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A mild steel wire of length $2l$ meter cross-sectional area $A \;m ^2$ is fixed horizontally between two pillars. A small mass $m \;kg$ is suspended from the mid point of the wire. If extension in wire are within elastic limit. Then depression at the mid point of wire will be .............
$\left(\frac{M g}{Y A}\right)^{1 / 3}$
$\left(\frac{M g}{1 A}\right)^{1 / 3}$
$\frac{M g}{2 Y A}$
$\left(\frac{M g l^3}{Y A}\right)^{1 / 3}$
Solution
Now, increave in tength $=2 P S-P R$
$=2 \sqrt{l^2+x^2}-2 l$.
$=2\left(\sqrt{\ell^2+x^2}-l\right)$.
$=2 l\left(\sqrt{1+x^2 /l^2}-1\right)$
$\therefore$ strain $=\frac{2 l\left[\sqrt{1+x^2/l^2}-1\right]}{2 l}$
$=\left(1+x^2 / l^2\right)^{1 / 2}-1$
$=1+\frac{1}{2} x^2 / L^2-1=\frac{x^2}{2l^2}$
Now, tension, $T=\frac{m g}{2 \cos \theta}$
$\cos \theta=\frac{x}{\left(1+\frac{x^2}{l^2}\right)^{1 / 2} l}$
$=\frac{x}{\left(1+\frac{1}{2} \frac{x^2}{l^2}\right) l}$
$=x / l(\because l>>x)$
$T=\frac{m g}{2 \cos \theta}$
$=\frac{m g l}{2 x}$
Stress $=T/ A=\frac{2 m g l}{2 A x}$
$y=\frac{m g l / 2 A x}{x^2 / 2 l^2}=\frac{m g l^3}{A x^3}$.
$\therefore x=l\left(\frac{m g}{\text { YA }}\right)^{1 / 3}$
