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. 0.5% decrease
B. 2% decrease
C. 0.5% increase
D. 1% increase

Explanation

To solve the problem, we need to analyze the changes in weight of a body when it is raised to a height \( h \) above the Earth's surface and when it is taken to a depth \( h \) in a mine.

### Step-by-Step Solution:

1. **Understanding Weight and Gravitational Acceleration**:
The weight \( W \) of a body is given by the formula:
\[ W = mg \]
where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity.

2. **Weight at Height \( h \)**:
When the body is raised to a height \( h \) above the Earth's surface, the gravitational acceleration \( g' \) at that height is given by:
\[ g' = g \left(1 - \frac{2h}{R}\right) \]
where \( R \) is the radius of the Earth.

3. **Change in Weight at Height \( h \)**:
The change in weight \( \Delta W \) when the body is raised to height \( h \) can be expressed as:
\[ \Delta W = W - W' = mg - mg' = mg - m \left(g \left(1 - \frac{2h}{R}\right)\right) = mg \frac{2h}{R} \]
The percentage decrease in weight is given as:
\[ \frac{\Delta W}{W} \times 100 = \frac{mg \frac{2h}{R}}{mg} \times 100 = \frac{2h}{R} \times 100 \]
Given that this decrease is 1%, we have:
\[ \frac{2h}{R} \times 100 = 1 \implies \frac{2h}{R} = 0.01 \implies 2h = 0.01R \implies h = \frac{0.01R}{2} \]

4. **Weight at Depth \( h \)**:
When the body is taken to a depth \( h \) in a mine, the gravitational acceleration \( g'' \) at that depth is given by:
\[ g'' = g \left(1 - \frac{h}{R}\right) \]

5. **Change in Weight at Depth \( h \)**:
The change in weight \( \Delta W' \) at depth \( h \) is:
\[ \Delta W' = W - W'' = mg - m \left(g \left(1 - \frac{h}{R}\right)\right) = mg - mg \left(1 - \frac{h}{R}\right) = mg \frac{h}{R} \]
The percentage decrease in weight at depth \( h \) is:
\[ \frac{\Delta W'}{W} \times 100 = \frac{mg \frac{h}{R}}{mg} \times 100 = \frac{h}{R} \times 100 \]

6. **Substituting \( h \)**:
From the earlier calculation, we found \( h = \frac{0.01R}{2} \). Substituting this into the percentage decrease at depth gives:
\[ \frac{h}{R} \times 100 = \frac{\frac{0.01R}{2}}{R} \times 100 = \frac{0.01}{2} \times 100 = 0.005 \times 100 = 0.5\% \]

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