A cylindrical cork of uniform density floats | vedclass

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(D) Let the cross-sectional area of the cork be $A$, its density be $\rho$, and its length be $L$. When the cork floats in a liquid of density $\rho_L

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$1$/$4$

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(D) Let the cross-sectional area of the cork be $A$, its density be $\rho$, and its length be $L$. When the cork floats in a liquid of density $\rho_L$, the submerged length $h$ is given by the condition of equilibrium: $A h \rho_L g = A L \rho g$, so $h = (\rho / \rho_L) L$.When the cork is depressed by a small distance $x$, the additional buoyant force acting on it is $F = -A x \rho_L g$. This force acts as a restoring force.The equation of motion is $m a = -A \rho_L g x$, where $m = A L \rho$ is the mass of the cork.Thus, $A L \rho (d^2x/dt^2) = -A \rho_L g x$, which gives $(d^2x/dt^2) + (\rho_L g / L \rho) x = 0$.This is the equation of simple harmonic motion with angular frequency $\omega = \sqrt{\rho_L g / L \rho}$.The time period is $T = 2\pi / \omega = 2\pi \sqrt{L \rho / \rho_L g}$.Since $T \propto 1/\sqrt{\rho_L}$, we have $T_1 / T_2 = \sqrt{\rho_2 / \rho_1}$.Given $T_1 = T$ and $T_2 = 2T$, we have $T / 2T = \sqrt{\rho_2 / \rho_1}$, which implies $1/2 = \sqrt{\rho_2 / \rho_1}$.Squaring both sides, we get $\rho_2 / \rho_1 = 1/4$.

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