Question

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

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

Answer 1

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

In$△ABD,\cos \theta =\frac{AB}{2R}$ or $AB=2R\cos \theta$
Along $AB,$ acceleration acting on ball $=g\cos \theta$
Using $s=ut+0.5a{t}^2$ Here $s=AB,a=g\cos \theta$ and $u=0$
So $AB=0.5g\cos \theta t^2$ or $t^2=\frac{2AB}{g\cos \theta }$ or $t^2=\frac{2\times 2R\cos \theta }{g\cos \theta }=\frac{4R}{g}$ or $t=2\left(\sqrt{\frac{R}{g}}\right)$