Exams › JEE Main › Physics
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:
- 5ML²/12
- 3ML²/7
- 2ML²/5
- 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.
Related JEE Main Physics questions
⚔️ Practice JEE Main Physics free + battle 1v1 →