Assuming that the earth is a sphere of uniform mass density, what is the percentage decreases in the weight of a body when taken to the end of the tunned 32 km below the surface of the earth?
(Radius of earth = 6400 km)

Options

A. `0.25%`
B. `0.5%`
C. `0.7%`
D. `1%`

Explanation

To find the percentage decrease in the weight of a body when taken to a depth of 32 km below the surface of the Earth, we can follow these steps:

### Step 1: Understand the relationship between weight and gravity
The weight of a body is given by the formula:
\[ W = mg \]
where \( W \) is the weight, \( m \) is the mass of the body, and \( g \) is the acceleration due to gravity.

### Step 2: Determine the acceleration due to gravity at the surface of the Earth
The acceleration due to gravity at the surface of the Earth is approximately:
\[ g = 9.8 \, \text{m/s}^2 \]

### Step 3: Calculate the new value of gravity at a depth of 32 km
When a body is taken to a depth \( d \) below the surface of the Earth, the acceleration due to gravity \( g' \) at that depth can be calculated using the formula:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
where \( R \) is the radius of the Earth.

Given:
- \( d = 32 \, \text{km} = 32 \times 10^3 \, \text{m} \)
- \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)

Substituting the values:
\[ g' = g \left(1 - \frac{32 \times 10^3}{6400 \times 10^3}\right) \]

### Step 4: Simplify the expression
Calculating the fraction:
\[ \frac{32 \times 10^3}{6400 \times 10^3} = \frac{32}{6400} = \frac{1}{200} \]

So,
\[ g' = g \left(1 - \frac{1}{200}\right) = g \left(\frac{199}{200}\right) \]

### Step 5: Calculate the percentage decrease in weight
The percentage decrease in weight can be calculated as:
\[ \text{Percentage Decrease} = \left(\frac{W - W'}{W}\right) \times 100 \]
where \( W' \) is the new weight at depth.

Since \( W' = mg' \):
\[ W' = m \left(g \frac{199}{200}\right) = \frac{199}{200} W \]

Now substituting into the percentage decrease formula:
\[ \text{Percentage Decrease} = \left(\frac{W - \frac{199}{200}W}{W}\right) \times 100 \]
\[ = \left(\frac{W - \frac{199}{200}W}{W}\right) \times 100 \]
\[ = \left(\frac{1 - \frac{199}{200}}{1}\right) \times 100 \]
\[ = \left(\frac{1}{200}\right) \times 100 \]
\[ = 0.5\% \]

### Final Answer
Thus, the percentage decrease in the weight of a body when taken to a depth of 32 km below the surface of the Earth is **0.5%**.

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