Two bodies of masses 5 kg and 20 kg are separated by a distance of 300 m. A body of mass 1 kg is placed between the two masses in a straight line. If the forces of attraction due to the two masses on the body of 1 kg mass are the same, the distance between the 1 kg body and the 5 kg body is ____________

Options

A. 100
B. 200
C. 300
D. 400

Explanation

To solve the problem, we need to find the distance between the 1 kg mass and the 5 kg mass when the gravitational forces acting on the 1 kg mass from the 5 kg and 20 kg masses are equal.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let the mass of the first body (m1) = 5 kg.
- Let the mass of the second body (m2) = 20 kg.
- Let the mass of the third body (m3) = 1 kg.
- The distance between the two masses (m1 and m2) = 300 m.
- Let the distance between the 1 kg mass and the 5 kg mass be \( x \).
- Therefore, the distance between the 1 kg mass and the 20 kg mass will be \( 300 - x \).

2. **Set Up the Equation for Gravitational Forces:**
- The gravitational force \( F \) between two masses is given by the formula:
\[ F = \frac{G \cdot m1 \cdot m2}{r^2} \]
- For the force between the 5 kg mass and the 1 kg mass:
\[ F_{5-1} = \frac{G \cdot 5 \cdot 1}{x^2} \]
- For the force between the 20 kg mass and the 1 kg mass:
\[ F_{20-1} = \frac{G \cdot 20 \cdot 1}{(300 - x)^2} \]

3. **Set the Forces Equal:**
- Since the forces are equal, we have:
\[ \frac{G \cdot 5 \cdot 1}{x^2} = \frac{G \cdot 20 \cdot 1}{(300 - x)^2} \]
- We can cancel \( G \) and \( 1 \) from both sides:
\[ \frac{5}{x^2} = \frac{20}{(300 - x)^2} \]

4. **Cross Multiply:**
- Cross multiplying gives:
\[ 5 \cdot (300 - x)^2 = 20 \cdot x^2 \]

5. **Expand and Simplify:**
- Expanding the left side:
\[ 5 \cdot (90000 - 600x + x^2) = 20x^2 \]
- This simplifies to:
\[ 450000 - 3000x + 5x^2 = 20x^2 \]
- Rearranging gives:
\[ 15x^2 + 3000x - 450000 = 0 \]

6. **Divide by 15:**
- To simplify, divide the entire equation by 15:
\[ x^2 + 200x - 30000 = 0 \]

7. **Factor the Quadratic Equation:**
- We need to factor the quadratic equation:
\[ (x + 300)(x - 100) = 0 \]
- This gives us two potential solutions for \( x \):
\[ x = 100 \quad \text{or} \quad x = -300 \]

8. **Select the Valid Solution:**
- Since distance cannot be negative, we discard \( x = -300 \).
- Therefore, the distance between the 1 kg mass and the 5 kg mass is:
\[ x = 100 \text{ meters} \]

### Final Answer:
The distance between the 1 kg body and the 5 kg body is **100 meters**.