Question

A frictionless wire $AB$ 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. $\sqrt{\frac{4R}{g{\cos }^2\theta }}$
• B. $\sqrt{\frac{4R{\cos }^2\theta }{g}}$
• C. $\sqrt{\frac{4R}{g}}$
• D. $\sqrt{\frac{g{R}^2}{\cos \theta }}$

Answer 1

The correct option is C $\sqrt{\frac{4R}{g}}$

The component of acceleration along $AB$ would be $gcos\theta$.
Also distance traveled by the ball will be the length of the chord $AB$ which is equal to $2Rcos\theta$
Using the equation $s=ut+\frac{1}{2}a{t}^2$, we get:
$2Rcos\theta =\frac{1}{2}gcos\theta t^2$
or $t=\sqrt{\frac{4R}{g}}$