Q. For spheres each of mass $M$ and radius $R$ are placed with their centers on the four corners $A,B,C\ $and $D$ of a square of side $b$. The spheres $A$ and $B$ are hollow and $C$ and $D$ are solids. The moment of inertia of the system about side $AD$ of square is
A) $\frac{8}{3}{MR}^{2} + \ {2Mb}^{2}$
B) $\frac{8}{5}{MR}^{2} + \ {2Mb}^{2}$
C) $\frac{32}{15}{MR}^{2} + \ {2Mb}^{2}$
D) $32{MR}^{2} + \ 4Mb^{2}$
Explanation
(c) Moment of inertia of a hollow sphere of
radius $R$ about the diameter passing
through $D$ is
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$I_{A} = \frac{2}{5}MR^{2}$ …(i)
Moment of inertia of solid sphere about diameter
$I_{B} = \frac{2}{5}MR^{2}$ …(ii)
$\therefore$ Moment of inertia of whole system about side
$AD = I_{A} + I_{D} + I_{B} + I_{C}$
$= \frac{2}{3}MR^{2} + \frac{2}{5}MR^{2} + \left( Mb^{2} + \frac{2}{3}MR^{2} \right) + \left( Mb^{2} + \frac{2}{5}MR^{2} \right)$
$= \frac{32}{15}MR^{2} + 2Mb^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.