JEE Main 2020PhysicsUnits and DimensionsEasyMCQ

JEE Main 2020 — Units and Dimensions Question with Solution

JEE Main 2020 (02 Sep Shift 1)

Question

If speed V, area A and force F are chosen as fundamental units, then the dimension of Young's modulus will be :

Choose an option

Show full solutionCorrect option: D
Correct answer
DFA-1V0

Step-by-step explanation

Y∝FaVbAc  Y=FA

MLT-2L2∝M1L1T-2aL1T-1bL2c

M1L-1T-2∝MaLa+b+2cT-2a-b

a + b + 2c = –1

–2a + b = –2

a = 1, b = 0, c = –1

Y=F1V0A-1

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About this question

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