JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2025
- Medium
- MCQ
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 :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(6 \%\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& d E=3 \alpha^2 d \alpha e^{-\beta t}+\alpha^3 e^{-\beta t}(-\beta) d t \\
& \frac{d E}{E}=\frac{3 d \alpha}{\alpha}+(-\beta) d t \\
& \frac{\Delta E}{E}=\frac{3 \Delta \alpha}{\alpha}+\beta \Delta t
\end{aligned}\)
(\%) \(\Delta t=1.6 \% \times 5=8 \%\)
\(\%\left(\frac{3 \Delta \alpha}{\alpha}\right)=3 \times 1.2 \%=3.6 \%\)
Putting values we get
\(\Rightarrow 3.6+8 \times 0.3=6 \%\)
& 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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