NCERT Exemplar Solutions Class 11 Physics Systems of Particles and Rotational Motion — Multiple Choice Questions Q1–13

NCERT Exemplar Problems, Class 11 Physics — Systems of Particles and Rotational Motion, MCQ I and MCQ II. All 13 questions solved.

The book numbers these questions 7.1 to 7.13; here they are Questions 1 to 13. The exemplar calls this Unit 7, while the rationalised NCERT textbook covers the same material in Chapter 6. Questions 1 to 8 are MCQ I (exactly one option is correct) and Questions 9 to 13 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Centre of mass. $\vec R = \dfrac{\sum m_i\vec r_i}{\sum m_i}$, or $\dfrac{\int x\,dm}{\int dm}$ for a continuous body. It shifts towards wherever mass is added.
• Angular momentum. $\vec L = \vec r\times\vec p$ depends on the point it is taken about, and it is conserved when the external torque is zero.
• Torque about an axis. A force has no torque about an axis if it is parallel to the axis or its line of action meets the axis.
• Moment of inertia. $I = \sum m_i r_i^2$, so it depends on how far the mass sits from the axis, and the parallel-axis theorem $I = I_{\text{cm}} + Md^2$ holds only for parallel axes.

Key insight. Nearly every question here is a statement about where mass or a force sits relative to a point or an axis. Draw the position vector first; the cross product then decides the direction.

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

Question 1

For which of the following does the centre of mass lie outside the body?

Solution. A pencil, a shotput and a dice are solid and symmetric, so their centres of mass lie at their geometric centres, inside the material. A bangle is a ring: by symmetry its centre of mass is at the centre of the circle, and there is no material there. So the centre of mass of a bangle lies outside the body.

Question 2

Which of the following points is the likely position of the centre of mass of the system shown in Fig. 7.1? (A hollow sphere of radius $R$ is half filled with sand; the upper half contains air.)

Solution. The hollow shell is symmetric, so its own centre of mass is at the centre B. Air is negligibly light. The sand fills the lower hemisphere, whose centre of mass lies $\tfrac{3R}{8}$ below B. The centre of mass of the whole system lies on the line joining these two, between B and the sand’s centre of mass. It is therefore pulled below B, but not right down to the bottom of the sphere, since the shell’s mass keeps it closer to B. Point C, just below the centre, is the likely position. A is in the wrong half, and D lies even below the sand’s own centre of mass, which no mixture with the shell can reach.

Question 3

A particle of mass $m$ is moving in the $yz$-plane with a uniform velocity $v$ with its trajectory running parallel to the $+$ve $y$-axis and intersecting the $z$-axis at $z = a$ (Fig. 7.2). The change in its angular momentum about the origin as it bounces elastically from a wall at $y = \text{constant}$ is:

Solution. The position of the particle is $\vec r = y\,\hat{\mathbf e}_y + a\,\hat{\mathbf e}_z$. Before the bounce $\vec v_i = v\,\hat{\mathbf e}_y$; after an elastic bounce from a wall perpendicular to $y$ it is $\vec v_f = -v\,\hat{\mathbf e}_y$. At the instant of the bounce the position is the same, so

$$\Delta\vec L = \vec r\times m(\vec v_f – \vec v_i) = (y\,\hat{\mathbf e}_y + a\,\hat{\mathbf e}_z)\times(-2mv\,\hat{\mathbf e}_y) = -2mva\,(\hat{\mathbf e}_z\times\hat{\mathbf e}_y) = 2mva\,\hat{\mathbf e}_x.$$

The $y$ part of $\vec r$ is parallel to the change in momentum and contributes nothing, which is why $y$ does not appear in the answer.

Question 4

When a disc rotates with uniform angular velocity, which of the following is not true?

Solution. Uniform angular velocity means the vector $\vec\omega$ is constant: its magnitude (speed of rotation), its direction (orientation of the axis) and its sign (sense of rotation) all stay the same. So (a), (b) and (c) are true. A constant $\vec\omega$ has zero rate of change, so the angular acceleration is zero, not non-zero. Statement (d) is the false one.

Question 5

A uniform square plate has a small piece Q of an irregular shape removed and glued to the centre of the plate leaving a hole behind (Fig. 7.3). The moment of inertia about the $z$-axis is then

Solution. The $z$-axis passes through the centre of the plate, perpendicular to it. The piece Q contributed $m_Q r^2$ to the moment of inertia when it sat at a distance $r$ from the axis. Glued at the centre, its distance from the axis is (nearly) zero, so its contribution falls to almost nothing. The rest of the plate is unchanged, so the total moment of inertia decreases.

