The energy of a system is given as E(t) = α³ e-β t, where t is the time and β = 0.3 s⁻¹. The errors in the measurement of α and t are 1.2% and 1.6%, respectively. At t = 5 s, maximum percentage error in the energy is: (a) 6% (b) 11.6% (c) 4% (d) 8.4%

Answer: (a). ln E = 3ln α - β t ⇒ (Δ E)/(E) = 3(Δ α)/(α) + β Δ t 3(Δ α)/(α) = 3(1.2%) = 3.6% Δ t = 1.6% of 5 s = 0.08 s, so βΔ t = 0.3 × 0.08 = 0.024 = 2.4% (Δ E)/(E) = 3.6% + 2.4% = 6% So the answer is (a) 6%.

The energy of a system is given as E(t) = α³ e-β t, where t is the… | JEE Main 2025 (23 Jan, Shift 2)

Q54JEE Main 2025 · 23 JAN · SHIFT 2Medium

Errors in Measurement

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

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Answer: (a)

$\ln E = 3\ln\alpha - \beta t \Rightarrow \frac{\Delta E}{E} = 3\frac{\Delta \alpha}{\alpha} + \beta\,\Delta t$
$3\frac{\Delta \alpha}{\alpha} = 3(1.2\%) = 3.6\%$
$\Delta t = 1.6\%$ of $5\text{ s} = 0.08\text{ s}$ , so $\beta\Delta t = 0.3 \times 0.08 = 0.024 = 2.4\%$
$\frac{\Delta E}{E} = 3.6\% + 2.4\% = 6\%$
So the answer is (a) $6\%$ .

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Errors in Measurement question from JEE Main 2025 (23 Jan, Shift 2)

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