Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 23 January, Morning Shift — Question 29

Four persons measure the length of a rod as 20.00 cm,19.75 cm,17.01 cm20.00 \mathrm{~cm}, 19.75 \mathrm{~cm}, 17.01 \mathrm{~cm} and 18.25 cm . The relative error in the measurement of average length of the rod is :

  1. Option A:

    0.24

  2. Option B:

    0.18

  3. Option C:

    0.06

    Correct
  4. Option D:

    0.08

Answer: C

Step-by-step solution

ℓmean =ℓ1+ℓ2+ℓ3+ℓ44\ell_{\text {mean }}=\frac{\ell_{1}+\ell_{2}+\ell_{3}+\ell_{4}}{4} ℓmean =20.00+19.75+17.01+18.254\ell_{\text {mean }}=\frac{20.00+19.75+17.01+18.25}{4} =18.75=18.75 Δℓmean =∣Δℓ1∣+∣Δℓ2∣+∣Δℓ3∣+∣Δℓ4∣4\Delta \ell_{\text {mean }}=\frac{\left|\Delta \ell_{1}\right|+\left|\Delta \ell_{2}\right|+\left|\Delta \ell_{3}\right|+\left|\Delta \ell_{4}\right|}{4} =1.25+1+1.74+0.54=1.12=\frac{1.25+1+1.74+0.5}{4}=1.12 So, relative error =Δℓmean ℓmean =1.1218.75=0.06=\frac{\Delta \ell_{\text {mean }}}{\ell_{\text {mean }}}=\frac{1.12}{18.75}=0.06

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Significant Figures and Error Analysis