$\begin{array}{1 1}(A)\;\large\frac{M\omega^2l^2}{3\pi r^2}\\(B)\;\large\frac{M\omega^2l^2}{2\pi r^2}\\(C)\;\large\frac{M\omega^2r^2}{3\pi l^2}\\(D)\;\large\frac{M\omega^2r}{2\pi l}\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Considering an element of $dx$ length.

Force on it =$\int\limits_x^l dm\omega^2Y$

$\Rightarrow \int_x^l \large\frac{M}{l}$$dy\omega^2y$

$\Rightarrow [\large\frac{M\omega^2}{l}\frac{Y^2}{2}]_x^l$

$\Rightarrow \large\frac{M\omega^2}{2l}$$(l^2-x^2)$

Let the extension in element $dx$ be $d\alpha$

As $Y=(\large\frac{f}{a})\frac{l}{\Delta l}$

$\Delta l=(\large\frac{f}{a})\frac{l}{Y}$

For element of $dx$ length

$d\alpha=\large\frac{M\omega^2(l^2-x^2)dx}{2laY}$

$\alpha=\int\limits_0^l \int \large\frac{M\omega^2}{2la}$$(l^2-x^2)dx$

$\;\;\;=\large\frac{M\omega^2}{2la}[$$l^3-\large\frac{l^3}{3}]$

$\;\;\;=\large\frac{M\omega^2l^2}{3a}$

As $a=\pi r^2$

$\;\;\;=\large\frac{M\omega^2l^2}{3\pi r^2}$

Hence (A) 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

...