A thin rod of length L is bent to form a semicircle. The mass of the rod is M. What will be the gravitational potential at the centre of the circle

The correct option is -πGM/L.

Let us find the gravitational potential at the center of the semicircle step-by-step.

Step 1: Relate the length of the rod to the radius of the semicircle
Let $R$ be the radius of the semicircle. A thin rod of length $L$ is bent to form this semicircle. The perimeter (length) of a semicircle of radius $R$ is given by:

$L=\pi R$

From this relation, we can express the radius $R$ in terms of $L$:

$R=\frac{L}{\pi }$

Step 2: Calculate the gravitational potential at the center
The gravitational potential $dV$ at the center of the semicircle due to an infinitesimal mass element $dm$ of the rod is:

$dV=-\frac{G·dm}{R}$

Since every mass element on the semicircle is at the same distance $R$ from the center, we find the total gravitational potential $V$ by integrating $dV$ over the entire length of the rod:

$V=���dV=\int -\frac{G·dm}{R}$

Since $G$ and $R$ are constants, they can be pulled out of the integral:

$V=-\frac{G}{R}\int dm$

The sum of all mass elements $\int dm$ is equal to the total mass of the rod, $M$:

$V=-\frac{GM}{R}$

Step 3: Substitute the value of radius R
Substitute $R=\frac{L}{\pi }$ into the potential equation:

$V=-\frac{GM}{\left(\frac{L}{\pi }\right)}$

Simplifying the expression, we get:

$V=-\frac{\pi GM}{L}$

Therefore, the gravitational potential at the centre of the semicircle is indeed $-\frac{\pi GM}{L}$.