If g on the surface of the earth is `9.8ms^(-2)`, its value of a depth of 3200 km,
(Radius of the earth =6400 km) is

Options

A. `9.8ms^(2)`
B. zero
C. `4.9ms^(-2)`
D. `2.45 ms^(-2)`

Explanation

To find the value of gravitational acceleration (g') at a depth of 3200 km below the Earth's surface, we can use the formula:

\[ g' = g \left(1 - \frac{x}{R}\right) \]

where:
- \( g' \) is the gravitational acceleration at depth,
- \( g \) is the gravitational acceleration at the surface (given as \( 9.8 \, \text{m/s}^2 \)),
- \( x \) is the depth (3200 km),
- \( R \) is the radius of the Earth (6400 km).

### Step-by-step solution:

1. **Identify the values:**
- Given \( g = 9.8 \, \text{m/s}^2 \)
- Depth \( x = 3200 \, \text{km} \)
- Radius of the Earth \( R = 6400 \, \text{km} \)

2. **Substitute the values into the formula:**
\[ g' = 9.8 \left(1 - \frac{3200}{6400}\right) \]

3. **Calculate the fraction:**
\[ \frac{3200}{6400} = \frac{1}{2} \]

4. **Substitute back into the equation:**
\[ g' = 9.8 \left(1 - \frac{1}{2}\right) \]

5. **Simplify the expression:**
\[ g' = 9.8 \left(\frac{1}{2}\right) \]

6. **Calculate \( g' \):**
\[ g' = \frac{9.8}{2} = 4.9 \, \text{m/s}^2 \]

### Final Answer:
The value of gravitational acceleration at a depth of 3200 km is \( 4.9 \, \text{m/s}^2 \).