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The length of a wire is 8cm when tension in it is 80N and is 15cm when tension in it 200N.The natural length of the wire is

$\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$
$l=\large\frac{80\times 8-200\times 15}{80-200}$
Hence (C) is the correct answer.
answered May 9, 2014 by sreemathi.v

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