The kinetic energy is given by,
K=21mv2
Differentiating the above equation,
dK=21(v2dm+2mvdv)
Dividing by K,
⇒KdK=mdm+v2dv
⇒1000dK=50.5+202×0.4
⇒dK=100+40=140
⇒KE=(1000±140)J
JEE Main 2023 — Physics Mechanics
A body of mass (5±0.5)kg is moving with a velocity of (20±0.4)ms−1. Its kinetic energy will be
Held on 13 Apr 2023 · Verified 6 Jul 2026.
(1000±0.14)J
(500±0.14)J
(500±140)J
(1000±140)J
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