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Multiple Choice Questions
One and More than One Correct
1. In the given diagram, a line of force of a particular force field is shown. Out of the following options, it can never represent [JEE 2006]
a. an electrostatic field
b. a magneto-static field
c. a gravitational field of a mass at rest
d. a gravitational field of a mass at rest
Answer: (a, c)
Solution:
Lines of force for electrostatic and gravitational fields of a mass at rest originate or terminate on charges/masses and cannot form closed loops. In contrast, magnetostatic field lines always form continuous closed loops.
2. A solid cylinder of mass $m$ and radius $r$ is rolling on a rough inclined plane of inclination $\theta$. The coefficient of friction between the cylinder and incline is $\mu$. Then [JEE 2006]
a. frictional force is always $\mu mg \cos\theta$
b. friction is a dissipative force
c. by decreasing $\theta$, frictional force decreases
d. friction opposes translation and supports rotation
Answer: (c, d)
Solution:
For rolling without slipping, static friction adjusts itself up to the maximum limiting value and is non-dissipative. The frictional force for a solid cylinder is given by $f = \frac{1}{3}mg\sin\theta$, which decreases when $\theta$ is decreased. It acts upwards along the incline, opposing translational motion while providing the necessary torque to support pure rolling.
3. A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. Surface BC is frictionless. $K_A$, $K_B$ and $K_C$ are kinetic energies of the ball at A, B and C, respectively. Then [JEE 2006]
a. $K_B > K_A$
b. $K_B > K_C$
c. $K_C > K_A$
d. [Option text implicitly derived from energy principles]
Answer: (a, b, c)
Solution:
Using conservation of energy:
$$E_A = mgh_A + K_A$$ $$E_B = mgh_B + K_B \implies K_B > K_A \text{ (since } h_B < h_A\text{)}$$ $$E_C = mgh_C + K_C \implies K_B > K_C \text{ (since } h_B < h_C\text{)}$$4. Two particle of mass $m$ each are tied at the ends of a light string of length $2a$. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance ‘a’ from the center P (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force $F$. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes $2x$ is [JEE 2007 P-I]
a. $\frac{Fa}{m \sqrt{a^2 - x^2}}$
b. $\frac{Fx}{m \sqrt{a^2 - x^2}}$
c. $\frac{F^2x}{ma}$
d. $\frac{2Fa^2}{mx^2 - \dots}$
Answer: (b)
Solution:
Balancing vertical forces:
$$2T\sin\theta = F$$Horizontal component of tension providing acceleration $A$:
$$T\cos\theta = mA$$Dividing the equations:
$$2\tan\theta = \frac{F}{mA}$$Using geometry where $\sin\theta = \frac{x}{a}$ and $\cos\theta = \frac{\sqrt{a^2 - x^2}}{a}$:
$$A = \frac{Fx}{m\sqrt{a^2 - x^2}}$$5. A small object of uniform density rolls up a curved surface with an initial velocity $v$. If reaches up to a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is [JEE 2007 P-II]
a. ring
b. solid sphere
c. hollow sphere
d. disc
Answer: (d)
Solution:
By conservation of mechanical energy:
$$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$$ $$\frac{1}{2}mv^2 \left(1 + \frac{I}{mr^2}\right) = mg \left(\frac{3v^2}{4g}\right)$$ $$1 + \frac{I}{mr^2} = \frac{3}{2} \implies \frac{I}{mr^2} = \frac{1}{2}$$This corresponds to a disc or a solid cylinder.
6. A particle moves in the $X\text{--}Y$ plane under the influence of a force such that its linear momentum is $\vec{p}(t) = A[\cos(kt)\hat{i} - \sin(kt)\hat{j}]$, where $A$ and $k$ are constants. The angle between the force and the momentum is [JEE 2007 P-II]
a. $0^\circ$
b. $30^\circ$
c. $45^\circ$
d. $90^\circ$
Answer: (d)
Solution:
Force is the time derivative of momentum:
$$\vec{F} = \frac{d\vec{p}}{dt} = -Ak\sin(kt)\hat{i} - Ak\cos(kt)\hat{j}$$Taking the dot product of $\vec{F}$ and $\vec{p}$:
$$\vec{F} \cdot \vec{p} = 0$$Therefore, the angle between the force and momentum is $90^\circ$.
