Question

A wire of radius $r$ and length 2$/$ is fixed at its ends a block of mass $m$ is suspended from the midpoint of the wire so that midpoint is lowered by a distance d from original position. If Young's modulus is Y then the value of $d$ is $(d$$<<1)$

• A. ${\left[\frac{mg{l}^3}{Y\pi r^2}\right]}^{1/3}$
• B. ${\left[\frac{2mg{l}^3}{Y\pi r^2}\right]}^{1/3}$
• C. ${\left[\frac{mg{l}^2}{Y\pi r^2}\right]}^{1/2}$
• D. ${\left[\frac{3mg{l}^{-1/2}}{Y\pi r^2}\right]}^{1/2}$

Answer 1

The correct option is A ${\left[\frac{mg{l}^3}{Y\pi r^2}\right]}^{1/3}$

$\begin{array}{c}\text{Given -}\\ \begin{array}{c}\text{radius}=r\\ \text{length}=2\ell \end{array}\end{array}$

$\begin{array}{c}\text{Taking vertical equilibrium.}\\ 2T\cos \theta =mg\\ \text{For half left part of wire.}\\ Y=\frac{Tl}{A\Delta l}\\ Y=\frac{mgl}{2\cos \theta \pi r^2\Delta l_1}\end{array}$

$\begin{array}{cc}\cos \theta \cdot \Delta l_1& =\frac{\mathrm{mgl}}{2Y\pi r^2}\\ \text{as}\frac{d}{x}\cdot (x-l)& =\frac{\mathrm{mgl}}{2Y\pi r^2}--\left(i\right)\end{array}$

$\begin{array}{c}{x}^2-l^2=d^2=(x-1)(x+l)\\ \text{putting in equation}-\left(i\right)\\ \frac{d}{x}\cdot \frac{d^2}{(x+1)}=\frac{mgl}{2Y\pi r^2}\end{array}$

$\begin{array}{c}\text{as given}\\ \begin{array}{cc}d& <<l\\ & x\cong l\end{array}\\ \frac{d^3}{2l^2}=\frac{mgl}{2Y\pi r^2}\\ d={\left(\frac{mg{l}^3}{Y\pi r^2}\right)}^{1{/}_3}\end{array}$