NCERT Exemplar Solutions Class 11 Physics Motion in a Plane — Multiple Choice Questions Q1–15

NCERT Exemplar Problems, Class 11 Physics — Motion in a Plane, MCQ I and MCQ II. All 15 questions solved.

The book numbers these questions 4.1 to 4.15; here they are Questions 1 to 15. Questions 1 to 10 are MCQ I (exactly one option is correct) and Questions 11 to 15 are MCQ II (one or more options may be correct). The exemplar calls this Unit 4; in the rationalised textbook it is Chapter 3. Four ideas carry the whole page:

• Vectors by components. Write each vector as $a\hat{\mathbf{i}}+b\hat{\mathbf{j}}$ and most questions about angles and signs become arithmetic.
• Dot and cross products. $\mathbf{A}\cdot\mathbf{B}=AB\cos\theta$ measures how far two vectors agree; $\mathbf{A}\times\mathbf{B}$ is perpendicular to both and has size $AB\sin\theta$.
• Projectiles. Horizontal and vertical motions are independent: $H=\dfrac{u^2\sin^2\theta}{2g}$, $T=\dfrac{2u\sin\theta}{g}$, $R=\dfrac{u^2\sin2\theta}{g}$.
• Constant speed is not constant velocity. When the speed is fixed the acceleration can only turn the velocity, so it is perpendicular to it.

Key insight. Before choosing an option, ask whether the statement must hold for every case the question allows. Questions 9 and 10 are both decided by a single counter-example, and that is exactly where the printed key goes wrong.

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

Question 1

The angle between $\mathbf{A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\mathbf{B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$ is

Solution. The dot product is $\mathbf{A}\cdot\mathbf{B}=(1)(1)+(1)(-1)=0$. Neither vector is zero (each has magnitude $\sqrt2$), so $\cos\theta=0$ and $\theta=90^\circ$. Geometrically, $\mathbf{A}$ points at $45^\circ$ above the $x$-axis and $\mathbf{B}$ at $45^\circ$ below it. An angle between two vectors is always taken between $0^\circ$ and $180^\circ$, so $-45^\circ$ is never a valid answer.

Question 2

Which one of the following statements is true?

Solution. A scalar is defined by how it behaves when the axes are rotated: it has a single value that does not depend on the orientation of the coordinate system. That is (d). The others confuse being a scalar with unrelated properties. Kinetic energy is a scalar but is not conserved in an inelastic collision, so (a) fails. Temperature in degrees Celsius and charge are scalars that can be negative, so (b) fails. Temperature varies from place to place in a room and is still a scalar, so (c) fails.

Question 3

Figure 4.1 shows the orientation of two vectors $\mathbf{u}$ and $\mathbf{v}$ in the $XY$ plane. If $\mathbf{u}=a\hat{\mathbf{i}}+b\hat{\mathbf{j}}$ and $\mathbf{v}=p\hat{\mathbf{i}}+q\hat{\mathbf{j}}$, which of the following is correct?

Solution. The sign of a component depends only on the direction of the vector, not on where it is drawn. $\mathbf{u}$ points up and to the right, so both $a>0$ and $b>0$. $\mathbf{v}$ is drawn from the head of $\mathbf{u}$, but what matters is that it points to the right (so $p>0$) and downward (so $q<0$). The downward part is much larger than the rightward part, which is why it is easy to misread $p$, but the arrow does lean to the right.

Question 4

The component of a vector $\mathbf{r}$ along $X$-axis will have maximum value if

Solution. The $x$-component is $r_x=r\cos\theta$, where $\theta$ is the angle with the positive $X$-axis. For a vector of fixed magnitude, $\cos\theta$ is largest, equal to $1$, at $\theta=0$. So $r_x=r$ is greatest when $\mathbf{r}$ lies along the positive $X$-axis. Along either $Y$ direction the $x$-component is zero, and at $45^\circ$ it is only $r/\sqrt2$.

Question 5

The horizontal range of a projectile fired at an angle of $15^\circ$ is $50$ m. If it is fired with the same speed at an angle of $45^\circ$, its range will be

Solution. With the same speed, $R=\dfrac{u^2\sin2\theta}{g}$ is proportional to $\sin2\theta$. At $15^\circ$, $\sin30^\circ=\tfrac12$, so $\dfrac{u^2}{g}=100$ m. At $45^\circ$, $\sin90^\circ=1$, so $$R_{45}=\frac{u^2}{g}=100\ \text{m}.$$ The tempting wrong step is to compare $\sin15^\circ$ with $\sin45^\circ$ instead of the doubled angles.

Question 6

Consider the quantities, pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are

Solution. Impulse is force multiplied by time (equal to the change in momentum), so it has the direction of the force: a vector. A plane area is treated as a vector whose direction is the normal to the surface; this is what lets us write flux as $\mathbf{E}\cdot\mathbf{A}$. Pressure is force per unit area but acts equally in all directions at a point in a fluid, so it is a scalar. Power, energy, gravitational potential, charge and temperature are all scalars.

