A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at a distance x the moment of inertia is given by $I = 2{x^2} - 12x + 27$. Then the x-coordinate of centre of mass is:
A. $x = 2$
B. $x = 0$
C. $x = 1$
D. $x = 3$

Hint: Here in this question, we have to find the coordinates of the centre of mass. When attempting to determine the centre of mass between two masses, multiplying the masses by their positions will yield the centre of mass. Following that, add these together and divide the total by the total of the individual masses. In this problem, we will differentiate the given expression of moment inertia with respect to $x$ and then put it equal to zero to find the x-coordinate of centre of mass.

Complete step by step solution:
As given in the question that a rigid body may be pivoted about any x-axis point. The moment of inertia is given by when it is hinged with the hinge at a distance of x,
$I = 2{x^2} - 12x + 27 \\ $
To find the x-coordinate of the centre of mass we know that, At the body's centre of mass, the moment of inertia is at its lowest level. By which we can say that,
$\dfrac{{dI}}{{dx}} = 0 \\ $
As by putting the value of moment of inertia in the above equation, we get the resultant as
$\dfrac{{d(2{x^2} - 12x + 27)}}{{dx}} = 0 \\ $
By doing differentiation with respect to x we get,
$4x - 12 = 0$
For the value of x, taking x at one side and others at the another we get the value of x as,
$x = 3$
Therefore, the correct answer for the coordinate of the centre of mass is $x = 3$.

Hence, the correct option is D.

Note: We must know that, A mass moving in translation has the same physical significance as the moment of inertia. When calculating inertia in translational motion, a body's mass is used. With an increase in mass, inertia rises. Additionally, a greater force will be needed to provide linear acceleration.