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A wire of r=1cm consist of 1m steel and 2m copper respectively.Total extension when a force applied is 1mm.Find the force is $Y_{cu}=1.1\times 10^{11}N/m^2,Y_{steel}=2\times 10^{11}N/m^2$.

$\begin{array}{1 1}(A)\;1.46\times 10^3N\\(B)\;1.2\times 10^3N\\(C)\;1.36\times 10^3N\\(D)\;\text{None}\end{array} $

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For steel wire
$a=\pi r_1^2$
$Y_1=(\large\frac{F}{a})\frac{l_1}{\Delta l_1}$
$\Delta l_1=(\large\frac{F}{a})\frac{l_1}{Y_1}$
Similarly for copper wire
$\Delta l_2=(\large\frac{F}{a})\frac{l_2}{Y_2}$
$\Delta l_1+\Delta l_2=\Delta l$
$\large\frac{F}{a}(\frac{l_1}{Y_1}+\frac{l_2}{Y_2})=$$\Delta l$
$F=\large\frac{10^{-3}\times \pi\times (10^{-2})^2\times 10^{11}}{(\large\frac{1}{2}+\frac{2}{1.1})}$
$\Rightarrow 1.36\times 10^3N$
Hence (C) is the correct answer.
answered May 7, 2014 by sreemathi.v

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