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Q.

The energy of a system as a function of time t is given as Et=A2e−αt, where α=0.2 s−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 E(t) at t = 5 s is

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answer is 4.

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Detailed Solution

E(t)=A2e−αtTake natural logarithm ln(E)=2ln(A)+(−αt)After differentiating dEE=2dAA+(αdt)Errors always add up for maximum error∴dEE=2dAA+α(dtt)×t Here, dAA=1.25%,dtt=1.5%,t=5s,α=0.2s−1∴dEE=(2×1.25%)+(0.2)×(1.5%)×5=4%

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