A thin 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 ?

Options

A. `-(GM)/(L)`
B. `-(GM)/(2piL)`
C. `-(piGM)/(2L)`
D. `-(piGM)/(L)`

Explanation

To find the gravitational potential at the center of a semicircular rod of length \( L \) and mass \( M \), we can follow these steps:

### Step 1: Relate the length of the rod to the radius of the semicircle
The length \( L \) of the rod is equal to the circumference of the semicircle. The circumference \( C \) of a full circle is given by \( C = 2\pi r \). Therefore, for a semicircle, the length can be expressed as:
\[ L = \pi r \]
From this equation, we can solve for the radius \( r \):
\[ r = \frac{L}{\pi} \]

### Step 2: Use the formula for gravitational potential
The gravitational potential \( V \) at a distance \( r \) from a mass \( M \) is given by the formula:
\[ V = -\frac{GM}{r} \]
where \( G \) is the universal gravitational constant.

### Step 3: Substitute the radius into the gravitational potential formula
Now that we have \( r \) in terms of \( L \), we can substitute \( r = \frac{L}{\pi} \) into the gravitational potential formula:
\[ V = -\frac{GM}{\frac{L}{\pi}} = -\frac{GM\pi}{L} \]

### Step 4: Final expression for gravitational potential
Thus, the gravitational potential at the center of the semicircle is:
\[ V = -\frac{GM\pi}{L} \]

### Summary
The gravitational potential at the center of the semicircle formed by the rod of length \( L \) and mass \( M \) is given by:
\[ V = -\frac{GM\pi}{L} \]