$\begin{aligned}
& d E=3 \alpha^2 d \alpha e^{-\beta t}+\alpha^3 e^{-\beta t}(-\beta) d t \
& \frac{d E}{E}=\frac{3 d \alpha}{\alpha}+(-\beta) d t \
& \frac{\Delta E}{E}=\frac{3 \Delta \alpha}{\alpha}+\beta \Delta t
\end{aligned}(%)\Delta t=1.6 % \times 5=8 %%\left(\frac{3 \Delta \alpha}{\alpha}\right)=3 \times 1.2 %=3.6 %Puttingvaluesweget\Rightarrow 3.6+8 \times 0.3=6 %$
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…
JEE Main 2025 — Physics Mechanics
2025mcqmedium
The energy of a system is given as E(t)=α3e−βt, 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 :
Official previous-year question
Held on 23 Jan 2025 · Verified 1 Jul 2026.
Options
- A
6%
- B
8.4%
- C
11.6%
- D
4%
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