The work done liftting a particle of mass 'm' from the centre of the earth to the surface of the earth is

Options

A. `-mgR`
B. `(1)/(2)mgR`
C. Zero
D. `mgR`

Explanation

To find the work done in lifting a particle of mass 'm' from the center of the Earth to its surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Gravitational Potential at the Center and Surface of the Earth**:
- The gravitational potential \( V \) at the center of the Earth is given by:
\[ V_1 = -\frac{3}{2} \frac{GM}{R} \]
- The gravitational potential \( V \) at the surface of the Earth is given by:
\[ V_2 = -\frac{GM}{R} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Calculating the Change in Gravitational Potential**:
- The change in gravitational potential \( \Delta V \) when moving from the center to the surface is:
\[ \Delta V = V_2 - V_1 \]
- Substituting the values:
\[ \Delta V = -\frac{GM}{R} - \left(-\frac{3}{2} \frac{GM}{R}\right) \]
- Simplifying this:
\[ \Delta V = -\frac{GM}{R} + \frac{3}{2} \frac{GM}{R} = \frac{GM}{2R} \]

3. **Calculating the Change in Potential Energy**:
- The change in potential energy \( \Delta U \) is given by:
\[ \Delta U = m \Delta V \]
- Substituting \( \Delta V \):
\[ \Delta U = m \cdot \frac{GM}{2R} = \frac{mGM}{2R} \]

4. **Relating Gravitational Acceleration to Potential Energy**:
- The gravitational acceleration \( g \) at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
- Rearranging gives:
\[ GM = gR^2 \]
- Substituting this into the expression for \( \Delta U \):
\[ \Delta U = \frac{m \cdot gR^2}{2R} = \frac{mgR}{2} \]

5. **Final Result**:
- Therefore, the work done in lifting the particle from the center of the Earth to its surface is:
\[ W = \frac{mgR}{2} \]

### Conclusion:
The work done in lifting a particle of mass 'm' from the center of the Earth to the surface of the Earth is \( \frac{mgR}{2} \).