Question 6

In problem 7.5, the CM of the plate is now in the following quadrant of $x$-$y$ plane,

Solution. Before the change the centre of mass was at the centre O. Moving the piece Q from the first quadrant to O is the same as removing mass from the first quadrant and adding it at O. Mass added at O does not shift the centre of mass; mass taken out of the first quadrant shifts it the opposite way, towards the third quadrant. Equivalently, the new plate is “full plate plus Q at O minus Q in quadrant I”, and the negative mass in quadrant I pushes the centre of mass into quadrant III.

Question 7

The density of a non-uniform rod of length $1$ m is given by $\rho(x) = a(1 + bx^2)$ where $a$ and $b$ are constants and $0 \le x \le 1$. The centre of mass of the rod will be at

Solution. For a rod the centre of mass is

$$x_{\text{cm}} = \frac{\int_0^1 x\,\rho\,dx}{\int_0^1 \rho\,dx} = \frac{a\left(\tfrac12 + \tfrac b4\right)}{a\left(1 + \tfrac b3\right)} = \frac{\tfrac{2+b}{4}}{\tfrac{3+b}{3}} = \frac{3(2+b)}{4(3+b)}.$$

A quick check: when $b \to 0$ the rod is uniform and the expression gives $\tfrac{6}{12} = 0.5$ m, the middle. Only option (a) passes that test.

Question 8

A merry-go-round, made of a ring-like platform of radius $R$ and mass $M$, is revolving with angular speed $\omega$. A person of mass $M$ is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is

Solution. No external torque acts about the axis, so angular momentum is conserved. Before the jump, $L = (MR^2 + MR^2)\omega = 2MR^2\omega$. The person jumps radially as seen from the round, so the jump adds no tangential velocity: just after leaving, the person still has the tangential speed $R\omega$ of the rim, carrying angular momentum $MR\cdot R\omega = MR^2\omega$. The round keeps the rest:

$$2MR^2\omega = MR^2\omega’ + MR^2\omega \;\Rightarrow\; \omega’ = \omega.$$

A radial push gives no torque about the axis, so the round neither speeds up nor slows down.

Question 9

Choose the correct alternatives:

Solution.

• (a) True. $\mathbf L$ and $\boldsymbol\omega$ are related through the moment of inertia, which depends on direction; for an unsymmetrical body they point in different directions.
• (b) False. Even about a fixed axis, $\mathbf L$ is parallel to $\boldsymbol\omega$ only if the body is symmetric about that axis. For an unsymmetrical body only the component of $\mathbf L$ along the axis equals $I\omega$; the rest is perpendicular to the axis and is held by the bearings.
• (c) True. $\mathbf p = m\mathbf v$ with $m$ a positive scalar, so they always point the same way.
• (d) False. In uniform circular motion $\mathbf a$ is perpendicular to $\mathbf v$; in a projectile’s path it is at a changing angle.

Question 10

Figure 7.4 shows two identical particles 1 and 2, each of mass $m$, moving in opposite directions with same speed $v$ along parallel lines. At a particular instant, $\mathbf r_1$ and $\mathbf r_2$ are their respective position vectors drawn from point A which is in the plane of the parallel lines. Choose the correct options: ($\odot$ represents a unit vector coming out of the page; $\otimes$ represents a unit vector going into the page.)

Solution. The magnitude of $\vec r\times\vec p$ is $p$ times the perpendicular distance from A to the line of motion, which is $d_1$ for particle 1 and $d_2$ for particle 2; the slanting lengths $r_1$, $r_2$ do not enter.

Take $x$ to the right, $y$ up and A as origin. Particle 1 is at $(x_1, -d_1)$ with momentum $(mv, 0)$, so $l_{1z} = x_1\cdot 0 – (-d_1)(mv) = +mvd_1$: out of the page. Particle 2 is at $(x_2, -d_2)$ with momentum $(-mv, 0)$, so $l_{2z} = -(-d_2)(-mv) = -mvd_2$: into the page.

• (a) True: $\mathbf l_1 = mvd_1\,\odot$. (The printed option has lost the $\odot$ symbol; the key reads it as intended.)
• (b) False: the magnitude is $mvd_2$, not $mv r_2$, and it points into the page.
• (c) False: angular momentum is not $mv$ times a sum of position vectors.
• (d) True: $\mathbf l = mvd_1\,\odot + mvd_2\,\otimes = mv(d_2 – d_1)\,\otimes$, since $d_2 > d_1$.

Question 11

The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it?

Solution. The torque of a force about an axis is the component along the axis of $\vec r\times\vec F$, where $\vec r$ is drawn from a point on the axis.

• (a) A force directed radially from a point on the axis has its line of action through the axis, so $\vec r\parallel\vec F$ and the torque is zero.
• (b) A force acting at a point on the axis has $\vec r = 0$ (taking that point), so zero torque.
• (c) A force parallel to the axis gives $\vec r\times\vec F$ perpendicular to the axis, so its component along the axis is zero.
• (d) Torques add as vectors; equal and opposite ones cancel.

