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