The energy of a system is given as $\mathrm{E}(\mathrm{t})=\alpha^{3} \mathrm{e}^{-\beta \mathrm{t}}$, 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 $\mathrm{t}=5 \mathrm{~s}$, maximum percentage error in the energy is :

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Units & Measurement

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 :

Answer: (C)

Given

E (t) = α^{3} e^{- β t}

, take logarithms:

ln E = 3 ln α - β t

For the maximum possible error, the individual contributions add:

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

With

\frac{Δ α}{α} = 1.2%

\frac{Δ t}{t} = 1.6%

β = 0.3 s^{- 1}

and

t = 5 s

(\frac{Δ E}{E})_{ma x} = 3 (1.2%) + (0.3) (5) (1.6%) = 3.6% + 2.4% = 6%

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option