Weight fo a body of a mass m decreases by 1% when it is raised to height h above the earth's surface. If the body is taken to depth h in a mine, change in its weight is

Options

A. `2%` decreases
B. `0.5%` decreases
C. `1%` increases
D. `0.5%` increases

Explanation

To solve the problem, we need to find the change in weight of a body of mass \( m \) when it is taken to a depth \( h \) in a mine, given that its weight decreases by 1% when raised to a height \( h \) above the Earth's surface.

### Step-by-Step Solution:

1. **Understanding Weight Change at Height \( h \)**:
- The weight of a body at the Earth's surface is given by \( W = mg \), where \( g \) is the acceleration due to gravity.
- When the body is raised to a height \( h \), the new weight \( W_1 \) is given by:
\[ W_1 = mg_1 \]
- The formula for \( g_1 \) (gravity at height \( h \)) is:
\[ g_1 = g \left(1 - \frac{2h}{R}\right) \]
where \( R \) is the radius of the Earth.

2. **Setting Up the Equation for Weight Decrease**:
- According to the problem, the weight decreases by 1%, so:
\[ W_1 = 0.99W \]
- Substituting for \( W_1 \):
\[ mg \left(1 - \frac{2h}{R}\right) = 0.99mg \]
- Dividing both sides by \( mg \):
\[ 1 - \frac{2h}{R} = 0.99 \]

3. **Solving for \( h \)**:
- Rearranging the equation gives:
\[ \frac{2h}{R} = 0.01 \]
- Therefore:
\[ h = \frac{0.01R}{2} = \frac{R}{200} \]

4. **Calculating Weight Change at Depth \( h \)**:
- When the body is taken to a depth \( h \), the new weight \( W_2 \) is given by:
\[ W_2 = mg_2 \]
- The formula for \( g_2 \) (gravity at depth \( h \)) is:
\[ g_2 = g \left(1 - \frac{h}{R}\right) \]
- Substituting \( h = \frac{R}{200} \):
\[ g_2 = g \left(1 - \frac{R/200}{R}\right) = g \left(1 - \frac{1}{200}\right) = g \left(0.995\right) \]

5. **Finding the New Weight**:
- Thus, the new weight \( W_2 \) becomes:
\[ W_2 = mg_2 = mg \cdot 0.995 = 0.995W \]

6. **Calculating the Change in Weight**:
- The change in weight is:
\[ \Delta W = W_2 - W = 0.995W - W = -0.005W \]
- To find the percentage change:
\[ \text{Percentage Change} = \frac{\Delta W}{W} \times 100 = \frac{-0.005W}{W} \times 100 = -0.5\% \]

### Final Answer:
The weight of the body decreases by **0.5%** when taken to a depth \( h \) in a mine.