Under the same load, wire A having length $5.0 \mathrm{~m}$ and cross section $2.5 \times 10^{-5} \mathrm{~m}^{2}$ stretches uniformly by the same amount as another wire B of length $6.0 \mathrm{~m}$ and a cross section of $3.0 \times 10^{-5}$ $\mathrm{m}^{2}$ stretches. The ratio of the Young's modulus of wire A to that of wire $B$ will be :
Solution
$\Delta \ell=\frac{F \ell}{S Y}$
<br/><br/>$F$ is same for both wire and $\Delta \ell$ is also same
<br/><br/>$$
\begin{aligned}
& \frac{\Delta \ell}{F}=\frac{\ell}{S Y} \\\\
& \Rightarrow \frac{\ell_{A}}{S_{A} Y_{A}}=\frac{\ell_{B}}{S_{B} Y_{B}} \\\\
& \Rightarrow \frac{5}{2.5 \times Y_{A}}=\frac{6}{3 \times Y_{B}} \\\\
& \Rightarrow \frac{Y_{A}}{Y_{B}}=1
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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