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(D) According to the parallel axis theorem,the moment of inertia $I$ about an axis at position $x$ is given by $I = I_{CM} + M(x - x_{CM})^2$,where $x

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$x = 3$

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(D) According to the parallel axis theorem,the moment of inertia $I$ about an axis at position $x$ is given by $I = I_{CM} + M(x - x_{CM})^2$,where $x_{CM}$ is the $x$-coordinate of the centre of mass.Expanding this,we get $I = M x^2 - 2 M x_{CM} x + (I_{CM} + M x_{CM}^2)$.Comparing this with the given equation $I = 2x^2 - 12x + 27$,we equate the coefficients of $x^2$ and $x$:Coefficient of $x^2$: $M = 2$.Coefficient of $x$: $-2 M x_{CM} = -12$.Substituting $M = 2$ into the second equation: $-2(2)x_{CM} = -12 \implies -4 x_{CM} = -12 \implies x_{CM} = 3$.Thus,the $x$-coordinate of the centre of mass is $3$.

$A$ rigid body can be hinged about any point on the $x$-axis. When it is hinged such that the hinge is at $x$,the moment of inertia is given by $I = 2x^2 - 12x + 27$. The $x$-coordinate of the centre of mass is:

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