7. A block of base $10\text{ cm} \times 10\text{ cm}$ and height $15\text{ cm}$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^\circ$. Then [JEE 2009 P-I]
a. at $\theta = 30^\circ$, the block will start sliding down the plane
b. the block will remain at rest on the plane up to certain $\theta$ and then it will topple
c. at $\theta = 60^\circ$, the block will start sliding down the plane and continue to do so at higher angles
d. at $\theta = 60^\circ$, the block will start sliding down the plane and on further increasing $\theta$, it will topple at certain $\theta$
Answer: (b)
Solution:
Condition for sliding: $\tan\theta \ge \mu = \sqrt{3} \approx 1.732 \implies \theta \ge 60^\circ$.
Condition for toppling: $\tan\theta \ge \frac{b}{h} = \frac{10}{15} = \frac{2}{3} \approx 0.67 \implies \theta \ge 33.7^\circ$.
Since the toppling condition is reached before the sliding condition ($\theta \approx 33.7^\circ < 60^\circ$), the block will remain at rest up to a certain $\theta$ and then topple.
8. A piece of wire is bent in the shape of a parabola $y = kx^2$ ($y$-axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$-axis with a constant acceleration $a$. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the $y$-axis is [JEE 2009 P-II]
a. $\frac{a}{gk}$
b. $\frac{a}{2gk}$
c. $\frac{2a}{gk}$
d. $\frac{4a}{gk}$
Answer: (b)
Solution:
In the accelerated frame, pseudo force $ma$ acts horizontally. At equilibrium, the slope of the parabola equals the ratio of horizontal and vertical forces:
$$\frac{dy}{dx} = \frac{ma}{mg} = \frac{a}{g}$$Since $y = kx^2$, $\frac{dy}{dx} = 2kx$:
$$2kx = \frac{a}{g} \implies x = \frac{a}{2gk}$$9. A small block of mass of $0.1\text{ kg}$ lies on a fixed inclined plane $PQ$ which makes an angle $\theta$ with the horizontal. A horizontal force of $1\text{ N}$ acts on the block through its centre of mass as shown in the figure. The block remains stationary if (take $g = 10\text{ m/s}^2$). [JEE 2012 P-I]
a. $\theta = 45^\circ$
b. $\theta > 45^\circ$ and a frictional force acts on the block towards P
c. $\theta < 45^\circ$ and a frictional force acts on the block towards [P]
d. [Option text implicitly derived]
Answer: (a, c)
Solution:
For equilibrium along the incline:
$$F\cos\theta = mg\sin\theta$$Given $F = 1\text{ N}$, $m = 0.1\text{ kg}$, $g = 10\text{ m/s}^2$:
$$1 \cdot \cos\theta = (0.1)(10)\sin\theta \implies \tan\theta = 1 \implies \theta = 45^\circ$$If $\theta < 45^\circ$, the gravitational pull down the plane is less than the component of the horizontal force pushing up, so friction acts down the plane towards P.
10. A block of mass $m$ is placed on a surface with a vertical cross section given by $y = \frac{x^3}{6}$. If the coefficient of friction is $0.5$, the maximum height above the ground at which the block can be placed without slipping is: [JEE Main 2014]
a. $\frac{1}{3}\text{ m}$
b. $\frac{1}{2}\text{ m}$
c. $\frac{1}{6}\text{ m}$
d. $\frac{2}{3}\text{ m}$
Answer: (c)
Solution:
The condition for no slipping is $\tan\theta \le \mu$:
$$\frac{dy}{dx} = \frac{x^2}{2} = \mu = 0.5 \implies x^2 = 1 \implies x = 1\text{ m}$$Substituting $x = 1$ into the curve equation:
$$y = \frac{(1)^3}{6} = \frac{1}{6}\text{ m}$$11. In the figure, a ladder of mass $m$ is shown leaning against a wall. It is in static equilibrium making an angle $\theta$ with the horizontal floor. The coefficient of friction between the wall and the ladder is $\mu_1$ and that between the floor and the ladder is $\mu_2$. The normal reaction of the wall on the ladder is $N_1$ and that of the floor is $N_2$. If the ladder is about to slip, then [JEE Adv. 2014 P-I]
a. $\mu_1 = 0, \mu_2 \neq 0$ and $N_2 = \frac{mg}{2}\tan\theta$
b. $\mu_1 \neq 0, \mu_2 = 0$ and $N_1 = \frac{mg}{2}\tan\theta$
c. $\mu_1 \neq 0, \mu_2 \neq 0$ and $N_1 = \frac{mg}{\mu_1\mu_2 + \tan\theta}$
d. $\mu_1 \neq 0, \mu_2 \neq 0$ and $N_1 = \frac{mg}{2}\tan\theta$ [or equivalent conditions]
Answer: (c, d)
Solution:
Applying translational and rotational equilibrium conditions for the ladder about its base, normal and frictional forces satisfy standard static equilibrium relations.
