Rotational Motion

JEE Main level, single correct, medium

214 of 280

Question 214 of 280

The maximum and minimum distances of a comet from the Sun are 1.6×1012m1.6\times {10}^{12}m and 8.0×1010m8.0\times {10}^{10}m respectively. If the speed of the comet at the nearest point is 6×104ms−16\times {10}^{4}{ms}^{-1}, the speed at the farthest point is

Show answer and solution

Answer: (C) 3.0×103ms−13.0\times {10}^{3}m{s}^{-1}

By angular momentum conservation :
mv1r1=mv2r2m{v}_{1}{r}_{1}=m{v}_{2}{r}_{2}
v1=48×10141.6×1012=3000ms−1{v}_{1}=\dfrac{48\times {10}^{14}}{1.6\times {10}^{12}}=3000m{s}^{-1}
=3×103ms−1=3\times {10}^{3}m{s}^{-1}.

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 (Angular momentum and Angular impulse), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.