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2 wire of length 1 : 2 and area in the ratio 2 : 1 is applied with equal force stretch equally .Find the ratio of modulus of rigidity

$\begin{array}{1 1}(A)\;4 : 1\\(B)\;2 : 1\\(C)\;1 : 1\\(D)\;1 : 4\end{array} $

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1 Answer

As $Y=(\large\frac{f}{a})\frac{l}{\Delta l}$
$\Rightarrow Y\propto \large\frac{l}{a}$
$\Rightarrow \large\frac{Y_1}{Y_2}=(\large\frac{l_1}{l_2})(\frac{a_2}{a_1})$
$\Rightarrow \large\frac{1}{2}\times \frac{1}{2}$
$\Rightarrow \large\frac{1}{4}$
Hence (D) is the correct answer.
answered May 7, 2014 by sreemathi.v

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