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

Asked in JEE Main 10th Jan 2nd Shift 2019 · Composite and cavity bodies

• (1) (209)/(15)MR²
• (2) (137)/(15)MR²
• (3) (152)/(15)MR²
• (4) (17)/(15)MR²

Step-by-step solution

Why the other options are wrong

• (1) comes from sanity check: each ball's centre sits 2R from the axis, so it contributes 2/5MR²+M(2R)²=(22)/5MR². Two of those plus the rod's (M(2R)²)/(12)=(MR²)/3 gives (137)/(15)MR².
• (3) comes from using the rod's end formula (mℓ²)/3 instead of the midpoint formula (mℓ²)/(12). The axis goes through the centre of the rod.
• (4) comes from leaving out the parallel-axis shift for the two balls: (MR²)/3+2·2/5MR²=(17)/(15)MR². Each ball's centre is 2R from the axis, and that shift dominates.

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