Question

A small ball of radius $2\text{cm}$ executes SHM in a hemispherical bowl of radius $42\text{cm}$. Find the time period of oscillation. Take $\pi =3.14$ and $g=10{\text{m/s}}^2$

• A. $2.5\text{s}$
• B. $5\text{s}$
• C. $1.25\text{s}$
• D. $0.75\text{s}$

Answer 1

The correct option is C $1.25\text{s}$

Given,
Radius of ball $\left(r\right)=2\text{cm}$
Radius of hemispherical bowl $\left(R\right)=42\text{cm}$
The oscillation of small spherical ball is analogous to the oscillation of a simple pendulum bob with length of string as, $l=R-r$

Hence time period of oscillation is,
$T=2\pi \sqrt{\frac{l}{g}}$
$T=2\pi \sqrt{\frac{R-r}{g}}$
$T=2\times 3.14\times \sqrt{\frac{0.4}{10}}$
$T=2\times 3.14\times 0.2$
$\therefore T=1.25\text{s}$