NCERT Exemplar Solutions Class 11 Physics Work, Energy and Power — Multiple Choice Questions Q1–21

NCERT Exemplar Problems, Class 11 Physics — Work, Energy and Power, MCQ I and MCQ II. All 21 questions solved.

The book numbers these questions 6.1 to 6.21; here they are Questions 1 to 21. Questions 1 to 18 are MCQ I (exactly one option is correct) and Questions 19 to 21 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Work needs displacement of the point where the force acts. A force perpendicular to the velocity (a magnetic force, a centripetal pull) or a force whose point of application does not move (the road under a skidding tyre, a stair under a foot) does no work.
• The work-energy theorem. The work done by the net force equals the change in kinetic energy, $W = \Delta K$, whatever the forces are.
• Mechanical energy is conserved only when every force that does work is conservative. Air drag, water drag and sliding friction turn mechanical energy into heat.
• Momentum survives every collision; kinetic energy survives only elastic ones. In an isolated system the total linear momentum is always conserved.

Key insight. Before calculating, ask which forces actually do work and whether they are conservative. That single question settles Questions 1, 4, 5, 6, 13, 15, 17 and 19 without any arithmetic.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,

Solution. The magnetic force on a moving charge is $q\,\mathbf{v}\times\mathbf{B}$, which is always perpendicular to the particle’s own velocity. Its power $\mathbf{F}\cdot\mathbf{v}$ is therefore zero at every instant, so it does no work on either particle and cannot change either kinetic energy. Option (a) is wrong because magnetic forces between two moving charges are not in general equal and opposite, and (d) is wrong because nothing makes them small; they simply do no work.

Question 2

A proton is kept at rest. A positively charged particle is released from rest at a distance $d$ in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time $t$, the work done on the two moving charged particles is

Solution. Both particles start at the same distance $d$ and feel the same initial repulsion $ke^2/d^2$. The positron is about $1836$ times lighter, so its acceleration is far larger and in the same time $t$ it moves much further from the fixed proton. The work done by the repulsion equals the drop in potential energy, $ke^2\left(\dfrac{1}{d} – \dfrac{1}{r}\right)$, which grows with the final separation $r$. The positron reaches a larger $r$, so more work is done on it. Option (d) is wrong because the stationary proton does not move, so no work is done on it at all.

Question 3

A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is

Solution. The man’s centre of mass starts at rest and ends at rest. To start it moving upward the net force must point up, so the reaction $N$ must exceed $mg$ at first. Once he stands still, $N = mg$. So $N$ varies, which rules out (a) and (b), and it cannot stay above $mg$ throughout, because then he would still be accelerating upward when he stood, which rules out (c).

Strictly, the centre of mass must also be brought to rest at the end of the rise, and in that short decelerating stage $N$ dips a little below $mg$ before settling. Of the four options, (d) is the one that describes the motion correctly, and it is the printed answer.

Question 4

A bicyclist comes to a skidding stop in $10$ m. During this process, the force on the bicycle due to the road is $200$ N and is directly opposed to the motion. The work done by the cycle on the road is

Solution. By Newton’s third law the cycle pushes the road with $200$ N. But work needs the point of application to move along the force, and the road does not move at all. The work done by the cycle on the road is therefore zero. The work done by the road on the cycle is a different quantity: $-200 \times 10 = -2000$ J, which is where the cycle’s kinetic energy went.

Question 5

A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?

Solution. Gravity is the only force, and it is conservative, so $K + U$ stays fixed while kinetic energy rises and potential energy falls by the same amount. The momentum of the body grows as $mgt$, because gravity is an external force on it, so (d) is false.

Question 6

During inelastic collision between two bodies, which of the following quantities always remain conserved?

Solution. In a collision the two bodies push on each other with equal and opposite internal forces, so the total momentum is unchanged whatever kind of collision it is. Inelastic means some kinetic energy is turned into heat, sound or permanent deformation, so neither the kinetic nor the mechanical energy is conserved, and the individual speeds certainly change.

Question 7

Two inclined frictionless tracks, one gradual and the other steep, meet at A from where two stones are allowed to slide down from rest, one on each track as shown in Fig. 6.1. Which of the following statement is correct?

The figure shows a triangle with apex A at height $h$ above the base BC. Track I is the long gentle slope AB at angle $\theta_1$ to the horizontal; track II is the short steep slope AC at angle $\theta_2 > \theta_1$.

