Mass M is divided into two parts xm and (1-x)m . For a given separation the value of x for which the gravitational attraction between the two pieces becomes maximum is

Options

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

Answer

To solve the problem of finding the value of \( x \) for which the gravitational attraction between the two masses becomes maximum, we can follow these steps:

### Step 1: Define the masses
Let the total mass \( M \) be divided into two parts:
- Mass \( m_1 = xM \)
- Mass \( m_2 = (1-x)M \)

### Step 2: Write the formula for gravitational force
The gravitational force \( F \) between the two masses is given by Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{r^2} \]
Substituting the expressions for \( m_1 \) and \( m_2 \):
\[ F = \frac{G (xM) ((1-x)M)}{r^2} \]
This simplifies to:
\[ F = \frac{G M^2 x(1-x)}{r^2} \]

### Step 3: Find the expression to maximize
To maximize the gravitational force, we need to maximize the product \( x(1-x) \).

### Step 4: Differentiate the expression
We differentiate \( F \) with respect to \( x \):
\[ \frac{dF}{dx} = \frac{G M^2}{r^2} \frac{d}{dx}(x(1-x)) \]
Using the product rule, we have:
\[ \frac{d}{dx}(x(1-x)) = 1 - 2x \]
Thus,
\[ \frac{dF}{dx} = \frac{G M^2}{r^2} (1 - 2x) \]

### Step 5: Set the derivative to zero
To find the maximum, we set the derivative equal to zero:
\[ 1 - 2x = 0 \]
Solving for \( x \):
\[ 2x = 1 \implies x = \frac{1}{2} \]

### Step 6: Conclusion
The value of \( x \) for which the gravitational attraction between the two pieces becomes maximum is:
\[ \boxed{\frac{1}{2}} \]