JEE Main 2025 Jan 23 Shift 2, Physics Q5: Units And Dimensions

JEE Main2025Jan 23, Shift 2Physics

The energy of a system is given as $E (t) = α^{3} e^{- β t}$ , 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
$4%$
B
$11.6%$
C
$6%$
D
$8.4%$

Solution

Take the natural log: $ln E = 3 ln α - β t$ .

Differentiate, taking magnitudes for the maximum (worst-case) error:

$(\frac{Δ E}{E})_{ma x} = 3 \cdot \frac{Δ α}{α} + β Δ t .$

$Δ t = t \cdot \frac{Δ t}{t} = 5 \times 0.016 = 0.08 s$ , so $β Δ t = 0.3 \times 0.08 = 0.024 = 2.4%$ .

Total: $3 (1.2%) + 2.4% = 3.6% + 2.4% = 6%.$

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JEE Main 2025 Jan 23 Shift 2, Physics Q5: Units And Dimensions

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