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(B) Let $a$ be the side of the cube,$\rho$ be the density of the cube,and $\sigma$ be the density of the mercury.The mass of the cube is $M = a^3 \rho

Vedclass · Accepted Answer

$2\pi \sqrt {\frac{{\rho \,a}}{{\sigma \,g}}} $

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(B) Let $a$ be the side of the cube,$\rho$ be the density of the cube,and $\sigma$ be the density of the mercury.The mass of the cube is $M = a^3 \rho$.In equilibrium,the weight of the cube is balanced by the buoyant force: $Mg = a^2 h \sigma g$,where $h$ is the depth of immersion.Thus,$a^3 \rho g = a^2 h \sigma g$,which gives $h = \frac{a \rho}{\sigma}$.When the cube is pushed down by a small vertical displacement $y$,the additional buoyant force provides the restoring force:$F_{restoring} = - (a^2 y) \sigma g$.The equation of motion is $M \frac{d^2y}{dt^2} = - (a^2 \sigma g) y$.Substituting $M = a^3 \rho$:$a^3 \rho \frac{d^2y}{dt^2} = - a^2 \sigma g y \implies \frac{d^2y}{dt^2} = - \left( \frac{\sigma g}{a \rho} \right) y$.Comparing this with the standard $S.H.M.$ equation $\frac{d^2y}{dt^2} = - \omega^2 y$,we get $\omega^2 = \frac{\sigma g}{a \rho}$.The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{a \rho}{\sigma g}}$.

$A$ brass cube of side $a$ and density $\rho$ is floating in mercury of density $\sigma$. If the cube is displaced a bit vertically,it executes $S.H.M.$ Its time period will be

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