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PhysicsNEET 2024Moment of inertia, theorems of axes

System of Particles and Rotational Motion

The moment of inertia of a thin rod about an axis passing through its midpoint and perpendicular to the rod is 2400 g cm22400\ \mathrm{g\,cm^2}. The length of the 400 g rod is nearly:

  1. A.17.5 cm
  2. B.20.7 cm
  3. C.72.0 cm
  4. D.8.5 cm✓

Solution

I=ML212⇒L2=12×2400400=72⇒L≈8.5I = \frac{ML^2}{12} \Rightarrow L^2 = \frac{12 \times 2400}{400} = 72 \Rightarrow L \approx 8.5 cm.

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