A body of mass 1 kg is placed between two bodies of masses 10 kg and 90 kg and they are arranged in a straight line. If the forces of attraction due to the two masses on the body of mass 1 kg are equal, then the ratio of the distance between the masses 10 kg and 1distance between the masses of 1 kg and 90 kg is____________

Options

A. ` 1: 1`
B. ` 2 : 3`
C. ` 1: 3`
D. ` 1 :9`

Explanation

To solve the problem, we need to find the ratio of the distances between the masses when the forces of attraction on the 1 kg mass due to the 10 kg and 90 kg masses are equal. Let's denote the distance between the 10 kg mass and the 1 kg mass as \( x \) and the distance between the 1 kg mass and the 90 kg mass as \( y \).

### Step-by-Step Solution:

1. **Understanding the Forces**:
The gravitational force \( F \) between two masses is given by Newton's law of gravitation:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between them.

2. **Setting Up the Forces**:
- The force \( F_1 \) exerted by the 10 kg mass on the 1 kg mass is:
\[ F_1 = \frac{G \cdot (10 \, \text{kg}) \cdot (1 \, \text{kg})}{x^2} \]
- The force \( F_2 \) exerted by the 90 kg mass on the 1 kg mass is:
\[ F_2 = \frac{G \cdot (90 \, \text{kg}) \cdot (1 \, \text{kg})}{y^2} \]

3. **Equating the Forces**:
Since the forces are equal, we can set \( F_1 = F_2 \):
\[ \frac{G \cdot (10) \cdot (1)}{x^2} = \frac{G \cdot (90) \cdot (1)}{y^2} \]
The \( G \) and \( 1 \) kg can be canceled from both sides:
\[ \frac{10}{x^2} = \frac{90}{y^2} \]

4. **Cross Multiplying**:
Cross multiplying gives us:
\[ 10y^2 = 90x^2 \]

5. **Simplifying the Equation**:
Dividing both sides by 10:
\[ y^2 = 9x^2 \]
Taking the square root of both sides:
\[ y = 3x \]

6. **Finding the Ratio**:
The ratio of the distances \( x \) and \( y \) is:
\[ \frac{x}{y} = \frac{x}{3x} = \frac{1}{3} \]

### Final Answer:
The ratio of the distance between the masses 10 kg and 1 kg to the distance between the masses 1 kg and 90 kg is:
\[ \frac{x}{y} = \frac{1}{3} \]