Solution. The tracks are frictionless, so each stone loses potential energy $mgh$ and gains the same kinetic energy: both arrive with speed $\sqrt{2gh}$. The times differ. Along a slope at angle $\theta$, the acceleration is $g\sin\theta$ and the length is $h/\sin\theta$, so

$$t = \sqrt{\frac{2h/\sin\theta}{g\sin\theta}} = \frac{1}{\sin\theta}\sqrt{\frac{2h}{g}}.$$

The steeper track has the larger $\sin\theta$, so stone II arrives first.

Question 8

The potential energy function for a particle executing linear SHM is given by $V(x) = \frac{1}{2}kx^2$ where $k$ is the force constant of the oscillator (Fig. 6.2). For $k = 0.5$ N/m, the graph of $V(x)$ versus $x$ is shown in the figure. A particle of total energy $E$ turns back when it reaches $x = \pm x_m$. If $V$ and $K$ indicate the P.E. and K.E., respectively of the particle at $x = +x_m$, then which of the following is correct?

Solution. The particle turns back at $x_m$, so its velocity there is zero and $K = 0$. Since $K + V = E$ at every point, $V = E$ at the turning point. That is exactly what the dotted line in the figure shows: the level $E$ meets the parabola at $\pm x_m$.

Question 9

Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed $V$ as shown in Fig. 6.3. If the collision is elastic, which of the following (Fig. 6.4) is a possible result after collision?

Fig. 6.3 shows ball 1 moving at speed $V$ towards balls 2 and 3, which touch each other. The four outcomes in Fig. 6.4 are: (a) ball 1 at rest, balls 2 and 3 moving together at $V/2$; (b) balls 1 and 2 at rest, ball 3 moving at $V$; (c) all three moving together at $V/3$; (d) balls 1, 2 and 3 moving at $V/1$, $V/2$ and $V/3$.

Solution. Test each outcome against both conservation laws. Initially $p = mV$ and $K = \tfrac12 mV^2$.

• (a) $p = 2m(V/2) = mV$, but $K = \tfrac12(2m)(V/2)^2 = \tfrac14 mV^2$. Energy is lost, so not elastic.
• (b) $p = mV$ and $K = \tfrac12 mV^2$. Both conserved.
• (c) $p = 3m(V/3) = mV$, but $K = \tfrac16 mV^2$. Not elastic.
• (d) $p = m(V + V/2 + V/3) = \tfrac{11}{6}mV$. Momentum is not even conserved.

Only (b) survives. Physically, each equal-mass elastic collision hands the whole velocity on, so ball 1 stops, ball 2 passes the velocity to ball 3, and ball 3 leaves at $V$.

Question 10

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

Solution. By the work-energy theorem, the work done by the net force is the change in kinetic energy, so there is no need to find the force. At $x = 0$, $v = 0$. At $x = 2$ m,

$$v^2 = a^2x^3 = 25 \times 8 = 200\ \text{m}^2\text{s}^{-2}, \qquad W = \tfrac12 m v^2 – 0 = \tfrac12 \times 0.5 \times 200 = 50\ \text{J}.$$

Question 11

A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in Fig. 6.5 correctly shows the displacement-time curve for its motion?

The four displacement-time graphs are: (a) a straight line through the origin; (b) a curve through the origin that bends upward ever more steeply; (c) a horizontal line; (d) a curve through the origin that rises steeply and then flattens.

Solution. Constant power means the kinetic energy grows at a steady rate: $\tfrac12 mv^2 = Pt$, starting from rest. So $v = \sqrt{2Pt/m} \propto t^{1/2}$, and integrating,

$$d = \int_0^t \sqrt{\frac{2P}{m}}\,t^{1/2}\,dt = \frac{2}{3}\sqrt{\frac{2P}{m}}\;t^{3/2}.$$

Displacement grows as $t^{3/2}$: it starts flat at the origin and curves upward with increasing slope, which is graph (b). The speed keeps rising, so the curve never flattens as in (d).

Question 12

Which of the diagrams shown in Fig. 6.6 most closely shows the variation in kinetic energy of the earth as it moves once around the sun in its elliptical orbit?

The four graphs are: (a) a constant K.E.; (b) a single hump that starts and ends at zero; (c) a sine curve that goes negative; (d) a curve that rises and falls gently about a positive mean value, never reaching zero.

Solution. In an elliptical orbit the earth moves fastest at perihelion and slowest at aphelion, so its kinetic energy varies, ruling out (a). But it never stops, so K.E. never reaches zero, ruling out (b), and kinetic energy can never be negative, ruling out (c). The earth’s orbit is nearly circular, so the variation is small about a positive mean value: graph (d).

Question 13

Which of the diagrams shown in Fig. 6.7 represents variation of total mechanical energy of a pendulum oscillating in air as function of time?

