JEE Main 2022PhysicsUnits and DimensionsEasyMCQ

JEE Main 2022 — Units and Dimensions Question with Solution

JEE Main 2022 (27 Jul Shift 2)

Question

An expression of energy density is given by u=αβsinαxkt, where α, β are constants, x is displacement, k is Boltzmann constant and t is the temperature. The dimensions of β will be

Choose an option

Show full solutionCorrect option: D
Correct answer
DM0L2T0

Step-by-step explanation

Given that energy density U=αβsinαxkt

As αxkt should be dimensionless, the dimensions of αx and kt should be same.

⇒αx=kt⇒α=ktx

Now,

U=αβ

⇒β=αU=ktxEV=EttxEV

⇒β=L3L=M0L2T0

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About this question

This is a previous-year question from JEE Main 2022, covering the Units and 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.