Answer:
Here, \[=\frac{\vartriangle
l}{2l}=\frac{{{x}^{2}}}{2{{l}^{2}}}\]\[\text{2 T cos }\!\!\theta\!\!\text{ =
mg}\]\[\text{T=}\frac{\text{mg}}{\text{2 cos }\!\!\theta\!\!\text{ }}\] \[\text{cos
}\!\!\theta\!\!\text{ =}\frac{\text{x}}{{{\left( {{\text{l}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}
\right)}^{\text{1/2}}}}\text{=}\frac{\text{x}}{\text{l}{{\left(
\text{1+}\frac{{{\text{x}}^{\text{2}}}}{{{\text{l}}^{\text{2}}}}
\right)}^{\text{1/2}}}}\text{=}\frac{\text{x}}{\text{l}\left(
\text{1+}\frac{\text{1}{{\text{x}}^{\text{2}}}}{\text{2}{{\text{l}}^{\text{2}}}}
\right)}\]
As, \[\text{As,xl,so1}\frac{\text{1}{{\text{x}}^{\text{2}}}}{\text{2}{{\text{l}}^{\text{2}}}}\]
\[1+\frac{1{{x}^{2}}}{2{{l}^{2}}}\approx
1\]\[\therefore \text{cos }\!\!\theta\!\!\text{ =}\frac{\text{x}}{l}\]
As, \[T=\frac{mg}{2\left( x/l
\right)}=\frac{mgl}{2x}\]
Fractional change in density
\[\text{=}\frac{\text{T}}{\text{A}}\text{=}\frac{\text{mgl}}{\text{2Ax}}\]
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