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← JEE Main 2022 · 29 Jun · Shift 1
PhysicsSingle correct+4 / -1Question 38 of 90

JEE Main 2022 · 29 Jun · Shift 1, Physics question 38

A particle of mass 500 gm is moving in a straight line with velocity v=bx5/2v=bx^{5/2}. The work done by the net force during its displacement from x=0x=0 to x=4x=4 m is : (Take b=0.25b=0.25 m−3/2^{-3/2} s−1^{-1}).

Choose one option
Answer and worked solution

Official answer: D

Work-energy. At x=4x=4: v=0.25×45/2=8v=0.25\times4^{5/2}=8 m s−1^{-1}, so W=12×0.5×64=16W=\frac12\times0.5\times64=16 J.

(D)