Work Energy Power

JEE Main level, single correct, medium

66 of 183

Question 66 of 183

A body of mass (5±0.5)kg(5\pm 0.5)kg is moving with a velocity of (20±0.4)ms−1.(20\pm 0.4)m{s}^{-1}. Its kinetic energy will be

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Answer: (D) (1000±140)J(1000\pm 140)J

The kinetic energy is given by,
K=12mv2K=\dfrac{1}{2}m{v}^{2}
Differentiating the above equation,
dK=12(v2dm+2mvdv)dK=\dfrac{1}{2}({v}^{2}dm+2mvdv)
Dividing by KK,
⇒dKK=dmm+2dvv\Rightarrow \dfrac{dK}{K}=\dfrac{dm}{m}+\dfrac{2dv}{v}
⇒dK1000=0.55+2×0.420\Rightarrow \dfrac{dK}{1000}=\dfrac{0.5}{5}+\dfrac{2\times 0.4}{20}
⇒dK=100+40=140\Rightarrow dK=100+40=140
⇒KE=(1000±140)J\Rightarrow KE=(1000\pm 140)J

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