Answer:
Here, \[=\frac{\vartriangle
l}{2l}=\frac{{{x}^{2}}}{2{{l}^{2}}}\]\[\text{2 T cos }\!\!\theta\!\!\text{ =
mg}\]
\[\text{T=}\frac{\text{mg}}{\text{2 cos
}\!\!\theta\!\!\text{ }}\]
\[\text{cos }\!\!\theta\!\!\text{
=}\frac{\text{x}}{{{\left(
{{\text{l}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}
\right)}^{\text{1/2}}}}\text{=}\frac{\text{x}}{\text{l}{{\left(
\text{1+}\frac{{{\text{x}}^{\text{2}}}}{{{\text{l}}^{\text{2}}}}
\right)}^{\text{1/2}}}}\text{=}\frac{\text{x}}{\text{l}\left(
\text{1+}\frac{\text{1}{{\text{x}}^{\text{2}}}}{\text{2}{{\text{l}}^{\text{2}}}}
\right)}\]\[\text{As,xl,so1}\frac{\text{1}{{\text{x}}^{\text{2}}}}{\text{2}{{\text{l}}^{\text{2}}}}\]
\[1+\frac{1{{x}^{2}}}{2{{l}^{2}}}\approx 1\]
From, \[\therefore \text{cos
}\!\!\theta\!\!\text{ =}\frac{\text{x}}{l}\]
\[T=\frac{mg}{2\left( x/l
\right)}=\frac{mgl}{2x}\]
\[\text{=}\frac{\text{T}}{\text{A}}\text{=}\frac{\text{mgl}}{\text{2Ax}}\]
\[\text{Y=}\frac{\text{Stress}}{\text{Strain}}\text{=}\frac{\text{mgl}}{\text{2Ax}}\text{
}\!\!\times\!\!\text{
}\frac{\text{2}{{\text{l}}^{\text{2}}}}{{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{mg}{{\text{l}}^{\text{3}}}}{\text{A}{{\text{x}}^{\text{3}}}}\]
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