The resistance R =frac V I where V =(200 pm 5) V and I=(20 pm 0.2) A , the percentage error in the measurement of R is :

Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 29 January, Shift 1 — Question 36

The resistance $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$ where $\mathrm{V}=(200 \pm 5) \mathrm{V}$ and $I=(20 \pm 0.2) A$ , the percentage error in the measurement of R is :

Option A:
$3.5 \%$
Correct
Option B:
$7 \%$
Option C:
$3 \%$
Option D:
$5.5 \%$

Answer: A

Step-by-step solution

$\mathrm{R}=\frac{\mathrm{V}}{1}$

According to error analysis

$\frac{\mathrm{dR}}{\mathrm{R}}=\frac{\mathrm{dV}}{\mathrm{V}}+\frac{\mathrm{dI}}{\mathrm{I}}$

$\frac{\mathrm{dR}}{\mathrm{R}}=\frac{5}{200}+\frac{0.2}{20}$

$\frac{\mathrm{dR}}{\mathrm{R}}=\frac{7}{200}$

$\% \operatorname{error} \frac{d R}{R} \times 100=\frac{7}{200} \times 100=3.5 \%$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Significant Figures and Error Analysis