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

Explanation

To solve the problem step by step, we will use the information given about the masses and the distances involved.

### Step 1: Define the masses
Let the mass of point A be \(4m\) and the mass of point B be \(3m\), based on the ratio \(4:3\).

### Step 2: Define the distances
Let the distance from point A to point C be \(x\). Therefore, the distance from point B to point C will be \(1 - x\) since the total distance between A and B is 1 meter.

### Step 3: Write the expressions for gravitational force
The gravitational force \(F_{AC}\) between A and C is given by:
\[ F_{AC} = \frac{G \cdot 4m \cdot M}{x^2} \]
The gravitational force \(F_{BC}\) between B and C is given by:
\[ F_{BC} = \frac{G \cdot 3m \cdot M}{(1 - x)^2} \]

### Step 4: Set up the equation based on the given condition
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 \(F_{BC}\):
\[ \frac{G \cdot 4m \cdot M}{x^2} = \frac{1}{3} \cdot \frac{G \cdot 3m \cdot M}{(1 - x)^2} \]

### Step 5: Simplify the equation
We can cancel \(G\), \(m\), and \(M\) 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 6: Cross-multiply to solve for \(x\)
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:
\[ 4x^2 - x^2 - 8x + 4 = 0 \]
This simplifies to:
\[ 3x^2 - 8x + 4 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 3\), \(b = -8\), and \(c = 4\):
\[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} \]
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
Thus, the distance of point C from point A is:
\[ \boxed{\frac{2}{3} \text{ meters}} \]