Mass `M=1` unit is divided into two parts `X` and `(1-X)`. For a given separation the value of `X` for which the gravitational force between them becomes maximum is

Options

A. `1/2`
B. `3/5`
C. 1
D. 2

Explanation

To solve the problem, we need to find the value of \( X \) that maximizes the gravitational force between two masses \( M_1 = X \) and \( M_2 = 1 - X \), where the total mass \( M = 1 \) unit. The gravitational force \( F \) between these two masses separated by a distance \( R \) is given by the formula:

\[ F = \frac{G \cdot M_1 \cdot M_2}{R^2} \]

### Step 1: Express the gravitational force in terms of \( X \)

Substituting \( M_1 \) and \( M_2 \):

\[ F = \frac{G \cdot X \cdot (1 - X)}{R^2} \]

### Step 2: Simplify the expression

We can rewrite the equation as:

\[ F = \frac{G}{R^2} \cdot (X - X^2) \]

### Step 3: Differentiate the force with respect to \( X \)

To find the maximum force, we need to differentiate \( F \) with respect to \( X \):

\[ \frac{dF}{dX} = \frac{G}{R^2} \cdot \left(1 - 2X\right) \]

### Step 4: Set the derivative equal to zero

To find the critical points, set the derivative equal to zero:

\[ 1 - 2X = 0 \]

### Step 5: Solve for \( X \)

Solving for \( X \):

\[ 2X = 1 \quad \Rightarrow \quad X = \frac{1}{2} \]

### Step 6: Conclusion

Thus, the value of \( X \) for which the gravitational force between the two masses becomes maximum is:

\[ X = \frac{1}{2} \]

### Summary

The two masses are \( M_1 = \frac{1}{2} \) and \( M_2 = \frac{1}{2} \).