Q. A frictionless 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) $\frac{2\sqrt{gR}}{g\cos\theta}$
B) $2\sqrt{gR}.\frac{\cos\theta}{g}$
C) $2\sqrt{\frac{R}{g}}$
D) $\frac{gR}{\sqrt{g\cos\theta}}$
Explanation
(c) Acceleration of the body along $AB$ is $g\cos\theta$
Distance travelled in time $t\sec{= AB = \frac{1}{2}}(g\ cos\ \theta)t^{2}$
From $\mathrm{\Delta}\ ABC,\ AB = 2R\cos\theta;2R\cos\theta = \frac{1}{2}g\cos{\theta t^{2}}$
$\Rightarrow t^{2} = \frac{4R}{g}$ or $t = 2\sqrt{\frac{R}{g}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.