Q. 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) $T \propto \frac{1}{\rho}$
B) $T \propto \frac{1}{\sqrt{m}}$
C) $T \propto \sqrt{\rho}$
D) $T \propto \frac{1}{\sqrt{A}}$
Explanation

(d) Let $l$ be the length of block immersed in liquid as shown in the figure. When the block is floating.
[image:14_image60.emf]
$\therefore mg = Al\rho g$
If the block is given vertical displacement $y$ then the effective restoring force is
$F = - \left\lbrack A(l + y)\rho g - mg \right) = - \left\lbrack A(l + y)\rho g - Al\rho g \right\rbrack = - (Al\rho g)y$
$i.e.,\ F \propto y.$ As this $F$ is directed towards its equilibrium position of block, so if the block is left free, it will execute simple harmonic motion.
Here inertia factor = mass of block $= m$
Spring factor $= A\rho g$
$\therefore Time\ period = T = 2\pi\sqrt{\frac{m}{A\rho g}}$
$i.e.\ T \propto \frac{1}{\sqrt{A}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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