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Multiple Choice Questions

One and More than One Correct

1. Two blocks A and B of masses $2m$ and $m$, respectively, are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of A and B, immediately after the string is cut, are respectively. [JEE 2006]

Answer: (b)

Solution:

$a_A = \frac{g}{2}$, $a_B = g$

2. Students I, II and III perform an experiment for measuring the acceleration due to gravity ($g$) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.

Least count for length $= 0.1\text{ cm}$

Least count for time $= 0.1\text{ s}$

If $E_{\text{I}}$, $E_{\text{II}}$ and $E_{\text{III}}$ are the percentage errors in $g$, i.e., $\left(\frac{\Delta g}{g} \times 100\right)$ for students, I, II and III, respectively, [JEE 2008 P-I]

a. $E_{\text{I}} = 0$

b. $E_{\text{I}}$ is minimum

c. $E_{\text{I}} = E_{\text{II}}$

d. $E_{\text{II}}$ is maximum

Answer: (b)

Solution:

$$T = 2\pi \sqrt{\frac{l}{g}} \Rightarrow g = 4\pi^2 \frac{l}{T^2}$$ $$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} = \frac{\Delta l}{l} + 2\left(\frac{\Delta t}{t}\right)$$

Greater the value of $t$, lesser the error.

Hence, fractional error in the $\text{I}^{\text{st}}$ observation is minimum.

3. A block B is attached to two unstretched springs $S_1$ and $S_2$ with spring constants $k$ and $4k$, respectively (see figure I). The other ends are attached to identical supports $M_1$ and $M_2$ not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance $x$ (figure II) and released. The block returns and moves a maximum distance $y$ towards wall 2. Displacements $x$ and $y$ are measured with respect to the equilibrium position of the block B. The ratio $y/x$ is: [JEE 2008 P-II]

a. 4

b. 2

c. $\frac{1}{2}$

d. $\frac{1}{4}$

Answer: (c)

Solution:

$$\frac{1}{2}kx^2 = \frac{1}{2}(4k)y^2 \Rightarrow y/x = \frac{1}{2}$$

4. A bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $V$ at position A is just sufficient to make it reach the point B. The angle $\theta$ at which the speed of the bob is half of that at A, satisfies [JEE 2008 P-II]

a. $\theta = \frac{\pi}{4}$

b. $\frac{\pi}{4} < \theta < \frac{\pi}{2}$

c. $\frac{\pi}{2} < \theta < \frac{3\pi}{4}$

d. $\frac{3\pi}{4} < \theta < \pi$

Answer: (d)

Solution:

$$\frac{1}{2}m\left(\frac{V}{2}\right)^2 = \frac{1}{2}mV^2 - mgL(1 - \cos\theta)$$ $$\cos\theta = -\frac{7}{8}$$

Hence, $\frac{3\pi}{4} < \theta < \pi$

5. The $x-t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t = \frac{4}{3}\text{ s}$ is: [JEE 2009 P-I]

a. $\frac{3\pi^2}{32}\text{ cm s}^{-2}$

b. $-\frac{\pi^2}{32}\text{ cm s}^{-2}$

c. $\frac{\pi^2}{32}\text{ cm s}^{-2}$

d. $-\frac{3\pi^2}{32}\text{ cm s}^{-2}$

Answer: (d)

Solution:

The given motion is represented by $x = 1\sin\left(\frac{\pi t}{4}\right)$

$$\frac{d^2x}{dt^2} = -\frac{\pi^2}{16}\sin\left(\frac{\pi t}{4}\right)$$

At $t = \frac{4}{3}\text{ sec}$,

$$\text{acceleration} = -\frac{3\pi^2}{32}\text{ cm s}^{-2}$$

6. A uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$. The springs are fixed to rigid supports as shown in the figure, and rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is: [JEE 2009 P-II]

a. $\frac{1}{2\pi}\sqrt{\frac{12k}{M}}$

b. $\frac{1}{2\pi}\sqrt{\frac{k}{M}}$

c. $\frac{1}{2\pi}\sqrt{\frac{6k}{M}}$

d. $\frac{1}{2\pi}\sqrt{\frac{24k}{M}}$

Answer: (c)

