question_answer

A block of mass m is placed on a smooth sphere of radius R. It slides when pushed slightly. At what distance h, from the top, will it leave the sphere?

A)  \[\frac{R}{4}\]                                 

B)  \[\frac{R}{3}\] 

C)   \[\frac{R}{2}\]                                

D)  \[R\]

Correct Answer: B

Solution :

                 Suppose the block will leave the sphere at point B, which is at a distance h from the sphere                        \[\frac{m{{v}^{2}}}{R}=mg\,\cos \theta N\] When block leaves the sphere at point B, the normal reaction N becomes zero \[\therefore \]  \[\frac{m{{v}^{2}}}{R}=mg\cos \theta \]                 \[\cos \theta =\frac{{{v}^{2}}}{Rg}\] From figure                 \[\cos \theta =\frac{R-h}{R}\] \[\therefore \]  \[\frac{R-h}{R}=\frac{{{v}^{2}}}{Rg}\]                 \[\frac{R-h}{R}=\frac{2gh}{Rg}\]               \[[\therefore \,{{v}^{2}}=2gh]\]                 \[h=\frac{R}{3}\]