$\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} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

$2l$ 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.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...