JEE Main 2020PhysicsRotational MotionAngular MomentummediumNumerical

JEE Main 2020 — Rotational Motion Question with Solution

From: JEE Main 2020 (Online) 5th September Evening Slot

Question

A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of move 0.1 kg moving in a straight line with velocity 80 m/s hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be _________. JEE Main 2020 (Online) 5th September Evening Slot Physics - Rotational Motion Question 134 English

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Show full solutionCorrect answer: 20
Correct answer
20

Step-by-step explanation

The given situation is shown in the following figure,

JEE Main 2020 (Online) 5th September Evening Slot Physics - Rotational Motion Question 134 English Explanation

Applying law of conservation of angular momentum about pivotal point,

..... (i)

Given, m = 0.1 kg, v = 80 ms1, M = 0.9 kg, l = 1 m

Substituting these values in eq. (i), we get

rad/s

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About this question

This is a previous-year question from JEE Main 2020, 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.