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 \[\cos \theta =\frac{R-h}{R}\] \[\therefore \] \[\frac{R-h}{R}=\frac{{{v}^{2}}}{Rg}\] \[\frac{R-h}{R}=\frac{2gh}{Rg}\] \[[\therefore \,{{v}^{2}}=2gh]\] \[h=\frac{R}{3}\]You need to login to perform this action.
You will be redirected in
3 sec