The energy of a system as a function of time t is given as Et=A2 exp −αt, where α=0.2s−1. The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of Et at t=5 s is
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answer is 4.
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Detailed Solution
E(t)=A2e−αt⇒logE=2logA−αtErrors always add up for maximum errorΔEE×100=2ΔAA×100+αΔtt×100×t⇒ΔEE×100=2×1.25+0.21.55=4%