Question
When a stress of $2 \times 10^6 N / m ^2$ is applied on a copper wire ( $Y =1^{11} N / m ^2$ ), what will be the percentage increase in its length?
✓
Answer
Given :
$\begin{aligned}
Y & =10^{11} N / m^2 \\
\text { Stress } & =\frac{F}{A}=2 \times 10^6 N / m^2 \\Y & =\frac{FL}{Al} \\
10^{11} & =2 \times 10^6 \times \frac{L}{l}
\end{aligned}$
$\therefore \quad \frac{l}{L}=\frac{2 \times 10^6}{10^{11}}=2 \times 10^{-5}$
For the percentage multiplying by 100 both sides
$\frac{1}{L} \times 100=2 \times 10^{-5} \times 100$
Percentage increase in length $\%$
$=2 \times 10^{-3}=0.002 \%$
$\begin{aligned}
Y & =10^{11} N / m^2 \\
\text { Stress } & =\frac{F}{A}=2 \times 10^6 N / m^2 \\Y & =\frac{FL}{Al} \\
10^{11} & =2 \times 10^6 \times \frac{L}{l}
\end{aligned}$
$\therefore \quad \frac{l}{L}=\frac{2 \times 10^6}{10^{11}}=2 \times 10^{-5}$
For the percentage multiplying by 100 both sides
$\frac{1}{L} \times 100=2 \times 10^{-5} \times 100$
Percentage increase in length $\%$
$=2 \times 10^{-3}=0.002 \%$
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