Question

A particle moves down the inclined surface and completes a vertical circle at the foot of the plane. Ratio of heights, $H$ and $h$ is :

• A. $5$
• B. $4$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$5$

Velocity at the bottom of the circle $=\sqrt{5gr}$
Velocity at the top of the circle $=\sqrt{gr}$
Using P.E = K.E., we get
$mgH=\frac{1}{2}m(\sqrt{5gr}{)}^2=\frac{1}{2}m(5gr)$ (i)
also $mgh=\frac{1}{2}m(\sqrt{gr}{)}^2=\frac{1}{2}mgr$ (ii)
Dividing (i) and (ii), we get
$\frac{H}{h}=5$