Question 7

In a two dimensional motion, instantaneous speed $v_0$ is a positive constant. Then which of the following are necessarily true?

Solution. Constant speed fixes only how fast distance along the path is covered: in any interval $\Delta t$ the path length is $v_0\,\Delta t$. That is (d). It says nothing about direction. In uniform circular motion, after one full revolution the displacement is zero, so the average velocity is zero, ruling out (a). The velocity keeps turning, so the average acceleration over half a revolution is not zero, ruling out (b). Displacements over equal intervals have equal size on a circle but different directions, so they are not equal vectors, ruling out (c).

Question 8

In a two dimensional motion, instantaneous speed $v_0$ is a positive constant. Then which of the following are necessarily true?

Solution. The acceleration is $\mathbf{a}=\dfrac{d\mathbf{v}}{dt}$. If the motion stays in a plane, every velocity vector lies in that plane, and so does the difference of any two of them; hence $\mathbf{a}$ lies in the plane of motion. That is (c). Constant speed only makes $\mathbf{a}$ perpendicular to $\mathbf{v}$, with size $v_0^2/r$ where $r$ is the local radius of curvature. A curved path at constant speed has $\mathbf{a}\neq0$, so (a) fails. The radius of curvature can be made as small as we like at a sharp bend, so $\mathbf{a}$ is not necessarily bounded, and (b) fails. Any curve, not only a circle, can be traced at constant speed, so (d) fails.

Question 9

Three vectors $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ add up to zero. Find which is false.

Solution. Since $\mathbf{A}+\mathbf{B}+\mathbf{C}=0$, we have $\mathbf{A}=-(\mathbf{B}+\mathbf{C})$, so $$\mathbf{A}\times\mathbf{B}=-(\mathbf{B}+\mathbf{C})\times\mathbf{B}=\mathbf{B}\times\mathbf{C}.$$ The three vectors close a triangle, so they always lie in one plane.

• (a) $(\mathbf{A}\times\mathbf{B})\times\mathbf{C}=(\mathbf{B}\times\mathbf{C})\times\mathbf{C}$. If $\mathbf{B}$ and $\mathbf{C}$ are not parallel, $\mathbf{B}\times\mathbf{C}$ is a non-zero vector perpendicular to $\mathbf{C}$, and the cross product of two perpendicular non-zero vectors is not zero. True.
• (b) $(\mathbf{A}\times\mathbf{B})\cdot\mathbf{C}=(\mathbf{B}\times\mathbf{C})\cdot\mathbf{C}=0$ always, because $\mathbf{B}\times\mathbf{C}$ is perpendicular to $\mathbf{C}$. The statement claims it is non-zero for non-parallel $\mathbf{B}$, $\mathbf{C}$, which is never so. False.
• (c) $\mathbf{A}\times\mathbf{B}$ is normal to the plane; crossing a normal with $\mathbf{C}$, which lies in the plane, gives a vector perpendicular to the normal, that is, a vector in the plane. True.
• (d) Since the left side is always zero, the condition can hold only if one of the vectors is zero, and then the other two are equal and opposite. If $\mathbf{A}$ or $\mathbf{B}$ is the zero vector, $C^2=A^2+B^2$ does follow; only the corner case $\mathbf{C}=0$ breaks it. The statement concerns degenerate triples only and is not the one the question is testing.

Printed key error. The key prints (c). But (c) is the standard property of a vector triple product: $(\mathbf{A}\times\mathbf{B})\times\mathbf{C}$ is perpendicular to $\mathbf{A}\times\mathbf{B}$, which is the normal, so it lies in the plane of $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$. The statement that is false for every such triple is (b), since the scalar triple product of three coplanar vectors is zero.

Question 10

It is found that $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}|$. This necessarily implies,

Solution. Square both sides: $$A^2+2\,\mathbf{A}\cdot\mathbf{B}+B^2=A^2\quad\Rightarrow\quad\mathbf{A}\cdot\mathbf{B}=-\frac{B^2}{2}\le0.$$ So (d) must hold. None of the others is forced. Take $\mathbf{A}=\hat{\mathbf{i}}$ and $\mathbf{B}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}$: then $\mathbf{A}+\mathbf{B}=\hat{\mathbf{j}}$ and $|\mathbf{A}+\mathbf{B}|=1=|\mathbf{A}|$, yet $\mathbf{B}\neq0$, and $\mathbf{A}$ and $\mathbf{B}$ are at $135^\circ$, neither antiparallel nor perpendicular.

Printed key error. The key prints (b). Antiparallel vectors can satisfy the condition (for example $\mathbf{B}=-2\mathbf{A}$), but they need not, as the example above shows, so (b) is not necessarily implied. The only statement that follows in every case is (d).

