The value of .g. at a particular point on the surface of the earth is 9.8 `ms^(2)`. Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass, the value of g at the same point will now be (assuming that the distance of the point from the centre of the earth does not change )

Options

A. `4.9ms^(-2)`
B. `3.1ms^(-2)`
C. `9.8ms^(-2)`
D. `19.6ms^(-2)`

Answer

To solve the problem, we need to analyze how the acceleration due to gravity (g) changes when the Earth shrinks uniformly to half its present size while maintaining its mass.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The initial value of acceleration due to gravity at a point on the surface of the Earth is given as \( g = 9.8 \, \text{m/s}^2 \).
- The Earth shrinks to half its radius (from \( R \) to \( \frac{R}{2} \)) but retains the same mass \( M \).

2. **Formula for Acceleration due to Gravity**:
- The formula for acceleration due to gravity at the surface of a spherical body is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

3. **Initial Conditions**:
- Initially, we have:
\[ g_{\text{initial}} = \frac{GM}{R^2} \]

4. **Final Conditions After Shrinking**:
- After the Earth shrinks to half its radius, the new radius becomes \( R' = \frac{R}{2} \).
- The mass remains the same, so we still have \( M \).
- The new acceleration due to gravity \( g' \) at the surface can be calculated as:
\[ g' = \frac{GM}{(R/2)^2} \]

5. **Calculating the New Value of g**:
- Substitute \( R' \) into the equation:
\[ g' = \frac{GM}{(R/2)^2} = \frac{GM}{\frac{R^2}{4}} = \frac{GM \cdot 4}{R^2} = 4 \cdot \frac{GM}{R^2} \]
- This means:
\[ g' = 4g_{\text{initial}} \]

6. **Final Calculation**:
- Since \( g_{\text{initial}} = 9.8 \, \text{m/s}^2 \):
\[ g' = 4 \cdot 9.8 \, \text{m/s}^2 = 39.2 \, \text{m/s}^2 \]

### Conclusion:
The value of \( g \) at the same point after the Earth shrinks to half its size while maintaining its mass will be \( 39.2 \, \text{m/s}^2 \).