The moment of inertia of a solid sphere,about | vedclass

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(D) According to the parallel axis theorem,the moment of inertia $I$ of a body about an axis at a distance $x$ from the parallel axis passing through

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(D) According to the parallel axis theorem,the moment of inertia $I$ of a body about an axis at a distance $x$ from the parallel axis passing through the center of mass is given by $I = I_{CM} + Mx^2$.For a solid sphere,the moment of inertia about its diameter (which passes through the center of mass) is $I_{CM} = \frac{2}{5}MR^2$.Substituting this into the parallel axis theorem,we get $I(x) = \frac{2}{5}MR^2 + Mx^2$.This equation is of the form $y = ax^2 + c$,where $a = M$ and $c = \frac{2}{5}MR^2$.Since $c > 0$,the graph is a parabola that does not pass through the origin,but starts at a positive value $I(0) = \frac{2}{5}MR^2$ on the $I$-axis and opens upwards. This corresponds to the graph shown in option $D$.

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

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