JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2025
- Medium
- Numerical
A physical quantity \(Q\) is related to four observables \(a, b, c, d\) as follows :
\(\mathrm{Q}=\frac{\mathrm{ab}{ }^4}{\mathrm{~cd}}\)
where, \(\mathrm{a}=(60 \pm 3) \mathrm{Pa} ; \mathrm{b}=(20 \pm 0.1) \mathrm{m} ; \mathrm{c}=(40 \pm 0.2) \mathrm{Nsm}^{-2}\) and \(\mathrm{d}=(50 \pm 0.1) \mathrm{m}\), then the percentage error in Q is \(\frac{x}{1000}\), where \(x=\) ________ .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 7700
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{Q}=\frac{\mathrm{ab}^4}{\mathrm{~cd}} \\ & \Rightarrow \frac{\Delta \mathrm{Q}}{\mathrm{Q}} \times 100=\left[\frac{\Delta \mathrm{a}}{\mathrm{a}}+4 \frac{\Delta \mathrm{~b}}{\mathrm{~b}}+\frac{\Delta \mathrm{c}}{\mathrm{c}}+\frac{\Delta \mathrm{d}}{\mathrm{d}}\right] \times 100 \\ & \Rightarrow \frac{\mathrm{x}}{1000}=\left[\frac{3}{60}+4\left(\frac{0.1}{20}\right)+\left(\frac{0.2}{40}\right)+\frac{0.1}{50}\right] \times 100 \\ & \Rightarrow \mathrm{x}=7700\end{aligned}\)
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