JEE Main 2022 · 29 July · evening · Single correct · +4 / −1
If momentum of a body is increased by 20%, then its kinetic energy increases by
- (a) 36%
- (b) 40%
- (c) 44%
- (d) 48%
Answer and solution
Answer (c) 44%
Solution
Physics sir · 3 steps
Write kinetic energy in terms of momentum: K goes as p squared, so square the factor 1.2.
- Link K and p. With p = mv, kinetic energy can be written as K = p²/2m. The mass does not change, so K is proportional to p².
K₂/K₁ = (p₂/p₁)² = (1.2)² = 1.44
- Find the increase. The new kinetic energy is 1.44 times the old one. The increase is 1.44 - 1 = 0.44 of the original value.
- So. Kinetic energy increases by 44%, which is option (c).
The idea to keep. For a fixed mass, K = p²/2m. Turn a percentage change into a factor, square it, then subtract 1. Do not double the percentage, because that rule only works for very small changes.
Answer from the NTA final answer key. Solution by PrepWiser.