If the earth were assumed to have uniform density and spherical symmetry, then the value of g in `ms^(-2)` halfway towards the centre of earth would be `[g=10m//s^2]`

Options

A. 0
B. 1.25
C. 5
D. 10

Answer

To find the value of acceleration due to gravity \( g' \) halfway towards the center of the Earth, we can use the formula for gravitational acceleration at a depth \( d \):

\[ g' = g \left(1 - \frac{d}{R}\right) \]

where:
- \( g \) is the acceleration due to gravity at the surface of the Earth,
- \( d \) is the depth below the surface,
- \( R \) is the radius of the Earth.

### Step-by-Step Solution:

1. **Identify Given Values**:
- The value of \( g \) at the surface of the Earth is given as \( 10 \, \text{m/s}^2 \).
- The depth \( d \) halfway to the center of the Earth is \( \frac{R}{2} \), where \( R \) is the radius of the Earth.

2. **Substitute the Depth into the Formula**:
- Since we are halfway to the center, we substitute \( d = \frac{R}{2} \) into the formula:
\[ g' = g \left(1 - \frac{d}{R}\right) = g \left(1 - \frac{R/2}{R}\right) \]

3. **Simplify the Expression**:
- The fraction simplifies as follows:
\[ g' = g \left(1 - \frac{1}{2}\right) = g \left(\frac{1}{2}\right) \]

4. **Substitute the Value of \( g \)**:
- Now, substitute \( g = 10 \, \text{m/s}^2 \):
\[ g' = 10 \left(\frac{1}{2}\right) = 5 \, \text{m/s}^2 \]

5. **Conclusion**:
- Therefore, the value of \( g \) halfway towards the center of the Earth is:
\[ g' = 5 \, \text{m/s}^2 \]