Question

A wire of cross section $4 \,mm^2$ is stretched by $0.1\, mm$ by a certain weight. How far (length) will the wire of same material and length but of area $8 \,mm^2$ stretch under the action of same force?

• A. 0.05 mm
• B. 0.10 mm
• C. 0.15 mm
• D. 0.20 mm

Answer 1

Answer: (A)

Given $A_1 = 4\, mm^2 = 4 \times 10^{-6} \, m^2$,
$\Delta L_{1}=0.1 \times10^{-3} m, A_{2 }=8 \times10^{-6} m^{2},$
$ Y_{2} = Y_{1}, L_{2}=L_{1},F_{2}=F_{1}$
$\Delta L_{1}=\frac{F_{1 }L_{1}}{A_{1}Y_{1}},\Delta L_{2} = \frac{F_{2}L_{2}}{A_{2}Y_{2}}$
$ \frac{\Delta L_{2}}{\Delta L_{1}}= \frac{A_{1}}{A_{2}}=\frac{4 \times10^{-6}}{8\times10^{-6}}=\frac{1}{2}$
$\therefore \, \Delta L_{2} =\frac{\Delta L_{1}}{2} = 0.05 \times 10^{-3} \, m = 0.05 mm$