Question

A friction less wire AB is fixed on a sphere of radius R. A very small spherical ball slips on this wire . The time taken by this ball to slip from A to B is:

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

Answer 1

The correct option is C 2$\sqrt{\frac{R}{g}}$

The correct option is C

Given,

A wire AB and a sphere of radius R

Acceleration of body along $AB=gcos\theta$

Distance traveled in time $tsec$ is:

$AB=\frac{1}{2}\left(gcos\theta \right)t^2$

Now,

From $△ABC,AB=2Rcos\theta$

$2Rcos\theta =\frac{1}{2}\left(gcos\theta \right)t^2$

$t^2=\frac{4R}{g}$

$t=2\sqrt{\frac{R}{g}}$

Thus, the time taken by this ball to slip from A to B is

$t=2\sqrt{\frac{R}{g}}$