JEE Main 2022PhysicsWork Power EnergyEasyMCQ

JEE Main 2022 — Work Power Energy Question with Solution

JEE Main 2022 (25 Jun Shift 2)

Question

For a particle in uniform circular motion, the acceleration a→ at any point PR,θ on the circular path of radius R is (when θ is measured from the positive x-axis and v is uniform speed):

Choose an option

Show full solutionCorrect option: C
Correct answer
C-v2Rcosθi^-v2Rsinθj^

Step-by-step explanation

Acceleration of a particle in uniform circular motion is directed towards the centre of the circle and its magnitude is v2R. The radius vector isr→=Rcosθi^+Rsinθj^

Therefore, a→=-v2Rr^=v2R−cosθi^−sinθj^

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About this question

This is a previous-year question from JEE Main 2022, covering the Work Power Energy 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.