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Two wires A and B of same material have radii in the ratio 4 : 1 and length on the ratio 2 : 1.The forces required to produce same change in length are in the ratio

$\begin{array}{1 1}(A)\;1 : 8\\(B)\;8 : 1\\(C)\;32 : 1\\(D)\;1 : 1\end{array} $

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As $Y=(\large\frac{F}{a})\frac{l}{\Delta l}$
$F=\large\frac{Ya\Delta l}{l}$
$\Rightarrow \large\frac{F_1}{F_2}=(\large\frac{a_1}{a_2})(\frac{l_2}{l_1})$
$\Rightarrow (\large\frac{r_1}{r_2})^2(\frac{l_2}{l_1})$
$\Rightarrow (\large\frac{4}{1})^2(\frac{1}{2})$
$\Rightarrow \large\frac{8}{1}$
Hence (B) is the correct answer.
answered May 8, 2014 by sreemathi.v

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