The value of `g` at a particular point is `10 m s^(-2)`. Suppose the earth shrinks uniformly to half of its present size without losing any mass. The value of `g` at the same point (assuming that the distance of the point from the centre of the earth does not change) will now be

Options

A. `5ms^(-2)`
B. `10ms^(-2)`
C. `3ms^(-2)`
D. `20 ms^(-2)`

Explanation

To solve the problem, we need to analyze how the acceleration due to gravity \( g \) is affected when the Earth shrinks uniformly to half of its present size without losing any mass.

### Step-by-Step Solution:

1. **Understanding the Formula for \( g \)**:
The formula for the acceleration due to gravity \( g \) at a distance \( r \) from the center of a planet is given by:
\[ g = \frac{GM}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( r \) is the radius of the planet.

2. **Initial Conditions**:
We are given that the initial value of \( g \) at a particular point is \( 10 \, \text{m/s}^2 \). This means:
\[ g = \frac{GM}{r^2} = 10 \, \text{m/s}^2 \]

3. **Effect of Shrinking the Earth**:
If the Earth shrinks uniformly to half of its present size, the new radius \( r' \) will be:
\[ r' = \frac{r}{2} \]
However, the mass \( M \) of the Earth remains unchanged.

4. **Calculating the New Value of \( g \)**:
We can substitute the new radius into the formula for \( g \):
\[ g' = \frac{GM}{(r')^2} = \frac{GM}{\left(\frac{r}{2}\right)^2} = \frac{GM}{\frac{r^2}{4}} = \frac{4GM}{r^2} \]
This shows that the new value of \( g' \) is four times the original value of \( g \):
\[ g' = 4g \]

5. **Substituting the Original Value of \( g \)**:
Since the original value of \( g \) is \( 10 \, \text{m/s}^2 \):
\[ g' = 4 \times 10 \, \text{m/s}^2 = 40 \, \text{m/s}^2 \]

### Final Answer:
Thus, the value of \( g \) at the same point after the Earth shrinks to half its size will be \( 40 \, \text{m/s}^2 \).

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