Work done by constant force in displacing the object by a distance $$\ell .$$
$$ = \frac{1}{2} \times {\text{Force }} \times \,{\text{extansion}} = \frac{{F\ell }}{2}$$

$$\eqalign{ & v \propto \frac{{\rho - {\rho _0}}}{\eta } \cr & \therefore \frac{{{v_2}}}{{{v_1}}} = \frac{{\rho - {\rho _{02}}}}{{\rho - {\rho _{01}}}} \times \frac{{{\eta _1}}}{{{\eta _2}}} = \frac{{7.8 - 1.2}}{{7.8 - 1}} \times \frac{{8.5 \times {{10}^{ - 4}} \times 10}}{{13.2}} \cr & = 6.25 \times {10^{ - 4}}\,cm{s^{ - 1}} \cr} $$

Young’s modulus of rubber,
$${Y_{{\text{rubber}}}} = \frac{F}{A} \times \frac{\ell }{{\Delta \ell }} \Rightarrow F = YA.\frac{{\Delta \ell }}{\ell }$$
On putting the values from question,
$$F = \frac{{5 \times {{10}^8} \times 25 \times {{10}^{ - 6}} \times 5 \times {{10}^{ - 2}}}}{{10 \times {{10}^{ - 2}}}} = 6250\,N$$
kinetic energy = potential energy of rubber
$$\eqalign{ & \frac{1}{2}m{v^2} = \frac{1}{2}F\Delta \ell \cr & v = \sqrt {\frac{{F\Delta \ell }}{m}} = \sqrt {\frac{{6250 \times 5 \times {{10}^{ - 2}}}}{{5 \times {{10}^{ - 3}}}}} = \sqrt {62500} \cr & = 250\,m/s \cr} $$

$$\eqalign{ & {\text{Bulk modulus,}}\,{\text{K }} = \frac{{{\text{volumetric stress}}}}{{{\text{volumetric strain}}}} \cr & {\text{K}} = \frac{{{\text{mg}}}}{{{\text{a}}\left( {\frac{{{\text{dV}}}}{{\text{V}}}} \right)}}\,\,\,\,\,\, \Rightarrow \frac{{{\text{dV}}}}{{\text{V}}}{\text{ = }}\frac{{{\text{mg}}}}{{{\text{Ka}}}}\,.....{\text{(i)}} \cr} $$
volume of sphere, $${\text{V}} = \frac{4}{3}\pi {{\text{R}}^{\text{3}}}$$
Fractional change in volume $$\frac{{{\text{dV}}}}{{\text{V}}} = \frac{{{\text{3dr}}}}{{\text{r}}}\,.....{\text{(ii)}}$$
Using equation (i) & (ii) $$\frac{{{\text{3dr}}}}{{\text{r}}} = \frac{{{\text{mg}}}}{{{\text{Ka}}}}$$
$$\therefore \,\,\frac{{{\text{dr}}}}{{\text{r}}} = \frac{{{\text{mg}}}}{{{\text{3Ka}}}}$$     (fractional decrement in radius)

Let $$V$$ be the volume of the material of which the cylinder is made. The cylinder is half immersed in water. Therefore the volume of water displaced because of the material of the cylinder is $$\frac{V}{2}.$$ Let $$h$$ be the total height of the cylinder. As the cylinder is half submerged therefore buoyant force
$$B = \frac{{V{\rho _\omega }g}}{2} + \frac{{hA{\rho _\omega }g}}{2}$$

where $$A$$ is the area of cross-section of the cylinder
The weight of the cylinder $$W = V{\rho _c}g$$
The weight of the water inside the cylinder $$ = h'A{\rho _\omega }g$$
For equilibrium,
$$\frac{{V{\rho _\omega }g}}{2} + \frac{{Ah{\rho _\omega }g}}{2} = V{\rho _c}g + = h'A{\rho _\omega }g$$
Here $${\rho _\omega } = 1$$
$$\eqalign{ & \therefore h' = \frac{h}{2} + \frac{V}{{2A}}\,\,\left[ {1 - 2{\rho _c}} \right] \cr & {\text{If }}{\rho _c} < 0.5\,\,{\text{then }}h' > \frac{h}{2} \cr & {\text{and if }}{\rho _c} > 0.5\,\,{\text{then }}h' < \frac{h}{2} \cr & {\text{if }}{\rho _c} = 0,\,\,{\text{ }}h' = \frac{h}{2} \cr} $$

$$\eqalign{ & {\text{Increase in surface energy}} = {\text{work done in area}} \times {\text{surface tension}} \cr & \because {\text{Increase in surface area,}} \cr & \Delta A = \left( {5 \times 4 - 4 \times 2} \right) \times 2\,\,\,\left( {\because {\text{film has two surfaces}}} \right) \cr & = \left( {20 - 8} \right) \times 2\,c{m^2} = 24\,c{m^2} = 24 \times {10^{ - 4}}{m^2} \cr} $$
So, work done, $$W = T \cdot \Delta A$$
$$\eqalign{ & 3 \times {10^{ - 4}} = T \times 24 \times {10^{ - 4}} \cr & \therefore T = \frac{1}{8} = 0.125\,N/m \cr} $$

Applying Bernoulli's theorem at point 1 and 2, difference in pressure energy between 1 and 2 = difference in kinetic energy between 1 and 2.
$$\eqalign{ & {\text{Hence,}}\,\,h\rho g + \frac{{mg}}{A} = \frac{1}{2}\rho {v^2} \cr & {\text{or}}\,v = \sqrt {2gh + \frac{{2mg}}{{\rho A}}} = \sqrt {2\left( {gh + \frac{{mg}}{{\rho A}}} \right)} \cr} $$

$$\eqalign{ & B = - \frac{{\Delta P}}{{\frac{{\Delta V}}{V}}} = \frac{{100 \times {{10}^5}}}{{\left( {\frac{{0.02}}{{100}}} \right)}} \cr & = 50 \times {10^9}N/{m^2}s \cr} $$

Let $$x$$ be the height of the water level from the hole $$v = \sqrt {2gx} = \frac{\ell }{t} = \frac{1}{t}\,......\left( {\text{i}} \right)$$
For vertical motion $$x = \frac{1}{2}g{t^2}\,......\left( {{\text{ii}}} \right)$$
Now solving (i) & (ii) we get
$$ \Rightarrow x = 0.25\,m\left( {{\text{i}}{\text{.e}}{\text{. level goes down from }}0.81\,m{\text{ to}}\,0.25\,m} \right)$$
Using equation of continuity
$$\sqrt 5 .\frac{{dx}}{{dt}} = \sqrt {2gx} .4 \times {10^{ - 4}}$$
On solving $$t = 1000\,s$$

$$\eqalign{ & \left( {P + \frac{{4\sigma }}{a}} \right)\left( {\frac{4}{3}\pi {a^3}} \right) + \left( {P + \frac{{4\sigma }}{b}} \right)\left( {\frac{4}{3}\pi {b^3}} \right) \cr & = \left( {P + \frac{{4\sigma }}{c}} \right)\left( {\frac{4}{3}\pi {c^3}} \right) \cr & {\text{or}}\,\,P\left[ {{a^3} + {b^3} - {c^3}} \right] = 4\sigma \left[ {{c^2} - {a^2} - {b^2}} \right] \cr & {\text{or}}\,\,\sigma = \frac{{P\left( {{c^3} - {a^3} - {b^3}} \right)}}{{4\left( {{a^2} + {b^2} - {c^2}} \right)}} \cr} $$