Rotational Motion

JEE Main level, single correct, medium

263 of 280

Question 263 of 280

A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsph{h}_{sph} and hcyl{h}_{cyl} on the inline. The ratio hsphhcyl\dfrac{{h}_{sph}}{{h}_{cyl}} is given by:

Show answer and solution

Answer: (C) 1415\dfrac{14}{15}

From conservation of mechanical energy,
For solid sphere: 12msphV2+12Isphω2=msphghsph\dfrac{1}{2}{m}_{sph}{V}^{2}+\dfrac{1}{2}{I}_{sph}{\omega }^{2}={m}_{sph}g{h}_{sph}
V22+V25=ghsph\dfrac{{V}^{2}}{2}+\dfrac{{V}^{2}}{5}=g{h}_{sph}
7V210=ghsph\dfrac{7{V}^{2}}{10}=g{h}_{sph} …(i)
For cylinder: 12mcylV2+12Icylω2=mcylghcyl\dfrac{1}{2}{m}_{cyl}{V}^{2}+\dfrac{1}{2}{I}_{cyl}{\omega }^{2}={m}_{cyl}g{h}_{cyl}
V22+V24=ghcyl\dfrac{{V}^{2}}{2}+\dfrac{{V}^{2}}{4}=g{h}_{cyl}
3V24=ghcyl\dfrac{3{V}^{2}}{4}=g{h}_{cyl} …(ii)
From (i) and (ii) hsphhcyl=1415\dfrac{{h}_{sph}}{{h}_{cyl}}=\dfrac{14}{15}

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 (Rolling without Slipping), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.