A) h=2R
B) \[h=\frac{R}{2}\]
C) h=R
D) \[h=\frac{R}{3}\]
Correct Answer: D
Solution :
From law of conservation of energy potential energy of fall gets converted to kinetic energy \[\therefore \] \[PE=KE\] \[mgh=\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow \] \[v=\sqrt{2gh}\] ?.. (i) Also, the horizontal component of force is equal centrifugal force. \[\therefore \] \[Mg\cos \theta =\frac{m{{v}^{2}}}{R}\] ?. (ii) From Eq. (i), \[v=\sqrt{2gh}\] \[\therefore \] \[mg\cos \theta =\frac{2mgh}{R}\] ?. (iii) From \[\Delta \,AOB,\] \[\cos \theta =\frac{2R-h}{R}\] \[\Rightarrow \] \[Mg\left( \frac{R-h}{R} \right)=\frac{2mgh}{R}\] \[\Rightarrow \] \[Mg\left( \frac{R-h}{R} \right)=\frac{2mgh}{R}\] \[\Rightarrow \] \[3h=R\] \[\Rightarrow \] \[h=\frac{R}{3}\]You need to login to perform this action.
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