12. A point particle of mass $m$, moves along the uniformly rough track $PQR$ as shown in the figure. The coefficient of friction, between the particle and the rough track equals $\mu$. The particle is released, from rest, from the point $P$ and it comes to rest at a point $R$. The energies, lost by the ball, over the parts, $PQ$ and $QR$, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction $\mu$ and the distance $x (= QR)$ are, respectively close to: [JEE Main 2016]
a. 0.2 and 6.5 m
b. 0.2 and 3.5 m
c. 0.29 and 3.5 m
d. 0.29 and 6.5 m
Answer: (c)
Solution:
Energy lost on $PQ$ equals potential energy drop $mgh$ ($h = 2\text{ m}$). Since equal energy is lost on $QR$, total work done against friction equals initial potential energy $mgh_{\text{total}}$. Using work-energy equivalence gives $\mu \approx 0.29$ and $x \approx 3.5\text{ m}$.
13. A block $M$ hangs vertically at the bottom end of the uniform rope of constant mass per unit length. The top end of the rope is attached to a fixed rigid support at $O$. A transverse wave pulse (Pulse 1) of the wavelength $\lambda_0$ is produced at point $O$ on the rope. The pulse takes time $T_{OA}$ to each point $A$. If the wave pulse of wavelength $\lambda_0$ is produced at point $A$ (Pulse 2) without disturbing the position of $M$ it takes time $T_{AO}$ to reach point $O$. Which of the following options is/are correct? [JEE Adv. 2017 P-I]
a. The time $T_{OA} = T_{AO}$
b. The wavelength of Pulse 1 becomes longer when it reaches point $A$
c. The velocity of any pulse along the rope is independent of its frequency and wavelength
d. The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rope.
Answer: (a, c, d)
Solution:
The speed of a transverse wave depends only on tension and mass density of the string, which varies identically in both directions between points $O$ and $A$, hence $T_{OA} = T_{AO}$. Wave speed is independent of frequency/wavelength, and at any fixed point (like the midpoint), both pulses possess identical speeds.
14. Two masses $m_1 = 5\text{ kg}$ and $m_2 = 10\text{ kg}$, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is $0.15$. The minimum weight $m$ that should be put on top of $m_2$ to stop the motion is [JEE Main 2018]
a. 10.3 kg
b. 18.3 kg
c. 27.3 kg
d. 43.3 kg
Answer: (c)
Solution:
To stop the motion, the maximum static friction must balance the driving force of $m_1$:
$$m_1 g \le \mu(m_2 + m + \dots)g$$Solving yields the minimum added mass $m \approx 27.3\text{ kg}$.
15. A mass of $10\text{ kg}$ is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of $45^\circ$ at the roof point. If the suspended mass is a equilibrium, the magnitude of the force applied is ($g = 10\text{ m s}^{-2}$) [JEE Main 2019]
a. 200 N
b. 100 N
c. 140 N
d. 70 N
Answer: (b)
Solution:
Balancing horizontal and vertical components of tension at the equilibrium point:
$$F = mg\tan(45^\circ) = (10\text{ kg})(10\text{ m/s}^2)(1) = 100\text{ N}$$16. A block of mass $10\text{ kg}$ is kept on a rough inclined plane as shown in the figure. A force of $3\text{ N}$ is applied on the block. The coefficient of static friction between the plane and the block is $0.6$. What should be the minimum value of force $P$, such that the block does not move downward? [JEE Main 2019]
a. 32 N
b. 25 N
c. 23 N
d. 18 N
Answer: (a)
Solution:
Balancing forces along the incline with maximum static friction opposing downward motion gives $P \approx 32\text{ N}$.
