question_answer

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

A) \[\frac{2\sqrt{\mathrm{R}}}{\sqrt{\mathrm{g}}\,\mathrm{cos}\,\theta }\]              

B) \[\frac{2\sqrt{\mathrm{R}}\,\mathrm{cos}\,\theta }{\mathrm{g}\,}\]

C) \[2\sqrt{\frac{\mathrm{R}}{\mathrm{g}\,}}\]                  

D) \[\frac{\sqrt{\operatorname{gR}}}{\cos \,\theta \,}\]

Correct Answer: C

Solution :

\[\frac{PR}{2R}=cos\,\,\theta \] \[PR=2\operatorname{R}\,\,cos\,\,\theta \]                              ?.(1) \[a=g\,\,cos\,\,\theta \] \[PR=\frac{1}{2}g\,\,cos\,\,\theta \,\,{{\operatorname{t}}^{2}}\]                  ?.(2) \[2R\,\,cos\,\theta =\frac{1}{2}g\,\,cos\,\theta \,\,{{\operatorname{t}}^{2}}\Rightarrow \operatorname{t}=2\sqrt{\frac{R}{g}}\]