Rotational Motion

JEE Main level, single correct, medium

48 of 280

Question 48 of 280

A system consists of two identical spheres each of mass 1.5kg1.5kg and radius 50cm50cm at the ends of a light rod. The distance between the centres of the two spheres is 5m.5m. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint?

Show answer and solution

Answer: (C) 19.05kgm219.05kg{m}^{2}


I=(25MR2+Mx2)×2=[25×1.5×(0.5)2+1.5×(2.5)2)]×2=[0.15+9.375]×2=19.05kgm2\left.I=\left(\dfrac{2}{5}M{R}^{2}+M{x}^{2}\right)\times 2=\left[\dfrac{2}{5}\times 1.5\times {\left(0.5\right)}^{2}+1.5\times {\left(2.5\right)}^{2}\right)\right]\times 2=[0.15+9.375]\times 2=19.05{kgm}^{2}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Rotational dynamics and rolling simulationDrag the sliders and watch the numbers in this question change.Revise Rotational Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Rotational Motion (Moment of inertia of rigid bodies), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.