If the radius of earth is reduced to 2%, keeping mass constant, then weight of the body on its surface

Options

A. decreases
B. increase
C. remains same
D. is more than that at pole

Answer

To solve the problem, we need to understand how the weight of a body on the surface of the Earth is affected by a change in the radius of the Earth while keeping its mass constant.

### Step-by-Step Solution:

1. **Understanding Weight and Gravitational Force**:
The weight \( W \) of a body on the surface of the Earth is given by the formula:
\[ W = F = \frac{G \cdot M \cdot m}{R^2} \]
where:
- \( W \) is the weight of the body,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the body,
- \( R \) is the radius of the Earth.

2. **Initial Setup**:
Let the initial radius of the Earth be \( R \) and the initial weight of the body be:
\[ W_1 = \frac{G \cdot M \cdot m}{R^2} \]

3. **Change in Radius**:
The radius of the Earth is reduced by 2%. Therefore, the new radius \( R' \) can be expressed as:
\[ R' = R - 0.02R = 0.98R \]

4. **Calculating New Weight**:
Now, we can calculate the new weight \( W_2 \) of the body using the new radius:
\[ W_2 = \frac{G \cdot M \cdot m}{(R')^2} = \frac{G \cdot M \cdot m}{(0.98R)^2} \]

5. **Simplifying the Expression**:
We can simplify \( W_2 \):
\[ W_2 = \frac{G \cdot M \cdot m}{0.9604R^2} = \frac{W_1}{0.9604} \]

6. **Comparing Weights**:
Since \( 0.9604 < 1 \), it follows that:
\[ W_2 > W_1 \]
This indicates that the weight of the body on the surface of the Earth increases when the radius is reduced by 2%.

### Conclusion:
Thus, the weight of the body on the surface of the Earth increases when the radius is reduced by 2%, keeping the mass of the Earth constant.