The bodies of masses 40 kg and 80 kg are at a distance of `0.15` m from each other . Two force of gravitation between the bodies is `1.0` mg wt. calculate the constant of gravitation . Take g = `10 ms^(-2)` .

Answer

To solve the problem, we need to calculate the gravitational constant \( G \) using the formula for gravitational force. The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:

\[ F = G \frac{m_1 m_2}{r^2} \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the first body, \( m_1 = 40 \, \text{kg} \)
- Mass of the second body, \( m_2 = 80 \, \text{kg} \)
- Distance between the two bodies, \( r = 0.15 \, \text{m} \)
- Gravitational force between the bodies, \( F = 1 \, \text{mg wt} = 1 \times 10^{-6} \, \text{N} \) (since \( g = 10 \, \text{m/s}^2 \))

2. **Substitute the values into the gravitational force formula:**
\[ 1 \times 10^{-6} = G \frac{40 \times 80}{(0.15)^2} \]

3. **Calculate \( 40 \times 80 \):**
\[ 40 \times 80 = 3200 \]

4. **Calculate \( (0.15)^2 \):**
\[ (0.15)^2 = 0.0225 \]

5. **Substitute these values back into the equation:**
\[ 1 \times 10^{-6} = G \frac{3200}{0.0225} \]

6. **Rearrange the equation to solve for \( G \):**
\[ G = 1 \times 10^{-6} \times \frac{0.0225}{3200} \]

7. **Calculate \( \frac{0.0225}{3200} \):**
\[ \frac{0.0225}{3200} = 7.03125 \times 10^{-6} \]

8. **Now multiply this result by \( 1 \times 10^{-6} \):**
\[ G = 1 \times 10^{-6} \times 7.03125 \times 10^{-6} = 7.03125 \times 10^{-12} \, \text{N m}^2/\text{kg}^2 \]

9. **Final result:**
\[ G \approx 7 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \]

### Summary:
The gravitational constant \( G \) is approximately \( 7 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).