JEE Main 2015PhysicsLaws of MotionMediumMCQ

JEE Main 2015 — Laws of Motion Question with Solution

JEE Main 2015 (11 Apr Online)

Question

A large number (n) of identical beads, each of mass m and radius r are strung on a thin smooth rigid horizontal rod of length L(L≫r) and are at rest at random positions. The rod is mounted between two rigid supports (see figure). If one of the beads is now given a speed v, the average force experienced by each support after a long time is (assume all collisions are elastic):

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Show full solutionCorrect option: B
Correct answer
Bmv2L-2nr

Step-by-step explanation

Average time for 1 collision to occur on 1 wall, 
=tavg=average distance travelledspeed=2L-2nrv

⇒tavg=2L-2nrv. Change in momentum in 1 collision =2mv. Average force on each support, favg=2mvtavg=2mv22L-2nr⇒favg=mv2L-2nr.

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About this question

This is a previous-year question from JEE Main 2015, covering the Laws of 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.