Each of these makes the net torque about the axis zero.

Question 12

Figure 7.5 shows a lamina in $x$-$y$ plane. Two axes $z$ and $z’$ pass perpendicular to its plane. A force $\mathbf F$ acts in the plane of lamina at point P as shown. Which of the following are true? (The point P is closer to $z’$-axis than the $z$-axis.)

Solution. Take $\hat{\mathbf k}$ along the axes (upwards in the figure). P lies to the right of the $z$-axis and to the left of the $z’$-axis, and $\mathbf F$ points away from the viewer across the lamina.

• (a) From the $z$-axis, the position vector of P points to the right and $\mathbf F$ points away; seen from above, this turns the lamina anticlockwise, so $\boldsymbol\tau$ is along $+\hat{\mathbf k}$. False.
• (b) From the $z’$-axis, the position vector of P points to the left; with the same $\mathbf F$ the turning is clockwise seen from above, so $\boldsymbol\tau’$ is along $-\hat{\mathbf k}$. True.
• (c) The magnitude is $F$ times the perpendicular distance from the axis to the line of $\mathbf F$. P is farther from $z$ than from $z’$, so the torque about $z$ is the larger. True. (The printed option says “than that about z axis”; the comparison intended is with $z’$.)
• (d) False. Torques about two different axes describe two different turning effects; adding them has no meaning.

Question 13

With reference to Fig. 7.6 of a cube of edge $a$ and mass $m$, state whether the following are true or false. (O is the centre of the cube.)

Solution. About any axis through its centre parallel to an edge, a cube has $I_x = I_y = I_z = \tfrac16 ma^2$.

• (a) False. The perpendicular-axis theorem $I_z = I_x + I_y$ holds only for a plane lamina. For the cube it would give $\tfrac13 ma^2 \ne \tfrac16 ma^2$.
• (b) True. The $z’$-axis runs along the edge AD, parallel to $z$, at distance $\sqrt{(a/2)^2 + (a/2)^2} = \tfrac{a}{\sqrt2}$ from it. The parallel-axis theorem gives $I’_z = I_z + m\left(\tfrac{a}{\sqrt2}\right)^2 = I_z + \tfrac{ma^2}{2}$.
• (c) False. The $z”$-axis runs along the face diagonal BG, which is not parallel to $z$, so the parallel-axis theorem cannot be used with $I_z$ as written. (Integrating directly, $I_{z”} = \tfrac16 ma^2 + m\left(\tfrac a2\right)^2 = I_z + \tfrac{ma^2}{4}$, because BG passes through the centre of a face, $\tfrac a2$ from O, and the cube’s moment of inertia is the same about every axis through O.)
• (d) True, by the symmetry of the cube.

Common mistakes

• Putting the centre of mass inside the body by habit in Question 1. A ring’s centre of mass is at an empty point.
• Choosing D in Question 2. The shell’s mass, centred at B, holds the combined centre of mass well above the sand’s own centre.
• Writing the answer to Question 3 with $y$ in it. The $y$ component of $\vec r$ is parallel to $\Delta\vec p$ and drops out of the cross product.
• Reading the quadrant of Q off the wrong side in Question 6. Q starts in quadrant I; the centre of mass moves to the opposite quadrant, III.
• Thinking the round speeds up in Question 8. A radial jump carries away exactly the angular momentum the person already had.
• Using $r_1$ and $r_2$ instead of $d_1$ and $d_2$ in Question 10. Only the perpendicular distance to the line of motion enters $|\vec r\times\vec p|$.
• Applying $I_z = I_x + I_y$ to a cube in Question 13. That theorem is for flat laminas only; and the parallel-axis theorem needs genuinely parallel axes.

Practise next

• Linearly varying density. A rod of length $L$ has linear density $\lambda(x) = \lambda_0\left(1 + \tfrac{x}{L}\right)$ for $0 \le x \le L$. Where is its centre of mass?

• Walking to the centre. A child of mass $50$ kg stands at the rim of a uniform disc-shaped platform of mass $200$ kg turning freely at $1$ rad/s. The child walks to the centre. What is the new angular speed?

• Exemplar Systems of Particles and Rotational Motion: Very Short, Short and Long Answer Questions — Questions 14 to 28, with rolling, discs in contact and centres of mass of half- and quarter-discs.
• NCERT textbook Chapter 6 — Systems of Particles and Rotational Motion for the theory of centre of mass, torque, angular momentum and moment of inertia.
• Exemplar Work, Energy and Power — the previous unit.
• Exemplar Gravitation — the next unit, where angular momentum conservation explains Kepler’s second law.