NEET miscellaneous MCQs & PYQs

Practice NEET miscellaneous Questions

Question 1:

moderate

Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to paths of the two particles. The phase difference is:

(2011 Mains)

Let displacement be $x = A\sin(\omega t + \phi)$. When they cross, $x = A/2$. $A/2 = A\sin(\phi) \Rightarrow \sin(\phi) = 1/2$. The two phases are $\pi/6$ and $5\pi/6$ (since moving in opposite directions). Phase difference $$= 5\pi/6 - \pi/6 = 4\pi/6 = 2\pi/3$$.

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Question 2:

moderate

A rectangular block of mass $m$ and area of cross-section $A$ floats in a liquid of density $\rho$. If it is given a small vertical displacement from equilibrium it undergoes with a time period $T$, then

(2006)

Restoring force on the block is $F = -(\rho A g)y$. Acceleration $a = F/m = -(\frac{\rho A g}{m})y$. This is SHM with $\omega^2 = \frac{\rho A g}{m}$. Time period $T = 2\pi\sqrt{\frac{m}{\rho A g}}$. Therefore, $T \propto \frac{1}{\sqrt{A}}$.

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Question 3:

moderate

The equations of two waves given as $x = a\cos(\omega t + \delta)$ and $y = a\cos(\omega t + \alpha)$, Where $\delta = \alpha + \pi/2$, then resultant wave represent:

(2000)

Substitute $\delta$: $x = a\cos(\omega t + \alpha + \pi/2) = -a\sin(\omega t + \alpha)$.\nCombining with $y = a\cos(\omega t + \alpha)$ yields $x^2 + y^2 = a^2$, a circle.\nThe direction of motion traces anti-clockwise as time progresses.

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Question 4:

easy

Two S.H.M.s with same amplitude and time period, when acting together in perpendicular directions with a phase difference of $\pi/2$, give rise to:

(1997)

Superposition of two mutually perpendicular SHMs of equal amplitude and a phase difference of $\pi/2$ produces a circular trajectory.\n$x = A\sin(\omega t)$ and $y = A\sin(\omega t + \pi/2) = A\cos(\omega t)$ gives $x^2 + y^2 = A^2$.

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Question 5:

moderate

The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi$ results in the displacement of the particle along:

(1990)

Let $x = A\sin(\omega t)$ and $y = B\sin(\omega t + \pi) = -B\sin(\omega t)$.\nThen $y/x = -B/A \implies y = -(B/A)x$.\nThis represents the equation of a straight line.

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Question 6:

moderate

Two pendulums of length $121 \text{ cm}$ and $100 \text{ cm}$ start vibrating in phase. At some instant, the two are at their means position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the means position is :

(2022)

$T \propto \sqrt{l}$. So $T_1/T_2 = \sqrt{121/100} = 11/10$. This gives $10 T_1 = 11 T_2$. The shorter pendulum ($T_2$) completes $11$ vibrations.

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Question 7:

easy

The amplitude of a S.H.M. reduces to $1/3$ in first $20 \text{ secs}$, then in first $40 \text{ sec.}$ its amplitude becomes:

(1999)

In damped S.H.M., amplitude at time $t$ is $A(t) = A_0 e^{-bt}$. At $t = 20 \text{ s}$, $A(20) = A_0 e^{-20b} = \frac{A_0}{3}$. At $t = 40 \text{ s}$, $A(40) = A_0 e^{-40b} = A_0 (e^{-20b})^2 = A_0 (\frac{1}{3})^2 = \frac{A_0}{9}$.

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