The four graphs are: (a) an oscillation of decreasing amplitude about $E = 0$; (b) a curve that stays nearly flat and then drops sharply to zero; (c) a curve that falls steeply at first and then ever more slowly towards zero; (d) an oscillation of constant amplitude about $E = 0$.

Solution. The total mechanical energy $K + U$ of a pendulum does not oscillate; the kinetic and potential parts swap back and forth, but their sum changes only through the work done by air drag. Drag always opposes motion, so it can only take energy away: $E$ falls all the time and never rises. The drag force grows with speed, so the rate of loss is largest when the swings are large and shrinks as the pendulum dies down. That gives a curve that drops quickly at first and then ever more slowly towards zero, which is graph (c). Graphs (a) and (d) let the energy go negative and rise again, which is impossible for a damped pendulum, and (b) would need the losses to grow as the motion dies away.

Printed key error. The key prints (d). Graph (d) is an undamped oscillation of constant amplitude about zero: it makes the energy rise, fall and go negative, which no pendulum’s total mechanical energy can do, in air or in vacuum (in vacuum $E$ would be a flat line). The correct answer is (c).

Question 14

A mass of $5$ kg is moving along a circular path of radius $1$ m. If the mass moves with $300$ revolutions per minute, its kinetic energy would be

Solution. $300$ revolutions per minute is $5$ revolutions per second, so $\omega = 2\pi \times 5 = 10\pi$ rad/s and $v = \omega r = 10\pi$ m/s. Then

$$K = \tfrac12 mv^2 = \tfrac12 \times 5 \times (10\pi)^2 = 250\pi^2\ \text{J}.$$

Question 15

A raindrop falling from a height $h$ above ground, attains a near terminal velocity when it has fallen through a height $(3/4)h$. Which of the diagrams shown in Fig. 6.8 correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground?

The four graphs plot energy against time. (a) P.E. rises as a straight line while K.E. falls from a high value. (b) P.E. falls from its starting value $h$ along a smooth curve to zero, while K.E. rises from zero and levels off at a steady value, marked with a dotted line at the level $h/4$. (c) P.E. rises and K.E. falls, both as straight lines. (d) K.E. and P.E. are two crossing arcs, the P.E. arc going up and coming back down.

Solution. The drop falls, so its potential energy must decrease all the way to zero at the ground. That alone rules out (a) and (c), where P.E. increases, and (d), where it rises before falling. In (b) the kinetic energy grows from zero while the drop accelerates and then levels off, because at terminal velocity the speed, and so the K.E., stops changing while the P.E. keeps falling. The energy lost after that goes into work against air drag. Graph (b) is the only one with both features.

Question 16

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

Solution. Air resistance is neglected, so mechanical energy is conserved and the angle of projection does not matter:

$$K_{\text{ground}} = \tfrac12 mu^2 + mgh = \tfrac12 \times 10 \times 1^2 + 10 \times 10 \times 1.5 = 5 + 150 = 155\ \text{J}.$$

Question 17

Which of the diagrams in Fig. 6.9 correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity?

The four graphs plot K.E. against depth: (a) a straight line through the origin; (b) a curve rising from the origin and levelling off to a constant value; (c) an S-shaped curve that starts flat, rises steeply and then levels off; (d) a curve that rises ever more steeply.

Solution. The slope of K.E. against depth is the net force on the sphere, $dK/dx = F_{\text{net}}$. At the start the sphere is at rest, so there is no drag and the net force (weight minus buoyancy) is at its largest: the curve leaves the origin with its steepest slope. As the sphere speeds up, the drag grows and the net force falls, so the slope keeps decreasing. At terminal velocity the net force is zero and the K.E. stays constant. A curve that starts steep and flattens out is graph (b). Graph (c) starts flat, which would need zero net force at the start.

Question 18

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

Solution. $126$ km/h $= 126 \times \dfrac{5}{18} = 35$ m/s. The bat is held fixed and the collision is elastic, so the ball rebounds with the same speed. The change in its momentum is

$$\Delta p = m(v – (-v)) = 2mv = 2 \times 0.15 \times 35 = 10.5\ \text{kg m s}^{-1}.$$

The bat exerts this impulse in $0.001$ s, so the average force is $10.5/0.001 = 1.05 \times 10^4$ N, and the batsman must supply an equal force to hold the bat still.

Question 19

A man, of mass $m$, standing at the bottom of the staircase, of height $L$ climbs it and stands at its top.

Solution. The man starts at rest and ends at rest, so his kinetic energy does not change. By the work-energy theorem the total work done by all forces on him is zero: (b) is true and (a) is false.

Gravity acts downward while he rises $L$, so it does $-mgL$ of work, not $+mgL$: (c) is false.