Solution:

Restoring torque:

$$\tau = -2 \times k\left(\frac{L\theta}{2}\right)\left(\frac{L}{2}\right) = -I\frac{d^2\theta}{dt^2}$$ $$\frac{ML^2}{12}\frac{d^2\theta}{dt^2} + \frac{kL^2}{2}\theta = 0$$ $$\omega = \sqrt{\frac{6k}{M}}$$

7. The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$. The amplitude of the point P is [JEE 2009 P-II]

a. $\frac{k_2}{k_1}A$

b. $\frac{k_1}{k_2}A$

c. $\frac{k_1}{k_1+k_2}A$

d. $\frac{k_2}{k_1+k_2}A$

Answer: (d)

Solution:

$$k_1x_1 = k_2x_2$$ $$x_1 + x_2 = A \Rightarrow x_2 = \frac{k_1}{k_1+k_2}A$$

8. A wooden block performs SHM on a frictionless surface with frequency, $v_0$. The block carries a charge $+Q$ on its surface. If now a uniform electric field $\vec{E}$ is switched-on as shown, then the SHM of the block will be [JEE 2011 P-II]

a. on the same frequency and with shifted mean position

b. of the same frequency and with the same mean position

c. of changed frequency and with shifted mean position

d. of changed frequency and with the same mean position

Answer: (a)

9. A particle of mass $m$ is attached to one end of a mass-less spring of force constant $k$, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t = 0$ with an initial velocity $u_0$. When the speed of the particle is $0.5u_0$, it collides elastically with a rigid wall. After this collision, [JEE Adv. 2013 P-II]

a. the speed of the particle when it returns to its equilibrium position is $u_0$

b. the time at which the particle passes through the equilibrium position for the first time is $t = \pi\sqrt{\frac{m}{k}}$

c. the time at which the maximum compression of the spring occurs is $t = \frac{3}{4}\pi\sqrt{\frac{m}{k}}$

d. the time at which the particle passes through the equilibrium position for the second time is $t = \frac{5}{3}\pi\sqrt{\frac{m}{k}}$

Answer: (a, d)

Solution:

Energy of particle and spring remains conserved.

10. The amplitude of a damped oscillator decreases to $0.9$ times its original magnitude in $5\text{ s}$. In another $10\text{ s}$ it will decrease to $\alpha$ times its original magnitude, where $\alpha$ equals [JEE Main 2013]

a. $0.7$

b. $0.81$

c. $0.729$

d. $0.6$

Answer: (c)

Solution:

$$A = A_0 e^{-kt} \Rightarrow 0.9A_0 = A_0 e^{-5k}$$ $$\alpha = e^{-15k} = (0.9)^3 = 0.729$$

11. An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $M$. The piston and the cylinder have equal cross sectional area $A$. When the piston is in equilibrium, the volume of the gas is $V_0$ and its pressure is $P_0$. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency [JEE Main 2013]

a. $\frac{1}{2\pi}\sqrt{\frac{\gamma P_0 A}{MV_0}}$

b. $\frac{1}{2\pi}\sqrt{\frac{A^2 \gamma P_0}{M V_0}}$

c. $\frac{1}{2\pi}\sqrt{\frac{\gamma P_0 A^2}{MV_0}}$

d. $\frac{1}{2\pi}\sqrt{\frac{M V_0}{\gamma P_0 A}}$

Answer: (c)

Solution:

$$\omega = \sqrt{\frac{\gamma P_0 A^2}{MV_0}} \Rightarrow f = \frac{1}{2\pi}\sqrt{\frac{\gamma P_0 A^2}{MV_0}}$$

12. A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is [JEE Adv. 2014 P-II]

a. always radially outwards

b. always radially inwards

c. radially outwards initially and radially inwards later

d. radially inwards initially and radially outwards later

Answer: (d)

Solution:

Initially bead is applying radially inward normal force. During motion at an instant, $N = 0$, after that $N$ will act radially outward.

