Question

A frictionless wire is fixed between A and B inside a sphere of radius R. A bead slips along the wire. The time taken by the bead to slip from A to B will be:

• A. $2\sqrt{R/g}$
• B. $gR/\sqrt{gcos\theta }$
• C. $\frac{2\sqrt{gr}}{gcos\theta }$
• D. $\frac{2\sqrt{gRcos\theta }}{g}$

Answer 1

The correct option is A $2\sqrt{R/g}$

When the bead travels from A to B,
$AB=2Rcos\theta$ .............. from geometry.
Also, initial velocity is zero and the component of acceleration acting along AB is $gcos\theta$.
$\therefore$ Using kinematic equations,
$s=ut+\frac{1}{2}a{t}^2$
$\therefore 2Rcos\theta =\frac{1}{2}gcos\theta t^2$
$\therefore \frac{4R}{g}=t^2$
$\therefore t=2\sqrt{\frac{R}{g}}$