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%`, increase
C. `0.5%`, increase
D. `2%`, decrease

Answer

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

### Step 1: Understand the change in weight at height \( h \)

When a body of mass \( m \) is raised to a height \( h \) above the Earth's surface, the weight of the body decreases. The weight \( W \) at height \( h \) can be expressed as:

\[ W' = mg' = mg(1 - \frac{2h}{R_e}) \]

where \( g' \) is the acceleration due to gravity at height \( h \) and \( R_e \) is the radius of the Earth.

### Step 2: Calculate the percentage decrease in weight at height \( h \)

The percentage decrease in weight when the body is raised to height \( h \) is given by:

\[ \text{Percentage decrease} = \frac{W - W'}{W} \times 100\% \]

Substituting \( W' \):

\[ \text{Percentage decrease} = \frac{mg - mg(1 - \frac{2h}{R_e})}{mg} \times 100\% \]

This simplifies to:

\[ \text{Percentage decrease} = \frac{mg \cdot \frac{2h}{R_e}}{mg} \times 100\% = \frac{2h}{R_e} \times 100\% \]

Given that this decrease is 1%, we can set up the equation:

\[ \frac{2h}{R_e} \times 100\% = 1\% \]

### Step 3: Solve for \( \frac{h}{R_e} \)

From the equation above, we can solve for \( \frac{h}{R_e} \):

\[ \frac{2h}{R_e} = 0.01 \quad \Rightarrow \quad \frac{h}{R_e} = 0.005 \]

### Step 4: Analyze the change in weight at depth \( h \)

When the body is taken to a depth \( h \) in a mine, the effective acceleration due to gravity \( g' \) at depth \( h \) can be expressed as:

\[ g' = g(1 - \frac{h}{R_e}) \]

### Step 5: Calculate the percentage change in weight at depth \( h \)

The weight at depth \( h \) is:

\[ W'' = mg' = mg(1 - \frac{h}{R_e}) \]

The percentage change in weight when taken to depth \( h \) is:

\[ \text{Percentage change} = \frac{W - W''}{W} \times 100\% \]

Substituting \( W'' \):

\[ \text{Percentage change} = \frac{mg - mg(1 - \frac{h}{R_e})}{mg} \times 100\% \]

This simplifies to:

\[ \text{Percentage change} = \frac{mg \cdot \frac{h}{R_e}}{mg} \times 100\% = \frac{h}{R_e} \times 100\% \]

### Step 6: Substitute \( \frac{h}{R_e} \)

Using \( \frac{h}{R_e} = 0.005 \):

\[ \text{Percentage change} = 0.005 \times 100\% = 0.5\% \]

### Conclusion

Thus, when the body is taken to a depth \( h \) in a mine, the change in its weight is a decrease of **0.5%**.