Question

A $2m{m}^2$ cross-sectional area wire is stretched by 4 mm by a certain weight.If the same material wire of cross-sectional area $8m{m}^2$ is stretched by the same weight, the stretched length is

• A. 2 mm
• B. 0.5 mm
• C. 1 mm
• D. 1.5 mm

Answer 1

Answer: (C)

1 mm

From Hooke's Law , $Y=\frac{FL}{A\Delta L}$
$⟹$ $\frac{\Delta L^{\prime }}{\Delta L}=\frac{A}{A^{\prime }}$ (For same $F,L$ and $Y$)
Given : $\Delta L=4mm$ $A=2m{m}^2$ and $A^{\prime }=8m{m}^2$
$\therefore$ $\frac{\Delta L^{\prime }}{4}=\frac{2}{8}$
$⟹\Delta L^{\prime }=1mm$