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). $( Y_{At }= 2\times10^{11}Nm^{-2} ,Y_{steet}= 7\times 10^{11} Nm^{-2})$

• 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 $e \, \alpha \frac{1}{\gamma}$
$\frac{e_1}{e_2} = \frac{\gamma_2}{\gamma_1} = \frac{2 times 10^{11}}{7 \times 10^{11}} = \frac{20}{7} \, e_1 = \frac{20}{7} e_2$
substituting in $e_1 + e_2 = 2.7 mm$
$\frac{20}{7} e_1 + e_2 = 2.7 \, mm \, \frac{27e_2}{7} = 2.7 \, mm $
$e_2 = 0.7 \,mm$
$e_1 = \frac{20}{7} e_2 = \frac{20 }{7} \times 0.7 =2.0 \,mm$