A particle of mass m was transferred from the centre of the base of a uniform hemisphere of mass M and radius R into infinity.
What work was performed in the process by the gravitational force exerted on the particle by the hemisphere?

Options

A. a) `(GMm)/(R)`
B. b) `-(1)/(2)(GMm)/(R)`
C. c) `-(3)/(2)(GMm)/(R)`
D. d) `-(3)/(4)(GMm)/(R)`

Explanation

To find the work performed by the gravitational force exerted on a particle of mass \( m \) when it is transferred from the center of the base of a uniform hemisphere of mass \( M \) and radius \( R \) to infinity, we can follow these steps:

### Step 1: Determine the Gravitational Potential at the Center of the Hemisphere

The gravitational potential \( V \) at a point due to a mass \( M \) is given by the formula:
\[ V = -\frac{GM}{r} \]
However, since we are dealing with a hemisphere, we need to consider the symmetry and the contribution of the mass distribution. For a uniform hemisphere, the potential at the center can be derived from the potential of the entire sphere and then halved.

The potential at the center of a full sphere of mass \( M \) and radius \( R \) is:
\[ V_{\text{sphere}} = -\frac{3GM}{2R} \]
Thus, for a hemisphere, the potential at the center is:
\[ V_{\text{hemisphere}} = -\frac{3GM}{4R} \]

### Step 2: Calculate the Change in Potential Energy

The change in potential energy \( \Delta U \) when moving the mass \( m \) from the center of the hemisphere to infinity (where the potential is zero) is given by:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} \]
At infinity, the potential energy \( U_{\text{final}} = 0 \). The initial potential energy \( U_{\text{initial}} \) at the center of the hemisphere is:
\[ U_{\text{initial}} = m \cdot V_{\text{hemisphere}} = m \left(-\frac{3GM}{4R}\right) = -\frac{3mGM}{4R} \]
Thus, the change in potential energy is:
\[ \Delta U = 0 - \left(-\frac{3mGM}{4R}\right) = \frac{3mGM}{4R} \]

### Step 3: Calculate the Work Done by the Gravitational Force

The work done \( W \) by the gravitational force is equal to the negative of the change in potential energy:
\[ W = -\Delta U = -\left(\frac{3mGM}{4R}\right) = -\frac{3mGM}{4R} \]

### Final Answer

The work performed by the gravitational force exerted on the particle by the hemisphere when it is moved to infinity is:
\[ W = \frac{3mGM}{4R} \]

---