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Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 5 April, Evening Shift — Question 2

In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as 10V and 5 A, respectively. The least counts of voltmeter and ammeter are 500mV500 \mathrm{mV} and 200mA200 \mathrm{mA}, respectively. The estimated error in the resistance measurement is Ω\Omega

  1. Option A:

    0.25

  2. Option B:

    2

  3. Option C:

    2.5

  4. Option D:

    0.18

    Correct

Answer: D

Step-by-step solution

R=V/I=10/5=2ΩR = V/I = 10/5 = 2\Omega. ΔR/R=ΔV/V+ΔI/I=(0.5/10)+(0.2/5)=0.05+0.04=0.09\Delta R/R = \Delta V/V + \Delta I/I = (0.5/10)+(0.2/5)=0.05+0.04=0.09, so ΔR=0.09×2=0.18Ω\Delta R = 0.09\times2 = 0.18\Omega.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Significant Figures and Error Analysis

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