The value of `g` (acceleration due to gravity) at earth's surface is `10 ms^(-2)`. Its value in `ms^(-2)` at the centre of the earth which is assumed to be a sphere of radius `R` metre and uniform mass density is

Options

A. `5`
B. `10/R`
C. `10/(2R)`
D. zero

Explanation

To find the value of `g` (acceleration due to gravity) at the center of the Earth, we can use the formula that describes how gravity varies with depth below the Earth's surface.

### Step-by-Step Solution:

1. **Understanding the Formula**:
The formula for the acceleration due to gravity at a depth `d` below the surface of the Earth is given by:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
where:
- \( g' \) is the acceleration due to gravity at depth \( d \),
- \( g \) is the acceleration due to gravity at the surface,
- \( R \) is the radius of the Earth.

2. **Identifying the Depth at the Center**:
At the center of the Earth, the depth \( d \) is equal to the radius \( R \) of the Earth. Therefore, we can substitute \( d = R \) into the formula.

3. **Substituting the Values**:
Substituting \( d = R \) into the formula gives:
\[ g' = g \left(1 - \frac{R}{R}\right) \]

4. **Simplifying the Expression**:
Simplifying the equation:
\[ g' = g \left(1 - 1\right) = g \cdot 0 \]
Thus, we find:
\[ g' = 0 \]

5. **Conclusion**:
Therefore, the value of acceleration due to gravity \( g' \) at the center of the Earth is:
\[ g' = 0 \, \text{ms}^{-2} \]

### Final Answer:
The value of `g` at the center of the Earth is \( 0 \, \text{ms}^{-2} \).

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