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. 100%
B. 400%
C. 300%
D. 200%
Answer: Option C
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%.
Explanation:
Methane (CH₄) is primary component. LPG is mainly propane and butane. CNG is compressed methane. This relates to everyday physics and chemistry.
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