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(A) Let the first cylinder be displaced downwards by a distance $x$. Then the second cylinder will be displaced upwards by the same distance $x$ becau

Vedclass · Accepted Answer

$T = 2\pi \sqrt {\frac{{2v}}{{gs}}\,\frac{{{\rho _0}}}{{\left( {{\rho _1} + {\rho _2}} \right)}}} $

Vedclass · Answer

(A) Let the first cylinder be displaced downwards by a distance $x$. Then the second cylinder will be displaced upwards by the same distance $x$ because they are connected by a string over pulleys.The change in the buoyant force on the first cylinder is $\Delta F_1 = \rho_1 s x g$ (upward force decreases as it moves down,but here the net restoring force is considered).When the first cylinder moves down by $x$,the buoyant force on it increases by $\rho_1 s x g$ (acting upwards). When the second cylinder moves up by $x$,the buoyant force on it decreases by $\rho_2 s x g$ (acting upwards).The net restoring force on the system is $F_{net} = -(\rho_1 s x g + \rho_2 s x g) = -(\rho_1 + \rho_2) s g x$.The total mass of the system is $2m = 2\rho_0 v$.Using Newton's second law,$2m a = F_{net} \implies 2\rho_0 v a = -(\rho_1 + \rho_2) s g x$.$a = -\left[ \frac{(\rho_1 + \rho_2) s g}{2 \rho_0 v} \right] x$.Comparing this with the $SHM$ equation $a = -\omega^2 x$,we get $\omega = \sqrt{\frac{(\rho_1 + \rho_2) s g}{2 \rho_0 v}}$.The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{2 \rho_0 v}{(\rho_1 + \rho_2) s g}}$.Since $m = \rho_0 v$,we have $T = 2\pi \sqrt{\frac{2m}{(\rho_1 + \rho_2) s g}}$. Substituting $m = \rho_0 v$,we get $T = 2\pi \sqrt{\frac{2 \rho_0 v}{(\rho_1 + \rho_2) s g}}$.

Column $I$	Column $II$
$(A)$ The object moves on the $x$-axis under a conservative force such that its speed $v = c_1 \sqrt{c_2 - x^2}$,where $c_1, c_2 > 0$.	$(p)$ The object executes simple harmonic motion.
$(B)$ The object moves on the $x$-axis such that its velocity $v = -kx$,where $k > 0$.	$(q)$ The object does not change its direction.
$(C)$ An object is attached to a spring in an elevator accelerating upwards with constant acceleration $a$. The motion is observed from the elevator.	$(r)$ The kinetic energy of the object keeps on decreasing.
$(D)$ The object is projected vertically upwards with speed $2 \sqrt{GM_e / R_e}$.	$(s)$ The object can change its direction only once.

List-$I$	List-$II$
$A$. Velocity is maximum	$I$. Acceleration is maximum
$B$. $K.E.$ is $\left(\frac{3}{4}\right)^{\text{th}}$ of total energy	$II$. At mean position
$C$. $P.E.$ is $\left(\frac{3}{4}\right)^{\text{th}}$ of total energy	$III$. At half of the amplitude
$D$. Acceleration is maximum	$IV$. At $\frac{\sqrt{3}}{2}$ times the amplitude

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