions Developer is basically a Software Engineer with specialisation in cloud computing. He or she possesses a solid understanding of cloud systems including their operations, deployment with security and efficiency with no little downtime.

2 Jobs Available

CRM Technology Consultant

A Customer Relationship Management Technology Consultant or CRM Technology Consultant is responsible for monitoring and providing strategy for performance improvement with logged calls, performance metrics and revenue metrics. His or her role involves accessing data for team meetings, goal setting analytics as well as reporting to executives.

2 Jobs Available

IT Manager

Career as IT Manager requires managing the various aspects of an organization's information technology systems. He or she is responsible for increasing productivity and solving problems related to software and hardware. While this role is typically one of the lower-level positions within an organisation, it comes with responsibilities related to people and ownership of systems.

2 Jobs Available

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

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

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

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

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

NCERT solutions for class 11 physics chapter wise

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

Work done

Kinetic energy

Potential Energy

Work-Energy Theorem

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

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

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