Question

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

Answer 1

The Correct Option is C

Acceleration of body along $AB$ is $g \cos \theta$ Distance travelled in time $t \,sec = AB =\frac{1}{2}( g \cos \theta) t ^{2}$ From $\triangle ABC , AB =2 R \cos \theta$ Thus, $2 R \cos \theta=\frac{1}{2} g \cos \theta t^{2}$ $\Rightarrow t^{2}=\frac{4 R}{g} \Rightarrow t=2 \sqrt{\frac{R}{g}}$