The energy of a system is given as $\mathrm{E}(\mathrm{t})={\alpha }^{3}{e}^{-\beta t}$, where $t$ is the time and $\beta =0.3{\mathrm{s}}^{-1}$. The errors in the measurement of $\alpha$ and $t$ are$1.2%$ and $1.6%$, respectively. At $t=5\mathrm{s}$, maximum percentage error in the energy is : [JEE Main 2025, 23 Jan (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: a) $6%$. $E(t)={\alpha }^{3}{e}^{-\beta t}\ dE(t)=3{\alpha }^{2}{e}^{-\beta t}d\alpha -\beta {\alpha }^{3}{e}^{-\beta t}dt\ \frac{dE(t)}{E(t)}=\frac{3{\alpha }^{2}{e}^{-\beta t}d\alpha }{E(t)}+\frac{\beta {\alpha }^{3}{e}^{-\beta t}dt}{E(t)}\ \frac{\Delta E}{E}=\frac{3\Delta \alpha }{\alpha }+\beta \Delta t\ \frac{\Delta t}{t}\times 100=1.6\ \frac{\Delta t}{5}=0.016\Rightarrow \Delta t=0.08$ $\frac{\Delta E}{E}=3\times 0.012+0.3\times 0.08\ \frac{\Delta E}{E}=0.036+0.024\Rightarrow \frac{\Delta E}{E}=0.06\ %E=6%$

The energy of a system is given as \(\mathrm{E}(\mathrm{t})={\alpha }^{3}{e}^{-\beta t}\), where \(t\) is the time and \…

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