The weight of a body at earth's surface is W. At a depth halfway to the centre of the earth , it will be (assuming uniform density in the earth )

Options

A. (a)W
B. (b)`W/2`
C. (c)`W/4`
D. (d)`W/8`

Answer

To solve the problem of finding the weight of a body at a depth halfway to the center of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We know that the weight of a body at the Earth's surface is \( W \).
- We need to find the weight of the same body at a depth of \( \frac{R}{2} \) (where \( R \) is the radius of the Earth).

2. **Using the Formula for Gravitational Acceleration at Depth**:
- The formula for the acceleration due to gravity \( g' \) at a depth \( d \) is given by:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
- Here, \( g \) is the acceleration due to gravity at the surface, and \( d \) is the depth.

3. **Substituting the Depth**:
- Since we are looking for the weight at a depth of \( \frac{R}{2} \), we substitute \( d = \frac{R}{2} \):
\[ g' = g \left(1 - \frac{\frac{R}{2}}{R}\right) = g \left(1 - \frac{1}{2}\right) = g \left(\frac{1}{2}\right) \]

4. **Finding the Weight at Depth**:
- The weight \( W' \) of the body at this depth can be expressed as:
\[ W' = m \cdot g' \]
- Substituting \( g' \) from the previous step:
\[ W' = m \cdot \left(g \cdot \frac{1}{2}\right) = \frac{1}{2} \cdot (m \cdot g) = \frac{W}{2} \]

5. **Conclusion**:
- Therefore, the weight of the body at a depth halfway to the center of the Earth is:
\[ W' = \frac{W}{2} \]

### Final Answer:
The weight of the body at a depth halfway to the center of the Earth will be \( \frac{W}{2} \).