Question

Two wires of equal cross section, but one made up of steel and the other copper, are joined end to end. When the combination is kept under tension, the elongations in the two wires are found to be equal. If $Y_{steel}=$ 2.0 x ${10}^{11}N{m}^{-2}$ and $Y_{copper}=$ 1.1 x 10${}^{11}N{m}^{-2}$, the ratio of the lengths of the two wires is :

• A. $20:11$
• B. $11:20$
• C. $5:4$
• D. $4:5$

Answer 1

Answer: (A)

$20:11$

$Y=\frac{F/A}{△l/l}$ (standard result)

$△l=\frac{Fl}{AY}$

$△l_S=\frac{F{L}_s}{A{Y}_s}$ (for different cases different subscripts are used)

$△l_c=\frac{F{l}_c}{A{Y}_c}$

$Byproblem△l_S=△l_c$

$\frac{F{l}_s}{A{Y}_s}=\frac{F{l}_c}{A{Y}_c}$

$\frac{l_S}{l_c}=\frac{Y_s}{Y_c}=\frac{2\times {10}^{11}}{1.1\times {10}^{11}}=\frac{20}{11}$