The radius (r), length (l) and resistance (R) of a metal wire was measured in the laboratory as r =(0.35 pm 0.05) cm R =(100 pm 10) ohm l=(15 pm 0.2) cm The percentage error in resistivity of the material of the wire is :

Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 1 February, Shift 1 — Question 41

The radius (r), length $(l)$ and resistance (R) of a metal wire was measured in the laboratory as $\mathrm{r}=(0.35 \pm 0.05) \mathrm{cm}$ $\mathrm{R}=(100 \pm 10)$ ohm $l=(15 \pm 0.2) \mathrm{cm}$ The percentage error in resistivity of the material of the wire is :

Option A:
$25.6 \%$
Option B:
$39.9 \%$
Correct
Option C:
$37.3 \%$
Option D:
$35.6 \%$

Answer: B

Step-by-step solution

$\rho = {\rm{R}}\frac{\rho }{\ell }$

$\begin{array}{*{20}{r}}{}&{\frac{{{\rm{\Delta }}\rho }}{\rho } = \frac{{{\rm{\Delta R}}}}{{\rm{R}}} + 2\frac{{{\rm{\Delta r}}}}{{\rm{r}}} + \frac{{{\rm{\Delta }}\ell }}{\ell }}\\{}&{\; = \frac{{10}}{{100}} + 2 \times \frac{{0.05}}{{0.35}} + \frac{{0.2}}{{15}}}\\{}&{\; = \frac{1}{{10}} + \frac{2}{7} + \frac{1}{{75}}}\\{}&{\frac{{{\rm{\Delta }}\rho }}{\rho } = 39.9{\rm{\% }}}\end{array}$

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

Answer: B

Step-by-step solution

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