A) \[mg/2\]
B) \[mg\]
C) \[mg/4\]
D) \[mg/8\]
Correct Answer: C
Solution :
Let the tension in the string be T and the acceleration of the centre of mass of the rod downwards be a. Then, \[mg-T=ma\] Again, \[\frac{mgl}{2}=I\,\,\alpha =\frac{m{{l}^{2}}}{3}\times \frac{a}{l/2}\] \[\Rightarrow \] \[a=\frac{3}{4}g\] \[\therefore \] \[T=\frac{mg}{4}\]You need to login to perform this action.
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