QJEEAdv15Ph6eea: JEE Advanced 2015 Physics - Class 11 JEEAdv15Ph6eea Solution: The energy of a system as a function of time t is given as E(t) …

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

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

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

$= 2.5 + 1.5$

$= 4\%$