question_answer

A thin rod of length L is bent to form a semicircle. The mass of the rod is M. The gravitation potential at the center of semi-circle would be

A) \[-\frac{GM}{L}\]                            

B) \[-\frac{GM}{2\pi L}\]

C) \[-\frac{\pi GM}{2L}\]                   

D) \[-\frac{\pi GM}{L}\]

Correct Answer: D

Solution :

\[\pi R=L\] \[\Rightarrow \]\[R=\frac{L}{\pi }\]Potential at centre is\[V=-\frac{GM}{R}\] \[=-\frac{GM}{L/\pi }=\frac{\pi GM}{L}\]