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Saturday, 31 October

JEE Main 2021 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 25 July 2021, Shift 2, Physics Q21

Question 21 of 90 in this shift, Physics question 21 of 30, Section B.

A force of F=(5y+20)j^F=(5y+20)\hat{j} N acts on a particle. The workdone by this force when the particle is moved from yy=0 m to yy=10 m is __________ J.

Official answer

450

NTA final key.