13. The period of oscillation of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$. Measured value of $L$ is $20.0\text{ cm}$ known to $1\text{ mm}$ accuracy and time for $100$ oscillations of the pendulum is found to be $90\text{ s}$ using a wrist watch of $1\text{s}$ resolution. The accuracy in the determination of $g$ is: [JEE Main 2015]

a. $2\%$

b. $3\%$

c. $1\%$

d. $5\%$

Answer: (b)

Solution:

$$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} = \frac{0.1}{20} + 2\left(\frac{1}{90}\right) = 0.005 + 0.022 = 0.027 \approx 3\%$$

14. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement $d$. Which one of the following represents these correctly? (Graph are schematic and not drawn to scale) [JEE Main 2015]

Answer: (b)

Solution:

For simple harmonic motion, potential energy goes as $\frac{1}{2}mx^2\omega^2$. The maximum potential energy being $\frac{1}{2}mA^2\omega^2$. Kinetic energy goes as $\frac{1}{2}m\omega^2(A^2 - x^2)$. Graph (b) fits these plots.

15. Two long current carrying thin wires, both with current $I$, are held by insulating threads of length $L$ and are in equilibrium as shown in the figure, with threads making an angle $\theta$ with the vertical. If wires have mass $\lambda$ per unit length then the value of $I$ is: ($g = \text{gravitational acceleration}$) [JEE Main 2015]

a. $\sqrt{\frac{\pi\lambda g L \sin\theta}{\mu_0 \cos\theta}}$

b. $\sqrt{\frac{2\pi\lambda g L \sin\theta}{\mu_0 \cos\theta}}$

c. $\sqrt{\frac{2\pi g L \tan\theta}{\mu_0}}$

d. $\sqrt{\frac{\pi\lambda g L \tan\theta}{\mu_0}}$

Answer: (b)

Solution:

Magnetic force per unit length $F = \frac{\mu_0 I^2}{2\pi x} = \frac{\mu_0 I^2}{2\pi (2L\sin\theta)}$

For equilibrium: $T\sin\theta = F$, $T\cos\theta = \lambda g L$

Solving gives $I = \sqrt{\frac{2\pi\lambda g L \sin\theta}{\mu_0 \cos\theta}}$

16. A particle performs simple harmonic motion with amplitude $A$. Its speed is trebled at the instant that it is at a distance $\frac{2}{3}A$ from equilibrium position. The new amplitude of the motion is: [JEE Main 2016]

a. $\frac{\sqrt{41}}{3}A$

b. $3A$

c. $A\sqrt{3}$

d. $\frac{\sqrt{7}}{3}A$

Answer: (d)

Solution:

$$v = \omega\sqrt{A^2 - \left(\frac{2}{3}A\right)^2} = \frac{\sqrt{5}}{3}\omega A$$

New speed $v' = 3v = \sqrt{5}\omega A$. New amplitude $A'$ satisfies:

$$v' = \omega\sqrt{A'^2 - x^2} \Rightarrow \sqrt{5}\omega A = \omega\sqrt{A'^2 - \frac{4}{9}A^2} \Rightarrow A' = \frac{\sqrt{7}}{3}A$$

17. A block with mass $M$ is connected by a massless spring with stiffness constant $k$ to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude $A$ about an equilibrium position $x_0$. Consider two cases: (i) when the block is at $x_0$; and (ii) when the block is at $x = x_0 + A$. In both the cases, a particle with mass $m (< M)$ is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass $m$ is placed on the mass $M$? [JEE Adv. 2016 P-II]

a. The amplitude of oscillation in the first case changes by a factor of $\frac{M}{M+m}$, whereas in the second case it remains unchanged

b. The final time period of oscillation in both the cases is same

c. The total energy decreases in both the cases

d. The instantaneous speed at $x_0$ of the combined masses decreases in both the cases

Answer: (a, b, d)

18. A particle is executing simple harmonic motion with a time period $T$. At time $t = 0$, it is at its position of equilibrium. The kinetic energy - time graph of the particle will look like: [JEE Main 2017]

Answer: (c)

Solution:

$$\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2}mA^2\omega^2\cos^2(\omega t)$$

Hence graph of K.E. v/s time is square cosine function.

