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

The correct option is 1/2.

Let us analyze the problem step-by-step to find the value of $x$ that maximizes the gravitational attraction between the two pieces of mass.

Step 1: Define the masses of the two parts
The total mass is $M$ and it is divided into two pieces:
The mass of the first piece,
$m_1=xM$
The mass of the second piece,
$m_2=(1-x)M$

Step 2: Write the expression for gravitational force
According to Newton's law of universal gravitation, the gravitational force of attraction $F$ between two masses separated by a distance $r$ is given by:

$F=\frac{G{m}_1{m}_2}{r^2}$
where $G$ is the universal gravitational constant.

Substituting the values of $m_1$ and $m_2$ into the equation:

$F=\frac{G\left(xM\right)(1-x)M}{r^2}$

$F=\frac{G{M}^2}{r^2}(x-x^2)$

Step 3: Maximize the force with respect to x
For a given separation $r$, the term
$\frac{G{M}^2}{r^2}$
is constant. Thus, the force $F$ is maximum when the term
$(x-x^2)$
is maximum.

To find the maximum value, we differentiate the expression with respect to $x$ and set the derivative to zero:

$\frac{d}{dx}(x-x^2)=0$

$1-2x=0$

$2x=1$

$x=\frac{1}{2}$

Step 4: Verify the maximum condition
We can verify that this value of $x$ yields a maximum by taking the second derivative:

$\frac{d^2}{d{x}^2}(x-x^2)=-2$
Since the second derivative is negative ($-2<0$), the gravitational attraction is indeed maximized when
$x=\frac{1}{2}$.