Question

A particle slides from rest from the topmost point of a smooth chord of a vertical ring making an angle $\theta$ with the vertical. The time of descent is

• A. least for $\theta =0$ and maximum for $\theta ={45}^{\circ }$
• B. maximum for $\theta =0$
• C. least for $\theta ={45}^{\circ }$ and maximum for $\theta =0$
• D. independent of $\theta$

Answer 1

The correct option is D independent of $\theta$

Acceleration along the chord is
$a=g\sin \alpha =g\sin ({90}^{\circ }-\theta )=g\cos \theta$
Length of the chord is $l=2R\cos \theta$

Now, using $s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow l=0\times t+\frac{1}{2}g\cos \theta t^2$
$\Rightarrow 2R\cos \theta =\frac{1}{2}g\cos \theta t^2$
$\Rightarrow t=\sqrt{\frac{4R}{g}}$
This is independent of $\theta$.