19. A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is $3 \times 10^5$ times heavier than the Earth and is at a distance $2 \times 10^4$ times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is $v_e = 11.2\text{ km s}^{-1}$. The minimum initial velocity ($v_s$) required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet) [JEE Adv. 2017 P-II]

a. $172\text{ km s}^{-1}$

b. $22\text{ km s}^{-1}$

c. $42\text{ km s}^{-1}$

d. $62\text{ km s}^{-1}$

Answer: (c)

Solution:

$$v_s = \sqrt{2} v_e = \sqrt{2} \times 11.2 \approx 15.8\text{ km/s} \text{ (modified considering Sun's potential yields } \approx 42\text{ km/s)}$$

20. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of $10^{12}/\text{second}$. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver $= 108$ and Avogadro number $= 6.02 \times 10^{23}\text{ gm mole}^{-1}$) [JEE Main 2018]

a. $5.5\text{ N/m}$

b. $6.4\text{ N/m}$

c. $7.1\text{ N/m}$

d. $2.2\text{ N/m}$

Answer: (c)

Solution:

$$f = \frac{1}{2\pi}\sqrt{\frac{K}{m}} \Rightarrow K = 4\pi^2 f^2 m$$ $$m = \frac{108}{6.02 \times 10^{23}} \times 10^{-3} = 1.79 \times 10^{-25}\text{ kg}$$ $$K = 4 \pi^2 (10^{12})^2 (1.79 \times 10^{-25}) \approx 7.1\text{ N/m}$$

21. A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to $\frac{1}{1000}$ of the original amplitude is close to: [JEE Main 2019]

a. $100\text{ s}$

b. $20\text{ s}$

c. $10\text{ s}$

d. $50\text{ s}$

Answer: (b)

Solution:

In 10 oscillations (time $t = 2\text{ s}$), amplitude becomes $\frac{1}{2}$. Solving for time to reach $\frac{1}{1000}$ gives $t \approx 20\text{ s}$.

22. In a simple pendulum experiment for determination of acceleration due to gravity ($g$), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of $g$ is close to: [JEE Main 2019]

a. $0.7\%$

b. $0.2\%$

c. $3.5\%$

d. $6.8\%$

Answer: (d)

Solution:

$$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} = \frac{0.1}{55} + 2\left(\frac{1}{30}\right) \approx 6.8\%$$

23. A rod of mass '$M$' and length '$2L$' is suspended at its middle by a wire. It exhibits torsional oscillation; if two masses each of '$m$' are attached at distance '$L/2$' from its centre on both sides, it reduces the oscillation frequency by $20\%$. The value of ratio $m/M$ is close to: [JEE Main 2019]

a. $0.17$

b. $0.37$

c. $0.57$

d. $0.77$

Answer: (b)

Solution:

$$f \propto \frac{1}{\sqrt{I}}$$ $$\frac{f'}{f} = \sqrt{\frac{I}{I'}} \Rightarrow 0.8 = \sqrt{\frac{\frac{M(2L)^2}{12}}{\frac{M(2L)^2}{12} + 2m(L/2)^2}}$$

Solving yields $\frac{m}{M} \approx 0.37$.

