$\begin{array}{1 1}(A)\;11.5cm\\(B)\;10cm\\(C)\;19.67cm\\(D)\;13cm\end{array} $

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Let $l$ be the natural length

Then $\Delta l_1=l_1-l$

$\Delta l_2=l_2-l$

As $Y=(\large\frac{f}{a})\frac{l}{\Delta l}$

$\Rightarrow Y_a=\large\frac{T_1l}{\Delta l_1}$

Similarly $Y_a=\large\frac{T_2l}{\Delta l_2}$

$\Rightarrow \large\frac{T_1l}{\Delta l_1}=\frac{T_2l}{\Delta l_2}$

$\Rightarrow \large\frac{T_1}{l_1-l}=\frac{T_2}{l_2-l}$

$\Rightarrow T_1l_2-T_1l=T_2l_1-T_2l$

$\Rightarrow \large\frac{T_1l_2-T_2l_1}{T_1-T_2}$$=l$

$l=\large\frac{80\times 8-200\times 15}{80-200}$

$\;\;=\large\frac{3000-640}{120}$

$\;\;=\large\frac{2360}{120}$

$\;\;=\large\frac{118}{6}$

$\;\;=\large\frac{59}{3}$

$\;\;=19.67$cm

Hence (C) is the correct answer.

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