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: Determine the radius of the semicircle
The length of the rod \( L \) is equal to the circumference of the semicircle. The circumference \( C \) of a full circle is given by the formula:
\[ C = 2\pi r \]
For a semicircle, the length \( L \) is:
\[ L = \pi r \]
From this, 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:
\[ V = -\frac{GM}{r} \]
where \( G \) is the universal gravitational constant.

### Step 3: Substitute the radius into the potential formula
Now, we 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:
\[ V = -\frac{GM \pi}{L} \]

### Conclusion
The gravitational potential at the center of the semicircle is given by:
\[ V = -\frac{GM \pi}{L} \]

---