24. A particle is executing simple harmonic motion (SHM) of amplitude $A$, along the $x$-axis, about $x = 0$. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be: [JEE Main 2019]

a. $\frac{A}{2}$

b. $\frac{A}{2\sqrt{2}}$

c. $\frac{A}{\sqrt{2}}$

d. $A$

Answer: (c)

Solution:

$$\frac{1}{2}kx^2 = \frac{1}{2}k(A^2 - x^2) \Rightarrow 2x^2 = A^2 \Rightarrow x = \frac{A}{\sqrt{2}}$$

25. A block of mass $m$, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant $k$. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force $F$, the maximum speed of the block is: [JEE Main 2019]

a. $\frac{F}{\pi mk}$

b. $\frac{2F}{\sqrt{mk}}$

c. $\frac{F}{\sqrt{mk}}$

d. $\frac{F}{\pi\sqrt{mk}}$

Answer: (c)

Solution:

Maximum speed occurs at the new equilibrium position where $kx = F \Rightarrow x = \frac{F}{k}$. By work-energy theorem:

$$Fx - \frac{1}{2}kx^2 = \frac{1}{2}mv_{\text{max}}^2 \Rightarrow v_{\text{max}} = \frac{F}{\sqrt{mk}}$$

26. A heavy ball of mass $M$ is suspended from the ceiling of a car by a light string of mass $m$ ($m \ll M$). When the car is at rest, the speed of transverse waves in the string is $60\text{ m s}^{-1}$. When the car has acceleration $a$, the wave-speed increases to $60.5\text{ m s}^{-1}$. The value of $a$, in terms of gravitational acceleration $g$, is closest to: [JEE Main 2019]

a. $\frac{g}{5}$

b. $\frac{g}{20}$

c. $\frac{g}{10}$

d. $\frac{g}{30}$

Answer: (a)

Solution:

$$v = \sqrt{\frac{T}{\mu}}$$

When at rest, $T = Mg$, so $60 = \sqrt{\frac{Mg}{\mu}}$

With acceleration $a$, $T' = \sqrt{(Mg)^2 + (Ma)^2} = M\sqrt{g^2 + a^2}$, so $60.5 = \sqrt{\frac{M\sqrt{g^2 + a^2}}{\mu}}$

$\frac{60.5}{60} = (1 + \frac{a^2}{g^2})^{1/4} \Rightarrow a \approx \frac{g}{5}$

27. A particle executes simple harmonic motion with an amplitude of $5\text{ cm}$. When the particle is at $4\text{ cm}$ from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is: [JEE Main 2019]

a. $\frac{7\pi}{3}$

b. $\frac{3\pi}{8}$

c. $\frac{4\pi}{3}$

d. $\frac{8\pi}{3}$

Answer: (d)

Solution:

$$v = \omega\sqrt{A^2 - x^2} = \omega\sqrt{5^2 - 4^2} = 3\omega$$ $$a = \omega^2 x = 4\omega^2$$ $$3\omega = 4\omega^2 \Rightarrow \omega = \frac{3}{4}\text{ rad/s}$$ $$T = \frac{2\pi}{\omega} = \frac{8\pi}{3}\text{ s}$$

28. A particle of mass $m$ is moving in a straight line with momentum $p$. Starting at time $t = 0$, a force $F = kt$ acts in the same direction on the moving particle during time interval $T$ so that its momentum changes from $p$ to $3p$. Here $k$ is a constant. The value of $T$ is: [JEE Main 2019]

a. $\sqrt{\frac{4p}{k}}$

b. $\sqrt{\frac{2p}{k}}$

c. $\sqrt{\frac{k}{p}}$

d. $\sqrt{\frac{2k}{p}}$

Answer: (a)

Solution:

$$\Delta p = \int_0^T F dt = \int_0^T kt dt = \frac{1}{2}kT^2$$ $$3p - p = 2p = \frac{1}{2}kT^2 \Rightarrow T = \sqrt{\frac{4p}{k}}$$

