A physical quantity Q is related to four observables a, b, c, d as follows : Q=frac a 4 c d where, a =(60 pm 3) Pa ; b =(20 pm 0.1) m ; c=(40 pm 0.2) Nsm -2 and d =(50 pm 0.1) m , then the percentage error in Q is x/1000 , where x = .

Physics · Units, Dimensions & Error Analysis

JEE Main 2025 — 29 January, Evening Shift — Question 39

A physical quantity Q is related to four observables $a, b, c, d$ as follows : $Q=\frac{a^{4}}{c d}$ where, $\mathrm{a}=(60 \pm 3) \mathrm{Pa} ; \mathrm{b}=(20 \pm 0.1) \mathrm{m} ;$ $c=(40 \pm 0.2) \mathrm{Nsm}^{-2}$ and $\mathrm{d}=(50 \pm 0.1) \mathrm{m}$ , then the percentage error in $Q$ is $\frac{x}{1000}$ , where $\mathrm{x}=$ $\qquad$ .

Answer: 77

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{Q}=\frac{\mathrm{ab}^{4}}{\mathrm{~cd}}$

$\Rightarrow \frac{\Delta \mathrm{Q}}{\mathrm{Q}} \times 100=\left[\frac{\Delta \mathrm{a}}{\mathrm{a}}+4 \frac{\Delta \mathrm{~b}}{\mathrm{~b}}+\frac{\Delta \mathrm{c}}{\mathrm{c}}+\frac{\Delta \mathrm{d}}{\mathrm{d}}\right] \times 100$

$\Rightarrow \frac{\mathrm{x}}{1000}=\left[\frac{3}{60}+4\left(\frac{0.1}{20}\right)+\left(\frac{0.2}{40}\right)+\frac{0.1}{50}\right] \times 100$

$\Rightarrow \mathrm{x}=7700$

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Significant Figures and Error Analysis

Answer: 77

Step-by-step solution

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