Given: Diameter of water tap $$ = \frac{2}{{\sqrt \pi }}\,cm$$
$$\eqalign{ & \therefore {\text{Radius,}}\,r = \frac{1}{{\sqrt \pi }} \times {10^{ - 2}}\,m\,\,\,\,\frac{{dm}}{{dt}} = \rho AV \cr & \frac{{15}}{{5 \times 60}} = {10^3} \times \pi {\left( {\frac{1}{{\sqrt \pi }}} \right)^2} \times {10^{ - 4}}\;V \Rightarrow V = 0.05\,m/s \cr} $$
Reynold’s number, $${R_e} = \frac{{\rho Vr}}{n} = \frac{{{{10}^3} \times 0.5 \times \frac{2}{{\sqrt \pi }}{{10}^{ - 2}}}}{{{{10}^{ - 3}}}} \cong 5500$$

Pressure difference is largest between atmosphere and smaller bubbles. Hence radius of curvature $$\left( R \right)$$ is smallest.

According to question, the situation can be drawn as following.
Applying Archemedies principle
Weight of cylinder $$ = {\left( {{\text{upthrust}}} \right)_1} + {\left( {{\text{upthrust}}} \right)_2}$$
i.e. $$ALdg = \left( {1 - P} \right)LA\rho g + \left( {PLA} \right)n\rho g$$
$$\eqalign{ & \Rightarrow d = \left( {1 - P} \right)\rho + Pn\rho = \rho - P\rho + n\,P\rho \cr & = \rho + \left( {n - 1} \right)P\rho = \rho \left[ {1 + \left( {n - 1} \right)\rho } \right] \cr} $$

$$\eqalign{ & Y = \frac{{\frac{F}{\pi }{{\left( {2R} \right)}^2}}}{{\frac{{\Delta {\ell _1}}}{{2L}}}} = \frac{{\frac{F}{{\pi {R^2}}}}}{{\frac{{\Delta {\ell _2}}}{{2L}}}} \cr & \therefore \frac{{\Delta {\ell _2}}}{{\Delta {\ell _1}}} = 2 \cr} $$

$$\eqalign{ & m \times 10 = 2 \times 3 \times {10^{ - 2}} \times \frac{{10}}{{100}} \cr & {\text{or}}\,\,m = 6 \times {10^{ - 4}}\,kg = 6 \times {10^{ - 4}} \times {10^3}\,g = 0.6\,g \cr} $$

From continuity equation, velocity at cross-section (1) is more than that at cross-section (2). Hence $${P_1} < {P_2}.$$

According to Bernoulli’s theorem,
$$\eqalign{ & P + \frac{1}{2}\rho {v^2} = {P_0} + 0 \cr & {\text{So,}}\,\Delta P = \frac{1}{2}\rho {v^2} \cr & F = \Delta PA = \frac{1}{2}\rho {v^2}A \cr & = \frac{1}{2} \times 1.2 \times 40 \times 40 \times 250 = 2.4 \times {10^5}N\left( {{\text{upwards}}} \right) \cr} $$

Consider a small element $$dx$$ of radius $$r,$$
$$r = \frac{{2R}}{L}x + R$$
At equilibrium change in length of the wire
$$\int\limits_0^1 {dL} = \int {\frac{{Mg\,dx}}{{\pi {{\left[ {\frac{{2R}}{L}x + R} \right]}^2}y}}} $$
Taking limit from $$0$$ to $$L$$
$$\Delta L = \frac{{Mg}}{{\pi y}}\left[ { - \frac{1}{{\left[ {\frac{{2Rx}}{L} + R} \right]_0^L}} \times \frac{L}{{2R}}} \right[ = \frac{{MgL}}{{3\pi {R^2}y}}$$
The equilibrium extended length of wire $$ = L + \Delta L$$
$$\eqalign{ & = L + \frac{{MgL}}{{3\pi {R^2}Y}} \cr & = L\left( {1 + \frac{1}{3}\frac{{Mg}}{{3\pi Y{R^2}}}} \right) \cr} $$

$${v_0} = \sqrt {2gh} \Rightarrow {v_2} = \sqrt {2\;g\frac{h}{{\sqrt 2 }}} = \frac{{{v_0}}}{{\root 4 \of 2 }}$$

Weight of cylinder = Upthrust due to upper liquid + Upthrust due to lower liquid.
$$\eqalign{ & D\left( {\frac{A}{5} \times L \times g} \right) = d\left( {\frac{A}{5}} \right)\left( {\frac{3}{4}L} \right)g + 2d\left( {\frac{A}{5}} \right)\left( {\frac{L}{4}} \right) \times g \cr & \therefore D = \frac{{5d}}{4} \cr} $$