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ExamsJEE MainPhysics

The linear mass density of a thin rod AB of length L varies from A to B as λ(x) = λ0(1 + x/L), where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is:

  1. 5ML²/12
  2. 3ML²/7
  3. 2ML²/5
  4. 7ML²/18

Correct answer: 7ML²/18

Solution

With lambda = lambda0(1 + x/L), total mass M = integral_0^L lambda dx = 3*lambda0*L/2, so lambda0 = 2M/(3L). Moment of inertia about A: I = integral_0^L x^2*lambda dx = lambda0*(L^3/3 + L^3/4) = lambda0*7L^3/12 = 7ML^2/18.

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