Question

A ball of $1$ $gm$ rolls down an inclined plane and describes a circle of radius $10$ $cm$ in the vertical plane on reaching the bottom. The minimum height of the inclined plane is :

• A. $35$ $cm$
• B. $5\sqrt{2}cm$
• C. $25$ $cm$
• D. $5\sqrt{3}cm$

Answer 1

The correct option is A $35$ $cm$

$R=10cm.$ (given)
In case of Solid spherical ball.
For no slipping
V bottom $=\sqrt{5gR}$
(for completing vertical circular motion)

for finding this velocity we can apply apply Work Energy Theorem
$mgH=\frac{1}{2}m{V}^2+\frac{1}{2}\times \frac{2}{5}\times m{R}^2(\frac{V}{R}{)}^2$
$10H=\frac{7}{10}{V}^2$
$H=\frac{7}{100}\times 5\times 10\times \frac{10}{100}$
$H=0.35m$
$=35cm$