Explanation
Let, Natural length {{L}_{o}}
Stress = Young modulus x strain
\dfrac{F}{A}=Y\times \left( \dfrac{L-{{L}_{o}}}{{{L}_{o}}} \right)\ ......\ (1)
\dfrac{yF}{A}=Y\times \left( \dfrac{xL-{{L}_{o}}}{{{L}_{o}}} \right)\ ......\ (2)
Divide equation (2) by (1)
y=\dfrac{xL-{{L}_{o}}}{L-{{L}_{o}}}
\Rightarrow y\left( L-{{L}_{o}} \right)=\left( xL-{{L}_{o}} \right)
\Rightarrow {{L}_{o}}=\dfrac{\left( y-x \right)L}{y-1}
Natural length of wire is \dfrac{\left( y-x \right)L}{y-1}
The force exerted by 45\;{\rm{cm}} part of the rod on 15\;{\rm{cm}}art of the rod is given as,
F = {F_1} \times \frac{{15}}{{60}} + {F_2} \times \frac{{45}}{{60}}
F = 21 \times \frac{{15}}{{60}} + 45 \times \frac{{45}}{{60}}
F = 45 \times \frac{{15}}{{60}} + 21 \times \frac{{45}}{{60}}
F = 27\;{\rm{N}}
Given that,
Young’s modulus Y=6.6\times {{10}^{10}}\,N/{{m}^{2}}
Bulk modulus B=11\times {{10}^{10}}\,N/{{m}^{2}}
We know that,
Y=3K\left( 1-2\mu \right)
6.6\times {{10}^{10}}=3\times 11\times {{10}^{10}}-66\times {{10}^{10}}\mu
-\mu =\dfrac{\left( 6.6-33 \right)\times {{10}^{10}}}{66\times {{10}^{10}}}
\mu =0.4
Hence, the poisson’s ratio is 0.4
Young’s modulus is given by
Y=\dfrac{FL}{A\Delta l}
Y=\dfrac{1.5\times 10\times 156}{3.14\times 2.7\times {{10}^{-4}}\times 5\times {{10}^{-4}}}
Y=1.002\times {{10}^{-11}}\,N/{{m}^{2}}
\Delta {{l}_{1}}=0.01\,m
Length {{l}_{1}}=l
Length {{l}_{2}}=2l
Radius {{r}_{1}}=r
Radius {{r}_{2}}=2r
Y=\dfrac{Fl}{A\Delta l}
\Delta l=\dfrac{Fl}{\pi {{r}^{2}}Y}
\Delta l\propto \dfrac{l}{{{r}^{2}}}
Now, the diameter and lengths are doubled to the original wire is stretched by the same force
So,
\dfrac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}}=\dfrac{2l}{l}\times \dfrac{{{r}^{2}}}{4{{r}^{2}}}
\dfrac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}}=2\times \dfrac{1}{4}
\Delta {{l}_{2}}=\dfrac{0.01}{2}
\Delta {{l}_{2}}=0.005\,m
Hence, the elongation is 0.005\ m
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