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\[V=-\frac{GM}{R}\] \[=-\frac{GM}{L/\pi }=\frac{\pi GM}{L}\]You need to login to perform this action.
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