$$\eqalign{ & \frac{1}{\rho }\frac{{d\rho }}{{dt}} = {\text{ constant}} \cr & \therefore \frac{{4\pi {R^3}}}{{3m}}\frac{d}{{dt}}\,\left[ {\frac{m}{{\frac{4}{3}\pi {R^3}}}} \right] = {\text{ constant}} \cr & \therefore {R^3}\frac{d}{{dt}}\left( {{R^{ - 3}}} \right) = {\text{ constant}} \cr & \therefore {R^3}\left( { - 3{R^{ - 4}}} \right)\frac{{dR}}{{dt}} = {\text{ constant}} \cr & \therefore \left| {\frac{{dR}}{{dt}}} \right| \propto R \cr} $$

Young’s modulus of elasticity is
$$\eqalign{ & Y = \frac{{\frac{F}{A}}}{{\frac{{\Delta L}}{L}}} \cr & \therefore \Delta L = \frac{{FL}}{{AY}} \cr & {\text{So,}}\,\,\Delta L \propto \frac{L}{A} \cr & \therefore \frac{{\Delta {L_2}}}{{\Delta {L_1}}} = \frac{{{L_2}}}{{{L_1}}} \times \frac{{{A_1}}}{{{A_2}}} = \frac{2}{1} \times \frac{2}{1} = 4 \cr & \Delta {L_2} = 4 \times \Delta {L_1} = 4 \times 1 = 4\,cm \cr} $$

Young's modulus of steel is highest.

Given $$d = 2700\;m \Rightarrow \rho = {10^3}\;kg/{m^3}$$
Compressibility $$ = 45.4 \times {10^{ - 11}}$$   per pascal
The pressure at the bottom of ocean is given by $$p = \rho gd = {10^3} \times 10 \times 2700 = 27 \times {10^6}Pa$$
$$\eqalign{ & {\text{So,}}\,{\text{fractional}}\,{\text{compression}} = {\text{compressibility}} \times {\text{pressure}} \cr & = 45.4 \times {10^{ - 11}} \times 27 \times {10^6} = 1.2 \times {10^{ - 2}} \cr} $$

The wettability of a surface by a liquid depends primarily on angle of contact between the surface and the liquid.

Maximum possible strain $$ = \frac{{0.2}}{{100}}$$
$$\eqalign{ & \therefore A = \frac{F}{{Y \times {\text{strain}}}} = \frac{{{{10}^4} \times 100}}{{\left( {7 \times {{10}^9}} \right) \times 0.2}} \cr & = 7.1 \times {10^{ - 4}}\,{m^2} \cr} $$

$$\eqalign{ & {\text{Given:}}\,F = 100\,kN = {10^5}N \cr & Y = 2 \times {10^{11}}N{m^{ - 2}} \cr & {\ell _0} = 1.0\,m \cr & {\text{radius }}r = 10\,mm = {10^{ - 2}}\,m \cr & {\text{From formula, }}Y = \frac{{{\text{Stress}}}}{{{\text{Strain}}}} \cr & \Rightarrow {\text{Strain}} = \frac{{{\text{Stress}}}}{Y} = \frac{F}{{AY}} \cr & = \frac{{{{10}^5}}}{{\pi {r^2}Y}} = \frac{{{{10}^5}}}{{3.14 \times {{10}^{ - 4}} \times 2 \times {{10}^{11}}}} = \frac{1}{{628}} \cr & {\text{Therefore }}\% {\text{ strain}} = \frac{1}{{628}} \times 100 = 0.16\% \cr} $$

Equating the rate of flow, we have
$$\sqrt {\left( {2gy} \right)} \times {L^2} = \sqrt {\left( {2g \times 4y} \right)} \,\,\pi {R^2}$$
[Flow $$=$$ (area) × (velocity), velocity $$ = \sqrt {2gx} $$  ]
where $$x=$$  height from top
$$ \Rightarrow {L^2} = 2\pi {R^2}\,\,\,\, \Rightarrow R = \frac{L}{{\sqrt {2\pi } }}$$

Inflow rate of volume of the liquid = Outflow rate of volume of the liquid
$$\pi {R^2}V = n\pi {r^2}\left( v \right) \Rightarrow v = \frac{{\pi {R^2}V}}{{n\pi {r^2}}} = \frac{{V{R^2}}}{{n{r^2}}}$$

Viscous force = $$6\pi \eta rv$$
$$\eqalign{ & = 6\pi \times 18 \times {10^{ - 5}} \times 0.03 \times 100 \cr & = 101.73 \times {10^{ - 4}}dyne \cr} $$