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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