Question 11

Two particles are projected in air with speed $v_0$ at angles $\theta_1$ and $\theta_2$ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices

Solution. With the same speed, $H=\dfrac{v_0^2\sin^2\theta}{2g}$. For acute angles $\sin\theta$ increases with $\theta$, so $H_1>H_2$ means $\theta_1>\theta_2$: (a) is true. The time of flight $T=\dfrac{2v_0\sin\theta}{g}$ also increases with $\theta$, so (b) is true. The range depends on $\sin2\theta$, which rises up to $45^\circ$ and then falls, so $\theta_1>\theta_2$ tells us nothing about which range is larger: (c) is not necessarily true. Both particles start with the same speed; their total energies are equal if the masses are equal, and the masses are not given, so (d) cannot be ticked.

Question 12

A particle slides down a frictionless parabolic ($y=x^2$) track (A – B – C) starting from rest at point A (Fig. 4.2). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then

The figure shows the parabola with A high on the left branch at $x=-x_2$, the vertex B at the origin, and C on the right branch at $x=-x_0$ (printed so), lower than A. At C the particle leaves the track with speed $v_0$ at angle $\theta$ and follows a dotted projectile path up to its highest point P, which is drawn below the level of A.

Solution. The track is frictionless and there is no air resistance, so mechanical energy is conserved all the way from A to P: (c) is true. At P the particle still has its horizontal velocity, so it has kinetic energy; hence its potential energy is less than at A and P is lower than A, so (b) is false. P is higher than B, so the kinetic energy at P is less than at B, and (a) is false. The two stretches A to B and B to P are different paths at different speeds, so there is no reason for their times to be equal: (d) is false.

Question 13

Following are four different relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one(s):

Solution. Options (b) and (d) are the definitions of average velocity and average acceleration, so they hold for any motion. Option (a), the mean of the initial and final velocities, equals the average velocity only when the acceleration is constant; for motion in general it is wrong. Option (c) is not a valid relation at all: even for constant acceleration the displacement is $\tfrac12\left(\mathbf{v}(t_1)+\mathbf{v}(t_2)\right)(t_2-t_1)$, with a plus sign. So the incorrect ones are (a) and (c).

Question 14

For a particle performing uniform circular motion, choose the correct statement(s) from the following:

Solution. “Uniform” means constant speed, so (a) is true. The velocity is along the tangent, and the tangent to a circle is perpendicular to the radius, so (b) is true. The acceleration is $v^2/r$ directed towards the centre; as the particle moves round, the direction towards the centre turns with it, so (c) is true. The angular momentum about the centre is $\mathbf{L}=m\,\mathbf{r}\times\mathbf{v}$, which stays perpendicular to the plane of the circle with constant size $mvr$; its direction does not change, so (d) is false.

Question 15

For two vectors $\mathbf{A}$ and $\mathbf{B}$, $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}-\mathbf{B}|$ is always true when

Solution. Squaring, $$A^2+B^2+2\,\mathbf{A}\cdot\mathbf{B}=A^2+B^2-2\,\mathbf{A}\cdot\mathbf{B}\quad\Rightarrow\quad\mathbf{A}\cdot\mathbf{B}=0.$$ This holds when the vectors are perpendicular, (b), or when either vector is zero, (d). Equal magnitudes alone, (a), do not make the dot product zero. Parallel or antiparallel vectors of equal non-zero size, (c), give $|\mathbf{A}+\mathbf{B}|=2A$ and $|\mathbf{A}-\mathbf{B}|=0$ (or the reverse), which are unequal.

Common mistakes

• Reading the position of $\mathbf{v}$ instead of its direction in Question 3. A vector’s components depend on where it points, not where its tail is drawn.
• Comparing $\sin15^\circ$ with $\sin45^\circ$ in Question 5. The range depends on $\sin2\theta$, so the ratio is $\sin90^\circ/\sin30^\circ=2$.
• Calling pressure a vector in Question 6. It acts equally in all directions at a point; it is force per area, but the force direction comes from the surface, not from the pressure.
• Accepting the key’s (c) in Question 9. $(\mathbf{A}\times\mathbf{B})\times\mathbf{C}$ does lie in the plane; the statement that fails is (b), since the scalar triple product of coplanar vectors is zero.
• Settling for one example in Question 10. Antiparallel vectors can satisfy $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}|$, but “necessarily” asks what every case shares, which is $\mathbf{A}\cdot\mathbf{B}=-B^2/2\le0$.
• Ticking the range in Question 11. Range rises and then falls with angle, so a higher projectile need not go further.
• Thinking angular momentum turns in Question 14. It is perpendicular to the plane of the circle and fixed in direction.

Practise next

• A projectile fired at $30^\circ$ to the horizontal has a range of $40$ m. Fired with the same speed at $45^\circ$, what is its range?

• Two vectors have $|\mathbf{A}|=3$ and $|\mathbf{B}|=4$, and $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}-\mathbf{B}|$. Find $|\mathbf{A}+\mathbf{B}|$.

• Exemplar Motion in a Plane: Very Short and Short Answer Questions — the same ideas applied to circular motion, projectiles seen from moving cars, and vector matching.
• Exemplar Motion in a Plane: Long Answer Questions — projectiles on inclines, river crossing, rain and polar coordinates.
• NCERT textbook Chapter 3 — Motion in a Plane for the theory of vectors, projectiles and uniform circular motion.