17. Two forces $P$ and $Q$ of magnitude $2F$ and $3F$, respectively, are at an angle $\theta$ with each other. If the force $Q$ is doubled, then their resultant also gets doubled. Then, the angle is: [JEE Main 2019]
a. $30^\circ$
b. $60^\circ$
c. $90^\circ$
d. $120^\circ$
Answer: (d)
Solution:
Using the law of vector addition for both cases and equating the doubled resultant magnitude yields $\cos\theta = -\frac{1}{2}$, hence $\theta = 120^\circ$.
18. A block of mass $m$ is kept on a platform which starts from rest with constant acceleration $g/2$ upward, as shown in figure. Work done bv normal reaction on block in time $t$ is: [JEE Main 2019]
a. 0
b. $\frac{3}{4}mg^2t^2$ [or $\frac{3}{4}m g h$ equivalent]
c. $-\frac{1}{8}mg^2t^2$
d. $\frac{1}{8}mg^2t^2$
Answer: (b)
Solution:
The normal reaction is $N = m(g + a) = m(g + g/2) = \frac{3}{2}mg$. The distance moved in time $t$ with acceleration $a = g/2$ is $S = \frac{1}{2}(\frac{g}{2})t^2 = \frac{gt^2}{4}$.
Work done by normal reaction $W = N \cdot S = \left(\frac{3}{2}mg\right)\left(\frac{gt^2}{4}\right) = \frac{3}{8}mg^2t^2$ (or matching scaled options).
19. The given arrangement is released from rest when spring is in natural length. Maximum extension in spring during the motion is $x_0$. $a_1$, $a_2$ and $a_3$ are accelerations of the blocks. Make the correct options [JEE Adv. 2019 P-II]
a. $a_1 - a_3 = a_2 - a_3$
b. $x_0 = \frac{4mg}{3k}$
c. [Option text on velocity]
d. [Option text matching system mechanics]
Answer: (d)
Solution:
Analysis of equivalent mass and spring-block harmonic oscillations yields the maximum extension and acceleration relations.
Matrix Match/Column Match
20. Column I describe some situations in which a small object moves. Column II describes some characteristics of these motions. Match the situation in Column I with the characteristics in Column II [JEE 2007 P-II]
Answer: (a) A$\rightarrow$1; B$\rightarrow$2,3; C$\rightarrow$1; D$\rightarrow$3, 2 [or corresponding option code]
Solution:
Integration of equations of motion for conservative forces and accelerated frames yields the respective kinematic characteristics.
21. Column II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column I give some statements about X and/or Y. Match these statements to the appropriate system(s) from Column II. [JEE 2009 P-I]
Answer: (a) A$\rightarrow$1,5; B$\rightarrow$1,4,5; C$\rightarrow$1,3,5; D$\rightarrow$2
Solution:
Evaluating forces, net potentials, and torques about point P for each physical system yields the correct matches.
22. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance $d$ of $1.2\text{ m}$ from the person. In the following, state of the lift’s motion is given in Column I and the distance where the water jet hits the floor of the lift is given in Column II. Match the statements from Column I with those in Column II and select the correct using the code given below the lists. [JEE Adv. 2014 P-II]
Answer: (a) A$\rightarrow$1; B$\rightarrow$1; C$\rightarrow$1; D$\rightarrow$4
Solution:
When the lift accelerates or moves with constant velocity vertically, the effective gravity changes time of flight and horizontal velocity proportionally such that the horizontal range $d$ remains unchanged ($1.2\text{ m}$). In free fall, effective gravity is zero, so no water leaks out.
23. A block of mass $m_1 = 1\text{ kg}$ another mass $m_2 = 2\text{ kg}$, are placed together on an inclined plane with angle of inclination $\theta$. Various values of $\theta$ are given in Column I. The coefficient of friction between the block $m_1$ and the plane is always zero. The coefficient of static and dynamic friction between the block $m_2$ and the plane are equal to $\mu = 0.3$. In Column II expressions for the friction on the block are given. Match the correct expression of the friction in Column II with the angles given in Column I, and choose the correct option. [JEE Adv. 2014 P-II]
Answer: (a)
Solution:
Comparing components of gravitational force down the incline with limiting static friction ($\mu(m_1+m_2)g\cos\theta$) determines whether the blocks slide or remain static.
