Question

An aluminium wire and a steel wire of the same length and cross-section are joined end to end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in the length of the composite wire is 2.7 mm, then the increase in the length of each wire is {in mm). $(Y_{At}=2\times 10^{11}N{m}^{-2},Y_{steet}=7\times 10^{11}N{m}^{-2})$

• A. 1.7,1
• B. 1.2,1.4
• C. 1.5,1.2
• D. 2,0.7

Answer 1

Answer: (D)

Total increase in length, $e=e_1+e_2$
We know $e=\frac{FL}{AY}$
As $F,A,L$ are some for both the wires
So $e\alpha \frac{1}{\gamma }$
$\frac{e_1}{e_2}=\frac{{\gamma }_2}{{\gamma }_1}=\frac{2times10^{11}}{7\times 10^{11}}=\frac{20}{7}{e}_1=\frac{20}{7}{e}_2$
substituting in $e_1+e_2=2.7mm$
$\frac{20}{7}{e}_1+e_2=2.7mm\frac{27e_2}{7}=2.7mm$
$e_2=0.7mm$
$e_1=\frac{20}{7}{e}_2=\frac{20}{7}\times 0.7=2.0mm$