Explanation
$$ Y = \dfrac { F l }{ A \Delta l } $$
$$ \Delta l= \dfrac {F l }{ A Y} $$
$$\quad = \dfrac { 80 \times l }{ r { (\dfrac { 2 }{2 } ) }^{ 2 }\times { 10 }^{ -6 }\times 2\times { 10 }^{ 11 } } $$
$$\quad = 5.0 \times {10 }^{ -4 }m $$
$$\quad = 0.5\ mm$$
$$B=-V\dfrac{dP}{dV}$$
Thus, $$\Delta V=-\dfrac{pV}{B}$$ or $$V'-V=-\dfrac{pV}{B}$$
Or, $$V'=V(1-\dfrac{p}{B})$$
Now, $$\rho '=\dfrac{m}{V'}=\dfrac{m}{V(1-\dfrac{p}{B})}=\dfrac{\rho}{(1-\dfrac{p}{B})}$$
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