Weight of a body decreases by 1.5% , when it is raised to a height h above the surface of earth. When the same body is taken to same depth h in a mine, its weight will show

Options

A. 0.75% increase
B. 3.0 % decrease
C. 0.75 % decrease
D. 1.5% decrease

Answer

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

### Step-by-Step Solution:

1. **Understanding Weight Change at Height \( h \)**:
- The weight of a body at the surface of the Earth is given by \( W = mg \), where \( g \) is the acceleration due to gravity at the surface.
- When the body is raised to a height \( h \), the weight decreases by 1.5%. This means:
\[ W' = W - 0.015W = 0.985W \]
- Therefore, the new weight at height \( h \) can be expressed as:
\[ W' = mg' = 0.985mg \]
- From this, we can derive:
\[ g' = 0.985g \]

2. **Using the Formula for Gravity at Height \( h \)**:
- The formula for gravitational acceleration at a height \( h \) above the Earth's surface is:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
- Setting the two expressions for \( g' \) equal gives:
\[ \frac{g}{(1 + \frac{h}{R})^2} = 0.985g \]
- Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ \frac{1}{(1 + \frac{h}{R})^2} = 0.985 \]

3. **Solving for \( h \)**:
- Taking the reciprocal:
\[ (1 + \frac{h}{R})^2 = \frac{1}{0.985} \]
- Taking the square root:
\[ 1 + \frac{h}{R} = \sqrt{\frac{1}{0.985}} \approx 1.0076 \]
- Rearranging gives:
\[ \frac{h}{R} \approx 0.0076 \implies h \approx 0.0076R \]

4. **Understanding Weight Change at Depth \( h \)**:
- Now, when the body is taken to a depth \( h \), the gravitational acceleration at depth \( h \) is given by:
\[ g'' = g \left(1 - \frac{h}{R}\right) \]
- Substituting \( h \approx 0.0076R \):
\[ g'' = g \left(1 - 0.0076\right) = 0.9924g \]

5. **Calculating the Change in Weight**:
- The change in weight when the body is taken to depth \( h \):
\[ \Delta g = g'' - g = 0.9924g - g = -0.0076g \]
- This indicates a decrease in weight.

6. **Calculating Percentage Change**:
- The percentage change in weight is given by:
\[ \text{Percentage Change} = \left(\frac{\Delta g}{g}\right) \times 100 = \left(\frac{-0.0076g}{g}\right) \times 100 = -0.76\% \]

### Final Answer:
The weight of the body will decrease by approximately **0.76%** when taken to the same depth \( h \) in a mine.