JEE Main Physics question
If momentum is increased by 20%, then kinetic energy increases by
Answer: (B)
Solution
K = p²/2m. If p' = 1.2p, then K' = (1.2p)²/2m = 1.44K. Percentage increase = (K'−K)/K × 100 = 0.44 × 100 = 44%.
If momentum is increased by 20%, then kinetic energy increases by
K = p²/2m. If p' = 1.2p, then K' = (1.2p)²/2m = 1.44K. Percentage increase = (K'−K)/K × 100 = 0.44 × 100 = 44%.