A uniform rod of mass 5kg and length 1.2m is pivoted at its center. What is its moment of inertia about the pivot?
For a rod pivoted at its center: I = 112ML2. M = 5kg, L = 1.2m. I = 112×5×(1.2)2 = 5×1.4412 = 0.6kg m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.6 kg m². This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
