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What is tension in the wire of radius r and length 2l in the given situation if young's modulus is $Y$.

$\begin{array}{1 1}(A)\;\large\frac{\pi Yrd^2}{2l}\\(B)\;\large\frac{\pi Yr^2d^2}{l^2}\\(C)\;\large\frac{\pi Yr^2d^2}{2l^2}\\(D)\;\large\frac{\pi Yr^2d^2}{4l^2}\end{array} $

1 Answer

$2l$ is the original length of wire.
$\Delta l=2(\sqrt{l^2+d^2}-L)$
$\Rightarrow 2l((1+\large\frac{d^2}{l^2})^{1/2}$$-1)$
$\Rightarrow 2l(1+\large\frac{1}{2}\frac{d^2}{l^2}$$-1)$ as $\large\frac{d^2}{l^2}$ is very small
$\Rightarrow \large\frac{d^2}{l}$
$\large\frac{\Delta L}{L}=\frac{d^2/l}{2l}=\frac{d^2}{2l^2}$
As $Y=(\large\frac{F}{a})\frac{1}{\Delta L}$
$F=\large\frac{Y\times a\times \Delta L}{L}$
$\Rightarrow Y\times \pi r^2\large\frac{d^2}{2l^2}$
$\Rightarrow \large\frac{\pi Yr^2d^2}{2l^2}$
Hence (C) is the correct answer.
answered May 8, 2014 by sreemathi.v

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