Rotational Motion

JEE Main level, numerical answer, medium

10 of 280

Question 10 of 280

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is n cm\sqrt{n} \mathrm{~cm}. The value of nn is

Show answer and solution

Answer: 265

Using the parallel axis theorem for radius of gyration:
k2=kcm2+d2k^2 = k_{cm}^2 + d^2 where kcmk_{cm} is the radius of gyration about the center of mass and dd is the distance from the center.
For a solid sphere:
Icenter=25MR2I_{center} = \frac{2}{5}MR^2, so kcm2=25R2k_{cm}^2 = \frac{2}{5}R^2
With R=10R = 10 cm and d=15d = 15 cm:
k2=25(10)2+(15)2=40+225=265k^2 = \frac{2}{5}(10)^2 + (15)^2 = 40 + 225 = 265
Since k=nk = \sqrt{n}, we have n=265n = 265.

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.