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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 oscillation with a time period $T$. Then :-
A
$T \propto \sqrt{\rho}$
B
$T \propto \dfrac{1}{\sqrt{A}}$
C
$T \propto \dfrac{1}{\rho}$
D
$T \propto \dfrac{1}{\sqrt{m}}$
Detailed Solution
In equilibrium the weight of the block is balanced by the upthrust.
When the block is pushed down by a small distance $x$, an extra volume $Ax$ of liquid is displaced, giving an extra upthrust $Ax\rho g$.
Restoring force $= Ax\rho g = kx \Rightarrow k = A\rho g$
The restoring force is proportional to the displacement and opposite to it, so the motion is simple harmonic.
$T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{m}{A\rho g}}$
So $T \propto \sqrt{m}$, $T \propto \dfrac{1}{\sqrt{\rho}}$ and $T \propto \dfrac{1}{\sqrt{A}}$.
Among the given relations, the correct one is $T \propto \dfrac{1}{\sqrt{A}}$.
When the block is pushed down by a small distance $x$, an extra volume $Ax$ of liquid is displaced, giving an extra upthrust $Ax\rho g$.
Restoring force $= Ax\rho g = kx \Rightarrow k = A\rho g$
The restoring force is proportional to the displacement and opposite to it, so the motion is simple harmonic.
$T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{m}{A\rho g}}$
So $T \propto \sqrt{m}$, $T \propto \dfrac{1}{\sqrt{\rho}}$ and $T \propto \dfrac{1}{\sqrt{A}}$.
Among the given relations, the correct one is $T \propto \dfrac{1}{\sqrt{A}}$.