$\begin{array}{1 1}(A)\;.28mm\\(B)\;.33mm\\(C)\;.24mm\\(D)\;\text{None}\end{array} $

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For brass wire $f=3\times 9.8N$

So $\Delta l_1=(\large\frac{f_1}{a_1})\frac{l_1}{Y_1}$

$\Rightarrow \large\frac{3\times 9.8}{5\times 10^{-6}}\times \frac{1.5}{1\times 10^{11}}$m

$\Rightarrow 8.82\times 10^{-5}$m

$\Rightarrow 0.0882mm$

For steel wire

$f=(3+2)\times 9.8N$

So $\Delta l_1=(\large\frac{f_2}{a_2})\frac{l_2}{Y_2}$

$\Rightarrow \large\frac{5\times 9.8\times 2}{2\times 10^{-6}\times 2\times 10^{11}}$m

$\Rightarrow 24.5\times 10^{-5}m$

$\Rightarrow$ .245mm

$\Delta l=\Delta l_1+\Delta l_2$

$\Rightarrow .088+.245$

$\Rightarrow$ .333mm

Hence (B) is the correct answer.

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