Question

A sphere of radius $R$ is half submerged in liquid of density $\rho$. If the sphere is slightly pushed down and released, it oscillates, then what is the frequency of its oscillation ?

• A. $\frac{1}{2\pi }\sqrt{\frac{g}{2R}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{2g}{3R}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{g}{3R}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{3g}{2R}}$

Answer 1

The correct option is D $\frac{1}{2\pi }\sqrt{\frac{3g}{2R}}$

A sphere of radius $R$ is half submerged in a liquid of density $\rho$.
At equilibrium ,
Weight of sphere = Upthrust of liquid on sphere.


$V\sigma g=\frac{V}{2}\rho g,$
where $\sigma$ = density of sphere.
$\therefore \sigma =\frac{\rho }{2}......\left(1\right)$
From this position, the sphere is slightly pushed down. Upthrust of liquid on the sphere will increase and it will act as restoring force.
$\therefore$ Restoring force = Upthrust due to extra immersion
F = - (extra volume immersed) $\times \rho g$
$\frac{4}{3}\pi R^3\times \sigma \times a=-\pi R^2\rho gx$
$\Rightarrow a=-\frac{3g\rho }{4R\sigma }x$
or $a=-\frac{3g}{2R}x$, [From $\left(1\right)$]
$\Rightarrow a\propto -x$
Hence the motion is simple harmonic.

Comparing this with $a=-{\omega }^{2}x$ we get,
$\omega =\sqrt{\frac{3g}{2R}}$
$\therefore$ Frequency of oscillation $f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{3g}{2R}}$
Thus, option (d) is the correct answer.