Q. Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is :

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Solution:

For Ball using parallel axis theorem.
$I_{\text{ball}} = \frac{2}{5} MR^{2} +M\left(2R\right)^{2} $
$ =\frac{22}{5} MR^{2}$
2 Balls so $ \frac{44}{5}MR^{2} $
Irod = for rod $ \frac{M\left(2R\right)^{2}}{R} =\frac{MR^{2}}{3} $
$ I_{\text{system}} =I_{\text{Ball}} +I_{\text{rod}} $
$ = \frac{44}{5} MR^{2} + \frac{MR^{2}}{3} $
$ = \frac{137}{15} MR^{2} $

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