Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

$T \propto \sqrt{\rho}$

$T \propto \frac{1}{\sqrt{A}}$

$T \propto \frac{1}{\rho}$

$T \propto \frac{1}{\sqrt{m}}$

Step-by-Step Solution

Identify the Restoring Force: When the block is depressed by a small distance $x$ , an additional volume of liquid $V = Ax$ is displaced. According to Archimedes' Principle, this creates an additional upthrust (buoyant force) which acts as the restoring force. $F_{restoring} = -(\text{Weight of displaced liquid}) = -(Ax\rho)g$
Determine Spring Constant ( $k$ ): Comparing this to the standard SHM equation $F = -kx$ , we identify the effective force constant $k = A\rho g$ .
Apply Time Period Formula: The time period of a particle/body in SHM is given by $T = 2\pi \sqrt{\frac{m}{k}}$ . Substituting $k = A\rho g$ : $T = 2\pi \sqrt{\frac{m}{A\rho g}}$
Analyze Proportionality: From the derived formula, the relationship between time period $T$ and cross-sectional area $A$ is: $T \propto \frac{1}{\sqrt{A}}$ (Note: It is also proportional to $\sqrt{m}$ and inversely proportional to $\sqrt{\rho}$ ).

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