Question

A copper wire of length $2.4m$ and a steel wire of length $1.6m$, both the diameter $3mm$, are connected end to end. The ratio fo elongation of steel to the copper wires is then
$(Y_{copper}=1.2\times {10}^{11}N{m}^{-2},Y_{steel}=2\times {10}^{11}N{m}^{-2})$

• A. $\frac{5}{2}$
• B. $\frac{2}{5}$
• C. $\frac{3}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{2}{5}$

Given:

Length copper wire, $L_c=2.4m$

Length steel wire, $L_s=1.6m$

$(Y_{copper}=1.2\times {1011}^{-2}Nm,Y_{steel}=2\times {1011}^{-2}Nm)$

We know that,

Elongation $\Delta L=\frac{FL}{YA}$

So,

$\frac{\Delta L_S}{\Delta L_C}=\frac{L_S}{L_C}\times \frac{Y_C}{Y_S}$

Substitution the given values, we get

$\frac{{\Delta L}_S}{{\Delta L}_C}=\frac{1.6m}{2.4m}\times \frac{1.2\times {10}^{11}N{m}^{-2}}{2\times {10}^{11}N{m}^{-2}}=\frac{2}{5}$