Question

A solid sphere of density $\rho$ and radius R is floating in a liquid of density $\sigma$ with half its volume submerged. When the sphere pressed down slightly and released, it executes simple harmonic motion of time period T. If viscous effect is ignored, then

• A. $\sigma =2\rho$
• B. $\rho =2\sigma$
• C. $T=2\pi \sqrt{\frac{2R}{3g}}$
  
• D. $T=2\pi \sqrt{\frac{3R}{2g}}$

Answer 1

Answer: (A), (C)

The correct options are
A

$\sigma =2\rho$

C

$T=2\pi \sqrt{\frac{2R}{3g}}$

Mass of sphere is

$m=V\rho \dots \left(1\right)$

where V is the volume of the sphere. The volume of sphere under water = volume of water displaced =$\frac{V}{2}$ . If $\sigma$ is the density of water, the upthrust is
$U=\frac{1}{2}\left(V\sigma g\right)$

From the law of flotation, upthrust = weight of the sphere, i.e. U = mg or $frac12\left(V\sigma g\right)=V\rho g$

or $\sigma =2\rho \dots \left(2\right)$

If the sphere is pressed down through a small distance x (fig.), the volume of water displaced due to this pressing = volume of a disc of radius R (since the half the sphere is submerged) and thickness x which is $\pi R^{2}x\sigma g$. Hence the restoring force acting on the sphere (buoyant force) when it is released is given by
$F=–\pi R^2\sigma gx$

Therefore, the acceleration of the sphere is [use Eq. (1)]
$a=\frac{F}{m}=\frac{\pi R^2\sigma gx}{\frac{4}{3}\pi R^3\rho }=\frac{3g}{4}\left(\frac{\sigma }{\rho }\right)\frac{x}{R}$

Using Eq. (2) we get

$a=–\left(\frac{3g}{2R}\right)x=–{\omega }^{2}x\dots \left(3\right)$

which gives $T=2\pi \sqrt{\frac{2R}{3g}}$. Hence the correct choices are (a) and (c)