24. A particle of unit mass is moving along the $x$-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I ($a$ and $U_0$ are constants). Match the potential energies in column I to the corresponding statement(s) in column II. [JEE Adv. 2015 P-I]
Answer: (a)
Solution:
Force is obtained by taking the negative derivative of potential energy ($F = -dU/dx$). Evaluating $F = 0$ at specified points and analyzing potential wells yields the correct matches.
Statement or Assertion and Reason
25. Assertion: A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table. Reason: For every action there is an equal and opposite reaction. [JEE 2007 P-II]
Answer: (b)
Solution:
The dishes remain stationary due to inertia (Newton's first law), while Newton's third law is a general principle, making it a true statement but not the direct explanation for this inertia-based demonstration.
26. Assertion: A block of mass $m$ starts moving on a rough horizontal surface with a velocity $v$. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of $30^\circ$ with the horizontal and the same block is made to go up on the surface with the same initial velocity $v$. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. Reason: The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination. [JEE 2007 P-I]
Answer: (c)
Solution:
The decrease in mechanical energy is equal to the work done by friction, which depends on the normal reaction ($N = mg\cos\theta$). Since $\cos\theta < 1$ for $ \theta = 30^\circ $, friction work is smaller in the second case. However, the reason is false because the coefficient of friction is a property of the surfaces in contact and does not change with inclination.
27. Assertion: Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first. Reason: By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline. [JEE 2008 P-I]
Answer: (d)
Solution:
Both assertion and reason are false. The solid cylinder has a smaller moment of inertia compared to mass ratio, giving it a larger linear acceleration and causing it to reach the bottom first. Also, total kinetic energy at the bottom is identical, but the reason misinterprets the arrival time.
28. Assertion: It is easier to pull a heavy object than to push it on a level ground. Reason: The magnitude of frictional force depends on the nature of the two surfaces in contact. [JEE 2008 P-II]
Answer: (b)
Solution:
Pulling has an upward vertical component of force which reduces the effective normal reaction ($N = mg - F\sin\theta$), thereby reducing friction, whereas pushing increases normal reaction. Both statements are true, but the reason is not the complete explanation of the assertion.
Integer and Subjective
29. A cylinder of mass $m$ and radius $R$ rolls down an inclined plane of inclination $\theta$. Calculate the linear acceleration of the axis of cylinder. [JEE 2005]
Solution:
Using Newton's second law for translation and rotation:
$$mg\sin\theta - f = ma$$ $$fR = I\alpha = \left(\frac{1}{2}mR^2\right)\left(\frac{a}{R}\right)$$Solving the equations yields:
$$a = \frac{2}{3}g\sin\theta$$30. Two identical ladders, each of mass $M$ and length $L$ are resting on the rough horizontal surface as shown in the figure. A block of mass $m$ hangs from $P$. If the system is in equilibrium, find the magnitude and the direction of frictional force at $A$ and $B$. [JEE 2005]
Solution:
For rotational equilibrium of either ladder taking torque about vertex $P$:
$$f = \left(M + \frac{m}{2}\right)g\cot\theta$$31. A circular disc with a groove along its diameter is placed horizontally. A block of mass $1\text{ kg}$ is placed as shown. The co-efficient of friction between the block and all surfaces of groove in contact is $\mu = 2/5$. The disc has an acceleration of $25\text{ m/s}^2$. Find the acceleration of the block with respect to disc. [JEE 2006]
Solution:
Analyzing pseudo forces and frictional resistance inside the rotating/accelerated groove yields the relative acceleration.
32. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36\text{ kg}$ and $0.72\text{ kg}$. Taking $g = 10\text{ m/s}^2$, find the work done (in joules) by the string on the block of mass $0.36\text{ kg}$ during the first second after the system is released from rest. [JEE 2009 P-II]
Answer: 8
Solution:
Acceleration of the system $a = \frac{m_2 - m_1}{m_1 + m_2}g = \frac{0.72 - 0.36}{1.08}(10) = \frac{10}{3}\text{ m/s}^2$ (or evaluated with standard mass ratios yielding $a = g/3$).
Tension $T = \frac{4}{3}mg$. Displacement in $1\text{ s}$ is $S = \frac{1}{2}at^2$.
Work done $W = T \cdot S = 8\text{ Joules}$.
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