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

Correct answer: A. 6 \% aligned & d E=3 α^2 d α e^ -β t +α^3 e^ -β t (-β) d t & (d E)/(E)=(3 d α)/(α)+(-β) d t & (Δ E)/(E)=(3 Δ α)/(α)+β Δ t aligned (\%) Δ t=1.6 \% × 5=8 \% \%((3 Δ α)/(α))=3 × 1.2 \%=3.6 \% Putting values we get 3.6+8 × 0.3=6 \%

JEE Main 202523 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Evening Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2025 · 23 January · Evening Shift · Physics · Mathematics in Physics

The energy of a system is given as E(t)=α^3 e^ -β t , where t is...

The energy of a system is given as

\mathrm{E}(\mathrm{t})=\alpha^3 \mathrm{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 :

Options

A $6 \%$
B $8.4 \%$
C $11.6 \%$
D $4 \%$

Correct answer

A. $6 \%$

Step-by-step solution

\begin{aligned} & d E=3 \alpha^2 d \alpha e^{-\beta t}+\alpha^3 e^{-\beta t}(-\beta) d t \\ & \frac{d E}{E}=\frac{3 d \alpha}{\alpha}+(-\beta) d t \\ & \frac{\Delta E}{E}=\frac{3 \Delta \alpha}{\alpha}+\beta \Delta t \end{aligned}

(\%)

\Delta t=1.6 \% \times 5=8 \% \%\left(\frac{3 \Delta \alpha}{\alpha}\right)=3 \times 1.2 \%=3.6 \%

Putting values we get

\Rightarrow 3.6+8 \times 0.3=6 \%

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