Rotational Motion

JEE Main level, numerical answer, medium

32 of 280

Question 32 of 280

A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively (ksph:kcyl)\left({k}_{sph}:{k}_{cyl}\right) is 2:x2:\sqrt{x}. The value of xx is _______.

Show answer and solution

Answer: 5

Considering rotational axis as the diametrical axis for sphere and axis of cylinder. Then
K12=25R2{K}_{1}^{2}=\dfrac{2}{5}{R}^{2} and K22=12R2{K}_{2}^{2}=\dfrac{1}{2}{R}^{2}
∴K1K2=2/51/2=45∴\dfrac{{K}_{1}}{{K}_{2}}=\sqrt{\dfrac{2/5}{1/2}}=\sqrt{\dfrac{4}{5}}
⇒K1K2=25\Rightarrow \dfrac{{K}_{1}}{{K}_{2}}=\dfrac{2}{\sqrt{5}}
Comparing it with the given value in the question,
25=2x\dfrac{2}{\sqrt{5}}=\dfrac{2}{\sqrt{x}}
∴∴ x=5x=5

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.