The reaction from a step acts on his foot only while the foot is planted, and a planted foot does not move. No displacement of the point of application means no work: (d) is true. The energy that lifts him comes from his own muscles, an internal source, which is why the external forces can total zero work while his potential energy rises.

Question 20

A bullet of mass $m$ fired at $30^\circ$ to the horizontal leaves the barrel of the gun with a velocity $v$. The bullet hits a soft target at a height $h$ above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target?

Solution. Kinetic energy goes as $v^2$, so halving it multiplies the speed by $1/\sqrt2 \approx 0.71$, which is more than half: (b) is true and (a) is false.

The bullet emerges still moving, with a horizontal velocity component, so it does not fall vertically: (e) is false. But it now moves more slowly from the same point, and a projectile’s path is fixed by its position and velocity, so it follows a new, tighter parabola: (d) is true and (c) is false.

The lost half of the kinetic energy has gone into the target, as heat and deformation, so the internal energy of the target’s particles rises: (f) is true.

Question 21

Two blocks $M_1$ and $M_2$ having equal mass are free to move on a horizontal frictionless surface. $M_2$ is attached to a massless spring as shown in Fig. 6.10. Initially $M_2$ is at rest and $M_1$ is moving toward $M_2$ with speed $v$ and collides head-on with $M_2$.

The figure shows block $M_1 = m$ moving right with speed $v$ towards block $M_2 = m$, which carries a light spring on its left face.

Solution. At full compression the two blocks move with the same velocity. Momentum is conserved (no external horizontal force), so that common velocity is $v/2$, and the kinetic energy then is $2 \times \tfrac12 m(v/2)^2 = \tfrac14 mv^2$, half the original. Only the other half, $\tfrac14 mv^2$, is in the spring: (a) is false. Momentum is conserved at every instant, not just at the end: (b) is false.

A massless spring stores and returns all its energy, so the collision is elastic. For equal masses in a head-on elastic collision the velocities are exchanged: $M_1$ ends at rest and $M_2$ moves off with $v$. So (c) is true.

For (d): whether a collision is elastic is a property of the interaction between the two bodies, here the spring, which is conservative. Friction from the surface is an external force acting on each block whether or not they collide; it slows the blocks down, just as an external force spoils momentum conservation, but it does not turn the spring’s interaction into an inelastic one. So (d) does not follow. The printed key gives (c) alone.

Common mistakes

• Calling the $-2000$ J done on the cycle “work done by the cycle on the road” in Question 4. The road never moves, so the work done on it is zero.
• Assuming the stones take equal times in Question 7. Equal speeds come from energy; the times depend on the slope, and the steeper one wins.
• Accepting a pair of moving balls at $V/2$ in Question 9. It conserves momentum but loses half the kinetic energy, so it cannot be elastic.
• Thinking the total mechanical energy of a pendulum oscillates in Question 13. $K$ and $U$ oscillate; their sum can only drain away.
• Using the angle of throw in Question 16. Without air resistance, energy conservation gives the landing K.E. regardless of direction.
• Forgetting the rebound in Question 18. The ball reverses, so the change of momentum is $2mv$, not $mv$.
• Taking work by all forces as $mgL$ in Question 19. The man starts and ends at rest; the total work on him is zero.
• Saying all the K.E. is in the spring at full compression in Question 21. The blocks still move together at $v/2$, carrying half of it.

Practise next

• A body of mass $2$ kg moves along a straight line with velocity $v = 4\sqrt{x}$ (SI units). Find the work done by the net force on it between $x = 1$ m and $x = 4$ m.

Show answer

$48$ J. $K = \tfrac12 mv^2 = \tfrac12 \times 2 \times 16x = 16x$, so $W = \Delta K = 16(4 – 1) = 48$ J by the work-energy theorem.

• A ball of mass $0.2$ kg moving at $20$ m/s hits a rigidly held bat and rebounds straight back with the same speed. The contact lasts $0.002$ s. Find the average force on the bat.

Show answer

$4000$ N. The momentum change is $2mv = 2 \times 0.2 \times 20 = 8$ kg m/s, and $8/0.002 = 4000$ N.

• Exemplar Work, Energy and Power: Very Short and Short Answer Questions — the same ideas applied to power, collisions and energy graphs.
• Exemplar Work, Energy and Power: Long Answer Questions — friction on an incline, rolling balls, the rocket and colliding steel cubes.
• NCERT textbook Chapter 5 — Work, Energy and Power for the theory of the work-energy theorem, potential energy, power and collisions.
• Exemplar Laws of Motion — the previous unit, whose forces and impulse ideas carry over.