A mild steel wire of length 2l meter cross-se | vedclass

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Question

Now, increave in tength $=2 P S-P R$

$=2 \sqrt{l^2+x^2}-2 l$.

$=2\left(\sqrt{\ell^2+x^2}-l\right)$.

$=2 l\left(\sqrt{1+x^2 /l^2}-1\right)$

Vedclass · Accepted Answer

$\left(\frac{M g l^3}{Y A}\right)^{1 / 3}$

Vedclass · Answer

Now, increave in tength $=2 P S-P R$

$=2 \sqrt{l^2+x^2}-2 l$.

$=2\left(\sqrt{\ell^2+x^2}-light)$.

$=2 l\left(\sqrt{1+x^2 /l^2}-1ight)$

$	herefore$ strain $=\frac{2 l\left[\sqrt{1+x^2/l^2}-1ight]}{2 l}$

$=\left(1+x^2 / l^2ight)^{1 / 2}-1$

$=1+\frac{1}{2} x^2 / L^2-1=\frac{x^2}{2l^2}$

Now, tension, $T=\frac{m g}{2 \cos 	heta}$

$\cos 	heta=\frac{x}{\left(1+\frac{x^2}{l^2}ight)^{1 / 2} l}$

$=\frac{x}{\left(1+\frac{1}{2} \frac{x^2}{l^2}ight) l}$

$=x / l(\because l>>x)$

$T=\frac{m g}{2 \cos 	heta}$

$=\frac{m g l}{2 x}$

Stress $=T/ A=\frac{2 m g l}{2 A x}$

$y=\frac{m g l / 2 A x}{x^2 / 2 l^2}=\frac{m g l^3}{A x^3}$.

$	herefore x=l\left(\frac{m g}{	ext { YA }}ight)^{1 / 3}$

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 .............

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