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The Young's modulus of a material is $2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ and its elastic limit is $1 \times 10^8 \mathrm{~N} / \mathrm{m}^2$. For a wire of $1 \mathrm{~m}$ length of this material, the maximum elongation achievable is
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The correct answer is:
$0.5 \mathrm{~mm}$
Given,
Young's modulus of material $=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
Elastic limit or stress $=1 \times 10^\theta \mathrm{N} / \mathrm{m}^2$
Length of the wire $=1 \mathrm{~m}$
We know that,
$$
Y=\frac{\text { Stress }}{\text { Strain }} \Rightarrow Y=\frac{\text { Stress }}{\frac{\Delta L}{l}}
$$
Here, $l=$ original length of the wire
$\Delta L=$ charge in the length of the wire
$$
\begin{aligned}
\Delta L & =\frac{\text { Stress } \times l}{Y}=\frac{1 \times 10^\theta \times 1}{2 \times 10^{11}} \\
& =0.5 \times 10^{-3}=0.5 \mathrm{~mm}
\end{aligned}
$$
Young's modulus of material $=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
Elastic limit or stress $=1 \times 10^\theta \mathrm{N} / \mathrm{m}^2$
Length of the wire $=1 \mathrm{~m}$
We know that,
$$
Y=\frac{\text { Stress }}{\text { Strain }} \Rightarrow Y=\frac{\text { Stress }}{\frac{\Delta L}{l}}
$$
Here, $l=$ original length of the wire
$\Delta L=$ charge in the length of the wire
$$
\begin{aligned}
\Delta L & =\frac{\text { Stress } \times l}{Y}=\frac{1 \times 10^\theta \times 1}{2 \times 10^{11}} \\
& =0.5 \times 10^{-3}=0.5 \mathrm{~mm}
\end{aligned}
$$
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