JEE Main 2022PhysicsRotational MotionEasyNumerical

JEE Main 2022 — Rotational Motion Question with Solution

JEE Main 2022 (28 Jun Shift 1)

Question

The position vector of 1 kg object is r→=3i^-j^ m and its velocity v→=3j^+k^ m s-1. The magnitude of its angular momentum is x N m s, where x is

Enter your answer

Show full solutionCorrect answer: 91
Correct answer
91

Step-by-step explanation

Angular momentum of a particle is given by,

L→=r→×P→=r→×mv→ also m=1 kg.

⇒L→=3i^-j^×3j^+k^=9k^-3j^-i^ N m s

The magnitude of the angular momentum will be,

⇒L→=92+32+12=91 N m s

Therefore, the value of x=91.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Rotational Motion chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, covering the Rotational Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.