JEE Main 2019PhysicsMotion In Two DimensionsHardMCQ

JEE Main 2019 — Motion In Two Dimensions Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

A particle of mass m is moving along a trajectory given by
x=x0+a cosω1t
y=y0+b sinω2t
The torque, acting on the particle about the origin, at t=0 is:

Choose an option

Show full solutionCorrect option: A
Correct answer
A+my0aω12k^

Step-by-step explanation

x=x0+a cosω1t
y=y0+b sin ω2t
ax=-a ω12cos ω1t
ay=-b ω22sin ω2t
At t=0 ,
F→=ma→=-a ω12 i^
And
r→=x0+ a i^+y0 j^
Torque, τ=r→×F→
τ=x0+ a i^+y0 j^×-a ω12 i^
=my0 aω12 k^

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About this question

This is a previous-year question from JEE Main 2019, covering the Motion In Two Dimensions 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.