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2 wires of radius in ratio 1 : 2 have young modulus in ratio 2 : 3.Equal force is applied on them ,what is the ratio of elastic potential energy

$\begin{array}{1 1}(A)\;24 : 1\\(B)\;1 : 24\\(C)\;3 : 1\\(D)\;1 : 3\end{array} $

1 Answer

$\large\frac{a_1}{a_2}=(\frac{r_1}{r_2})^2=(\large\frac{1}{2})^2=\frac{1}{4}$
$\large\frac{Y_1}{Y_2}=\frac{2}{3}$
Energy density =$\large\frac{1}{2}\times \frac{(stress)^2}{Y}$
$\Rightarrow \large\frac{1}{2}(\frac{f}{a})^2\frac{1}{Y}$
$\Rightarrow \propto \large\frac{1}{a^2Y}$
$\large\frac{E_1}{E_2}=(\frac{a_2}{a_1})^2(\frac{Y_2}{Y_1})$
$\Rightarrow 16\times \large\frac{3}{2}$
$\Rightarrow \large\frac{24}{1}$
$\Rightarrow 24 : 1$
Hence (A) is the correct answer.
answered May 8, 2014 by sreemathi.v
 

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