The energy of a system as a function of time t is given as E(t)=A2 exp (−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=5s is
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answer is 4.00.
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Detailed Solution
Given :α=0.2s−1dAA×100=1.25dtt×100=1.50⇒dt×100=1.5t=1.5×5=7.5 Et=A2e−αtTaking log on both sides, we getlogE=2logA−αt⇒dEE=2dAA+αdt∴dEE×100=2dAA×100+αdt×100⇒dEE×100=21.25+0.27.5⇒dEE×100=2.5+1.5=4%