Two point masses A and B having masses in the ratio `4:3` are separated by a distance of 1m. When another point mass C of mass M is placed in between A and B, the force between A and C is `(1/3)^(rd)` of the force between B and C. Then the distance C from A is

Options

A. `2/3m`
B. `1/3m`
C. `1/4m`
D. `2/7m`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Define the masses and distances
Let the mass of A be \( M_A \) and the mass of B be \( M_B \). According to the problem, the ratio of the masses is given as:
\[ \frac{M_A}{M_B} = \frac{4}{3} \]
Let \( M_A = 4k \) and \( M_B = 3k \) for some constant \( k \).

### Step 2: Define the distance from A to C
Let the distance from A to C be \( x \) meters. Therefore, the distance from C to B will be:
\[ 1 - x \text{ meters} \]

### Step 3: Write the expressions for the gravitational forces
The gravitational force between A and C (\( F_{AC} \)) is given by:
\[ F_{AC} = \frac{G M_A M}{x^2} \]
The gravitational force between B and C (\( F_{BC} \)) is given by:
\[ F_{BC} = \frac{G M_B M}{(1 - x)^2} \]

### Step 4: Set up the ratio of the forces
According to the problem, the force between A and C is one third of the force between B and C:
\[ F_{AC} = \frac{1}{3} F_{BC} \]

Substituting the expressions for \( F_{AC} \) and \( F_{BC} \):
\[ \frac{G M_A M}{x^2} = \frac{1}{3} \cdot \frac{G M_B M}{(1 - x)^2} \]

### Step 5: Simplify the equation
We can cancel \( G \) and \( M \) from both sides:
\[ \frac{M_A}{x^2} = \frac{1}{3} \cdot \frac{M_B}{(1 - x)^2} \]

Substituting \( M_A = 4k \) and \( M_B = 3k \):
\[ \frac{4k}{x^2} = \frac{1}{3} \cdot \frac{3k}{(1 - x)^2} \]

### Step 6: Cancel \( k \) and simplify further
Cancelling \( k \) from both sides:
\[ \frac{4}{x^2} = \frac{1}{3} \cdot \frac{3}{(1 - x)^2} \]
This simplifies to:
\[ \frac{4}{x^2} = \frac{1}{(1 - x)^2} \]

### Step 7: Cross-multiply and rearrange
Cross-multiplying gives:
\[ 4(1 - x)^2 = x^2 \]
Expanding the left side:
\[ 4(1 - 2x + x^2) = x^2 \]
This simplifies to:
\[ 4 - 8x + 4x^2 = x^2 \]
Rearranging gives:
\[ 3x^2 - 8x + 4 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = -8, c = 4 \):
\[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} \]
Calculating the discriminant:
\[ x = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6} \]
This gives us two possible solutions:
\[ x = \frac{12}{6} = 2 \quad \text{(not valid since distance cannot exceed 1m)} \]
\[ x = \frac{4}{6} = \frac{2}{3} \]

### Final Answer
The distance from A to C is:
\[ \boxed{\frac{2}{3} \text{ meters}} \]