If the error in the measurement of the momentum of a particle is +100%, what will be the percentage error in the measurement of its kinetic energy? MCQ with Answer and Explanation

If the error in the measurement of the momentum of a particle is +100%, what will be the percentage error in the measurement of its kinetic energy?
A. 300%
B. 200%
C. 100%
D. 400%
Answer: Option A
Solution (By JKSSB Mock Tests)
Kinetic Energy (K) is given by K = p^2 / 2m. If momentum (p) increases by 100%, the new momentum is p' = p + 1.0p = 2p. The new kinetic energy is K' = (2p)^2 / 2m = 4(p^2 / 2m) = 4K. The increase is 4K - K = 3K. Percentage increase = (3K/K) * 100 = 300%.

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