Rankers Physics

Gravitational Potential: Practice Problem & Solution

A thin rod of length L is bent to form a semicircle. The mass of rod is M. What will be the gravitational potential at the centre of the circle
\[ \frac{-GM}{L}\]
\[ \frac{-GM}{2\pi L} \]
\[ \frac{- \pi GM}{2L}\]
\[ \frac{- \pi GM}{L} \]

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Gravitational Potential. Understanding the underlying formula is key to arriving at the correct answer below:

The gravitational potential at the center due to a semicircular rod is given by:

\[
V = - \frac{GM}{R}
\]

Where \( R \) is the radius of the semicircle, which is related to the length of the rod:

\[
L = \pi R \quad \Rightarrow \quad R = \frac{L}{\pi}
\]

Substituting \( R \) into the potential formula:

\[
V = - \frac{GM}{\frac{L}{\pi}} = - \frac{\pi GM}{L}
\]

Thus, the gravitational potential at the center is:

\[
V = - \frac{\pi GM}{L}
\]

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