Question

A particle slides from rest from the highest point of a vertical circle of radius $r$, along a smooth chord. Time of descent of the particle along the chord is

• A. $\sqrt{\frac{r}{4g}}$
• B. $\sqrt{2rg}$
• C. $\sqrt{\frac{2r}{g}}$
• D. $\sqrt{\frac{4r}{g}}$

Answer 1

Answer: (D)

$\sqrt{\frac{4r}{g}}$

Note, $H{H}^{\prime }=2H{O}^{\prime }=2r\cos a$
Using $s=1/2a{t}^2$, we get
$2r\cos a=1/2\left(g\cos \alpha \right)t^2$
or $t^2=\frac{4r}{g}$ or $t=\sqrt{\frac{4r}{g}}$