Two point objects of mass 2x and 3x are seperated by a distance r. keeping the distance fixed, how much mass should be transferred from 3x to 2x , so that gravitational force between them becomes maximum ?

Options

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

Answer

To solve the problem, we need to determine how much mass should be transferred from the object with mass \(3x\) to the object with mass \(2x\) in order to maximize the gravitational force between them.

### Step-by-Step Solution:

1. **Define the Masses After Transfer:**
Let's denote the mass transferred from the object with mass \(3x\) to the object with mass \(2x\) as \(y\).
- The new mass of the first object (initially \(3x\)) becomes \(3x - y\).
- The new mass of the second object (initially \(2x\)) becomes \(2x + y\).

2. **Write the Gravitational Force Equation:**
According to Newton's law of gravitation, the gravitational force \(F\) between two masses is given by:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
Substituting the new masses into the equation, we have:
\[ F = \frac{G \cdot (3x - y) \cdot (2x + y)}{r^2} \]
Since \(G\) and \(r^2\) are constants, we can simplify this to:
\[ F = k \cdot (3x - y)(2x + y) \]
where \(k = \frac{G}{r^2}\).

3. **Expand the Force Equation:**
Expanding \(F\):
\[ F = k \cdot (6x^2 + 3xy - 2xy - y^2) = k \cdot (6x^2 + xy - y^2) \]

4. **Differentiate the Force with Respect to \(y\):**
To find the maximum force, we need to differentiate \(F\) with respect to \(y\):
\[ \frac{dF}{dy} = k \cdot (x - 2y) \]

5. **Set the Derivative to Zero:**
To find the critical points, set the derivative equal to zero:
\[ x - 2y = 0 \]
Solving for \(y\):
\[ y = \frac{x}{2} \]

6. **Check the Second Derivative:**
To confirm that this point is a maximum, we check the second derivative:
\[ \frac{d^2F}{dy^2} = k \cdot (-2) \]
Since \(\frac{d^2F}{dy^2} < 0\), this indicates that the force is maximized at \(y = \frac{x}{2}\).

### Conclusion:
The amount of mass that should be transferred from the object with mass \(3x\) to the object with mass \(2x\) in order to maximize the gravitational force between them is:
\[ \boxed{\frac{x}{2}} \]