Mass particles of 1 kg each are placed along x-axus at `x=1, 2, 4, 8, .... oo`. Then gravitational force o a mass of 3 kg placed at origin is (G= universal gravitation constant) :-

Options

A. `4G`
B. `(4G)/3`
C. `2G`
D. `oo`

Answer

To solve the problem, we need to calculate the gravitational force acting on a mass of 3 kg placed at the origin due to an infinite series of 1 kg masses located at positions \( x = 1, 2, 4, 8, \ldots \).

### Step-by-Step Solution:

1. **Identify the Positions and Masses:**
The masses are placed at positions \( x = 1, 2, 4, 8, \ldots \). Each mass is 1 kg.

2. **Use the Gravitational Force Formula:**
The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:
\[ F = \frac{G m_1 m_2}{r^2} \]
Here, \( m_1 = 3 \) kg (the mass at the origin) and \( m_2 = 1 \) kg (the mass at position \( x \)). The distance \( r \) will be the position of the mass.

3. **Calculate the Force from Each Mass:**
The gravitational force due to the mass at \( x = 1 \):
\[ F_1 = \frac{G \cdot 3 \cdot 1}{1^2} = 3G \]
The gravitational force due to the mass at \( x = 2 \):
\[ F_2 = \frac{G \cdot 3 \cdot 1}{2^2} = \frac{3G}{4} \]
The gravitational force due to the mass at \( x = 4 \):
\[ F_3 = \frac{G \cdot 3 \cdot 1}{4^2} = \frac{3G}{16} \]
The gravitational force due to the mass at \( x = 8 \):
\[ F_4 = \frac{G \cdot 3 \cdot 1}{8^2} = \frac{3G}{64} \]
And so on for masses at \( x = 16, 32, \ldots \).

4. **Sum the Forces:**
The total gravitational force \( F \) acting on the 3 kg mass at the origin is the sum of all these forces:
\[ F = F_1 + F_2 + F_3 + F_4 + \ldots \]
This can be expressed as:
\[ F = 3G + \frac{3G}{4} + \frac{3G}{16} + \frac{3G}{64} + \ldots \]

5. **Factor Out Common Terms:**
We can factor out \( 3G \):
\[ F = 3G \left(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots \right) \]

6. **Recognize the Series:**
The series inside the parentheses is a geometric series where the first term \( a = 1 \) and the common ratio \( r = \frac{1}{4} \). The sum \( S \) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

7. **Calculate the Total Force:**
Substituting back into the force equation:
\[ F = 3G \cdot \frac{4}{3} = 4G \]

### Final Answer:
The gravitational force on the 3 kg mass placed at the origin due to the infinite series of 1 kg masses is:
\[ \boxed{4G} \]