JEE Advance - Physics (2015 - Paper 2 Offline - No. 4)

The energy of a system as a function of time t is given as E(t) = $${A^2}\exp \left( { - \alpha t} \right)$$, where $$\alpha = 0.2\,{s^{ - 1}}$$. 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
Answer
4

Explanation

$$E(t) = {A^2}{e^{ - \alpha t}}$$ ...... (i)

$$\alpha$$ = 0.2 s$$-$$1

$$\left( {{{dA} \over A}} \right) \times 100 = 1.25\% $$

$$\left( {{{dt} \over t}} \right) \times 100 = 1.50$$

$$ \Rightarrow (dt \times 100) = 1.5t = 1.5 \times 5 = 7.5$$

Differentiating on both sides of equation (i), we get

$$ d \mathrm{E}=(2 \mathrm{~A} d \mathrm{~A}) e^{-\alpha t}+\mathrm{A}^2 e^{-\alpha t}(-\alpha d t) $$

Dividing throughout by $\mathrm{E}=\mathrm{A}^2 e^{-\alpha t}$

$$ \frac{d \mathrm{E}}{\mathrm{E}}=\frac{2}{\mathrm{~A}} d \mathrm{~A}+\alpha d t $$

(Considering worst possible case)

$$\therefore$$ $$\left( {{{dE} \over E}} \right) \times 100 = 2\left( {{{dA} \over A}} \right) \times 100 + \alpha (dt \times 100)$$

$$ = 2(1.25) + 0.2(7.5)$$

$$ = 2.5 + 1.5$$

$$ = 4\% $$

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