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JEE Main 2019 (10 Jan, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2019 (10 Jan, Shift 2)

Q163JEE Main 2019 · 10 JAN · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

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:

Figure

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Answer: (b)

  1. Ball centres are R+R=2RR + R = 2R from the axis.
  2. Each ball: 25MR2+M(2R)2=225MR2\frac{2}{5}MR^2 + M(2R)^2 = \frac{22}{5}MR^2
  3. Rod: M(2R)212=MR23\frac{M(2R)^2}{12} = \frac{MR^2}{3}
  4. I=445MR2+13MR2=13715MR2I = \frac{44}{5}MR^2 + \frac{1}{3}MR^2 = \frac{137}{15}MR^2, so the answer is (b).
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