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Answer :
Answer: [tex]0.05\ mm[/tex]
Explanation:
Given
Cross-sectional area of wire [tex]A_1=4\ mm^2[/tex]
Extension of wire [tex]\delta l=0.1\ mm[/tex]
Extension in a wire is given by
[tex]\Rightarrow \delta l=\dfrac{FL}{AE}[/tex]
where, [tex]E=\text{Youngs modulus}[/tex]
[tex]\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)[/tex]
for same force, length and material
[tex]\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)[/tex]
Divide (i) and (ii)
[tex]\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm[/tex]
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