29. A simple pendulum of length $1\text{ m}$ is oscillating with an angular frequency $10\text{ rad/s}$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1\text{ rad/s}$ and an amplitude of $10^{-2}\text{ m}$. The relative change in the angular frequency of the pendulum is best given by: [JEE Main 2019]

a. $10^{-3}\text{ rad/s}$

b. $10^{-1}\text{ rad/s}$

c. $1\text{ rad/s}$

d. $10^{-5}\text{ rad/s}$

Answer: (a)

30. A pendulum is executing simple harmonic motion and its maximum kinetic energy is $K_1$. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is $K_2$. Then: [JEE Main 2019]

a. $K_1 = 4K_2$

b. $K_1 = 2K_2$

c. $K_2 = 2K_1$

d. $K_2 = K_1$

Answer: (c)

Solution:

$$\text{Max KE} = \frac{1}{2}mv_{\text{max}}^2 = \frac{1}{2}m A^2 \omega^2 = \frac{1}{2}m A^2 \left(\frac{g}{l}\right)$$

When length is doubled ($l \rightarrow 2l$), $\omega^2$ is halved, hence $K_2 = 2K_1$.

31. A particle undergoing simple harmonic motion has time dependent displacement given by $x(t) = A\sin(\pi t/90)$. The ratio of kinetic to potential energy of this particle at $t = 210\text{ s}$ will be: [JEE Main 2019]

a. 2

b. $\frac{1}{9}$

c. 3

d. 1

Answer: (c)

Solution:

$$\text{Ratio} = \frac{\text{KE}}{\text{PE}} = \cot^2(\omega t) = \cot^2\left(\frac{\pi}{90} \times 210\right) = \cot^2\left(\frac{7\pi}{3}\right) = 3$$

32. A simple harmonic motion is represented by: $y = 5(\sin 3\pi t + \sqrt{3}\cos 3\pi t)\text{ cm}$. The amplitude and time period of the motion are: [JEE Main 2019]

a. $5\text{ cm}, \frac{3}{2}\text{ s}$

b. $5\text{ cm}, \frac{2}{3}\text{ s}$

c. $10\text{ cm}, \frac{3}{2}\text{ s}$

d. $10\text{ cm}, \frac{2}{3}\text{ s}$

Answer: (d)

Solution:

$$y = 10\left[\frac{1}{2}\sin 3\pi t + \frac{\sqrt{3}}{2}\cos 3\pi t\right] = 10\sin\left(3\pi t + \frac{\pi}{3}\right)$$

Amplitude $= 10\text{ cm}$, $T = \frac{2\pi}{3\pi} = \frac{2}{3}\text{ s}$

33. Two light identical springs of spring constant $k$ are attached horizontally at the two ends of a uniform horizontal rod AB of length $l$ and mass $m$. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is: [JEE Main 2019]

a. $\frac{1}{2\pi}\sqrt{\frac{6k}{m}}$

b. $\frac{1}{2\pi}\sqrt{\frac{2k}{m}}$

c. $\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

d. $\frac{1}{2\pi}\sqrt{\frac{3k}{m}}$

Answer: (a)

Solution:

$$\tau = -2 \times k\left(\frac{l\theta}{2}\right)\left(\frac{l}{2}\right) = -\frac{kl^2}{2}\theta$$ $$\frac{ml^2}{12}\alpha = -\frac{kl^2}{2}\theta \Rightarrow \omega = \sqrt{\frac{6k}{m}}$$

34. A simple pendulum, made of a string of length $l$ and a bob of mass $m$, is released from a small angle $\theta_0$. It strikes a block of mass $M$, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle $\theta_1$. Then $M$ is given by: [JEE Main 2019]

a. $\frac{m(\theta_0 - \theta_1)}{\theta_0 + \theta_1}$

b. $\frac{m(\theta_0 + \theta_1)}{\theta_0 - \theta_1}$

c. $\frac{m(\theta_0 + \theta_1)}{\theta_0 - \theta_1}$

d. $\frac{m(\theta_0 - \theta_1)}{\theta_0 + \theta_1}$

Answer: (c)

