Young's modulus is determined by the equation given by Y=49000 frac m ell frac dyne cm 2 where M is the mass and ell is the extension of wire used in the experiment. Now error in Young modules(Y) is estimated by taking data from M -ell plot in graph paper. The smallest scale divisions are 5 g and 0.02 cm along load axis and extension axis respectively. If the value of M and ell are 500 g and 2 cm respectively then percentage error of Y is :

Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 8 April, Shift 1 — Question 39

Young's modulus is determined by the equation given by $Y=49000 \frac{\mathrm{~m}}{\ell} \frac{\text { dyne }}{\mathrm{cm}^{2}}$ where $M$ is the mass and $\ell$ is the extension of wire used in the experiment. Now error in Young modules(Y) is estimated by taking data from $\mathrm{M}-\ell$ plot in graph paper. The smallest scale divisions are 5 g and 0.02 cm along load axis and extension axis respectively. If the value of M and $\ell$ are 500 g and 2 cm respectively then percentage error of Y is :

Option A:
$0.2 \%$
Option B:
$0.02 \%$
Option C:
$2 \%$
Correct
Option D:
$0.5 \%$

Answer: C

Step-by-step solution

$\frac{\Delta \mathrm{Y}}{\mathrm{Y}}=\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{\Delta \ell}{\ell}$

\begin{aligned} & =\frac{5}{500}+\frac{0.02}{2}=0.01+0.01 \frac{\Delta Y}{Y} & =0.02 \Rightarrow \% \frac{\Delta Y}{Y}=2 \% \end{aligned}

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Significant Figures and Error Analysis

Answer: C

Step-by-step solution

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