Question

The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of $x$ from it is $I\left(x\right)$. Which one of the graphs represents the variation of $I\left(x\right)$ with $x$ correctly ?

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let $XX$ be the axis of rotation for solid sphere passing through it's centre of mass $\left(C\right)$, and $X{X}^{\prime }$ be the axis parallel to $XX$ and at a distance of $x$ from it.
$\Rightarrow$Taking $M$ and $R$ be the mass and radius of solid sphere, and Applying parallel axis theorem for moment of inertia about axis $X{X}^{\prime }$:


$I_{X{X}^{\prime }}=I_{\text{CM}}+M{x}^2...\left(i\right)$
$I_{XX}=I_{\text{CM}}=\frac{2}{5}M{R}^2$
$I\left(X{X}^{\prime }\right)=I\left(x\right)$ as given in question
Hence, from Eq $\left(i\right)$
$I\left(x\right)=\frac{2}{5}M{R}^2+M{x}^2...\left(ii\right)$
$\Rightarrow$since mass $M$ and radius $R$ for the solid sphere is a fixed quantity, the variable in Eq $\left(ii\right)$ are $I\left(x\right)$ and $x$, so it is analogous to following equation:
$(y=a{x}^2+b)$ $\to$ Equation of parabola
where $I\left(x\right)$ represents $y$ and constant $b=\frac{2}{5}M{R}^2$
$\therefore$ The graph of $I\left(x\right)$ versus $x$ will be parabola opening upwards, since the coefficient of $x^2$ is $M$ which is $+ve$, also the parabola won't start from origin$(0,0)$.
$\because$At $x=0$ putting in Eq $\left(ii\right)$ we get $I\left(x\right)\ne 0$
$\Rightarrow$ Option $\left(b\right)$ is correct.