The energy of a system as a function of time $t$ is given as $E (t) = A^{2} \exp (- α t)$ , 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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Detailed Solution

Energy of system is given as $E (t) = A^{2} e^{- α t}$
Taking natural log on both side gives
$\ln E = \ln A^{2} + \ln e^{- α t}$
$\Rightarrow \ln E = 2 \ln A - \ln α t$
Percentage error in measurement of E is given as
$\frac{Δ E}{E} \times 100 = 2 \frac{Δ A}{A} \times 100 + α Δ t \times 100$
$\Rightarrow \frac{Δ E}{E} \times 100 = 2 \times 1.25 + \frac{2}{10} \times 1.5 \times 5$
$\Rightarrow \frac{Δ E}{E} \times 100 = 2.25 + 1.5 = 4 %$