Question

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

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

Answer 1

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



$\frac{PC}{2R}=cos\theta PC=2Rcos\theta ...\left(1\right)a=gcos\theta PC=\frac{1}{2}gcos\theta t^2....\left(2\right)2Rcos\theta =\frac{1}{2}gcos\theta t^2\Rightarrow t=2\sqrt{\frac{R}{g}}$