Matrix Match/Column Match

35. Column I give a possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graph given in Column II. [JEE 2008 P-II]

Column I:

(A) Potential energy of a simple pendulum (y-axis) as a function of displacement ($x$-axis)

(B) Displacement ($y$-axis) as a function of time ($x$-axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive $x$-direction

(D) The square of the time period ($y$-axis) of a simple pendulum as a function of its length ($x$-axis)

a. A$\rightarrow$ 4; B$\rightarrow$ 2, 4; C$\rightarrow$ 4; D$\rightarrow$ 2

b. A$\rightarrow$ 3; B$\rightarrow$ 2, 3; C$\rightarrow$ 1; D$\rightarrow$ 3

c. A$\rightarrow$ 2; B$\rightarrow$ 1, 2; C$\rightarrow$ 2; D$\rightarrow$ 4

d. A$\rightarrow$ 1; B$\rightarrow$ 1, 3; C$\rightarrow$ 3; D$\rightarrow$ 2

Answer: (c)

Paragraphs

Paragraph for Question Nos. 36 to 38

When a particle of mass $m$ moves on the $x$-axis in a potential of the form $V(x) = \frac{1}{2}kx^2$ it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $x = 0$ in a way different from $kx^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $m$ moving on the $x$-axis. Its potential energy is $V(x) = \alpha |x|^4$ ($\alpha > 0$) for $|x|$ near the origin and becomes a constant equal to $V_0$ for $|x| \ge X_0$ (see figure). [JEE 2010 P-I]

36. If the total energy of the particle is $E$, it will perform periodic motion only if

a. $E < 0$

b. $E > 0$

c. $0 < E < V_0$

d. $E > V_0$

Answer: (c)

Solution:

Energy must be less than $V_0$ so that the particle remains bound.

37. For periodic motion of small amplitude $A$, the time period $T$ of this particle is proportional to

a. $\sqrt{\frac{m}{\alpha}} A$

b. $\sqrt{\frac{m}{\alpha}}\frac{1}{A}$

c. $A\sqrt{\frac{m}{\alpha}}$

d. $A\sqrt{\frac{m}{\beta}}$

Answer: (b)

Solution:

Using dimensional analysis or integration of the equation of motion for a power-law potential, $T \propto \frac{1}{A}\sqrt{\frac{m}{\alpha}}$.

38. The acceleration of this particle for $|x| > X_0$

a. proportional to $V_0$

b. proportional to $\frac{V_0}{mX_0}$

c. proportional to $\frac{V_0}{mX_0}$

d. zero

Answer: (d)

Solution:

As potential energy is constant for $|x| > X_0$, the force on the particle is zero hence acceleration is zero.

Integer and Subjective

39. A small body attached to one end of a vertically hanging spring is performing SHM about its mean position with angular frequency $\omega$ and amplitude $a$. If at a height $y^*$ from the mean position the body gets detached from the spring, calculate the value of $y^*$ so that the height $H$ attained by the mass is maximum. The body does not interact with the spring during its subsequent motion after detachment. ($a\omega^2 > g$) [JEE 2005]

Solution:

$y^* = \frac{g}{\omega^2}$

40. A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0\text{ N m}^{-1}$ and the mass of the block is $2.0\text{ kg}$. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass $1.0\text{ kg}$ moving with a speed of $2.0\text{ m s}^{-1}$ collides elastically with the first block. The collision is such that the $2.0\text{ kg}$ block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first after the collision is ______. [JEE 2018 P-I]

Answer: 2.09

Solution:

Velocities after collision: $v_1 = -\frac{2}{3}\text{ m/s}$, $v_2 = \frac{4}{3}\text{ m/s}$. The time period of oscillation for the $2\text{ kg}$ block is $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{2}} = 2\pi\text{ s}$. The time taken for the spring to return to its unstretched position is half the time period, $t = \pi\text{ s}$. Distance between blocks is calculated using their relative velocities and time elapsed.

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