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 determine the value of \( X \) that maximizes the gravitational force between two masses, \( X \) and \( 1-X \), separated by a distance \( r \).

### Step-by-Step Solution:

1. **Define the Masses**:
Let the first mass be \( X \) and the second mass be \( 1 - X \). The total mass is given as \( M = 1 \).

2. **Write the Gravitational Force Formula**:
The gravitational force \( F \) between the two masses is given by Newton's law of gravitation:
\[ F = \frac{G \cdot X \cdot (1 - X)}{r^2} \]
where \( G \) is the gravitational constant and \( r \) is the distance between the two masses.

3. **Simplify the Expression**:
Since \( G \) and \( r^2 \) are constants, we can focus on maximizing the expression \( F \) which is proportional to \( X(1 - X) \):
\[ F \propto X(1 - X) \]

4. **Differentiate the Expression**:
To find the maximum, we differentiate \( F \) with respect to \( X \):
\[ \frac{dF}{dX} = \frac{d}{dX}(X(1 - X)) = 1 - 2X \]

5. **Set the Derivative to Zero**:
To find the critical points, set the derivative equal to zero:
\[ 1 - 2X = 0 \]

6. **Solve for \( X \)**:
Rearranging gives:
\[ 2X = 1 \implies X = \frac{1}{2} \]

7. **Conclusion**:
The value of \( X \) that maximizes the gravitational force between the two masses is:
\[ \boxed{\frac{1}{2}} \]