Question

A spherical ball of radius $5.5\text{cm}$ is oscillating in a hemispherical bowl. If the time taken to complete one oscillation by the ball is $\frac{\pi }{2}s$, then find the radius of hemispherical bowl.

• A. $68\text{cm}$
• B. $60\text{cm}$
• C. $58\text{cm}$
• D. $32\text{cm}$

Answer 1

The correct option is A $68\text{cm}$

Given,
Radius of spherical ball $r=5.5\text{cm}=0.055\text{m}$
The oscillation of small spherical ball is analogous to the oscillation of a simple pendulum bob with length of string as, $l=R-r$

Let $R$ be the radius of hemispherical bowl
Hence time period of oscillation is,
$T=2\pi \sqrt{\frac{l}{g}}$
$T=2\pi \sqrt{\frac{R-r}{g}}$
Given, $T=\frac{\pi }{2}\text{s}$
Hence on equating both,
$\frac{\pi }{2}=2\pi \sqrt{\frac{R-0.055}{10}}$
$\Rightarrow \frac{R-0.055}{10}=\frac{1}{16}$
$\Rightarrow R=\frac{10}{16}+0.055$
$\therefore R=0.68\text{m}=68\text{cm}$