$$\eqalign{ & {\text{Strain}} = \frac{{\frac{F}{A}}}{\gamma }\& F = mg \cr & A = \pi {r^2} \cr} $$

Using the formula of the terminal velocity of a body falling through a viscous medium,
$$V = \frac{{2{r^2}\left( {\rho - \sigma } \right)}}{{9\eta }} \Rightarrow \eta = \frac{{2{r^2}\left( {\rho - \sigma } \right)g}}{{9v}}$$
Where $$\rho $$ is the density of material of body and $$\sigma $$ is the density of medium.
In case of the air bubble $$\rho = 1$$  and $$\sigma = 1.47 \times {10^3}\,kg/ms$$     and the air bubble rises up.
$$\eqalign{ & \eta = \frac{{2{r^2}\sigma g}}{{9V}} = \frac{{2 \times {{\left( {{{10}^{ - 2}}} \right)}^2} \times 1.47 \times {{10}^3} \times 9.8}}{{9 \times 0.21 \times {{10}^{ - 2}}}} \cr & = 1.52 \times {10^3}\,{\text{decapoise}} \cr & = 1.52 \times {10^4}\,{\text{poise}} \cr} $$

At equilibrium,
$$\eqalign{ & 2Tl = mg \cr & T = \frac{{mg}}{{2l}} = \frac{{1.5 \times {{10}^{ - 2}}}}{{2 \times 30 \times {{10}^{ - 2}}}} = \frac{{1.5}}{{60}} = 0.025\,N/m = 0.025N{m^{ - 1}} \cr} $$

According to ascent formula for capillary tube, $$h = \frac{{2T\cos \theta }}{{\rho gr}}$$
$$\therefore \frac{{\cos {\theta _1}}}{{{\rho _1}}} = \frac{{\cos {\theta _2}}}{{{\rho _2}}} = \frac{{\cos {\theta _3}}}{{{\rho _3}}}$$
$$\eqalign{ & {\text{Thus,}}\,\cos \theta \propto \rho \cr & \therefore {\rho _1} > {\rho _2} > {\rho _3} \cr & \therefore \cos {\theta _1} > \cos {\theta _2} > \cos {\theta _3} \cr & 0 \leqslant {\theta _1} < {\theta _2} < {\theta _3} < \frac{\pi }{2} \cr} $$

Air flows just above the roof and there is no air flow just below the roof inside the room. Therefore $${v_1} = 0$$  and $${v_2} = v.$$   Applying Bernaulli’s theorem at the points inside and outside the roof, we obtain.
$$\left( {\frac{1}{2}} \right)\rho v_1^2 + \rho g{h_1} + {P_1} = \left( {\frac{1}{2}} \right)\rho v_2^2 + \rho g{h_2} + {P_2}.$$
Since $${h_1} = {h_2} = h,{v_1} = 0\,{\text{and}}\,{v_2} = {v_1}$$
$${P_1} = {P_2} + \frac{1}{2}\rho {v^2} \Rightarrow {P_1} - {P_2} = \Delta P = \frac{1}{2}\rho {v^2}.$$
Since the area of the roof is $$A,$$ the aerodynamic lift exerted on it $$ = F = \left( {\Delta P} \right)A$$
$$ \Rightarrow F = \frac{1}{2}\rho A{v^2}$$
$$\eqalign{ & {\text{where}}\,\rho = {\text{density}}\,{\text{of}}\,{\text{air}} = 1.3\,kg/{m^3} \cr & A = 20\,{m^2},v = 29.44\,m/\sec . \cr & \Rightarrow F = \left\{ {\frac{1}{2} \times 1.3 \times 20 \times {{\left( {29.44} \right)}^2}} \right\}N \cr & = 1.127 \times {10^4}\,N. \cr} $$

$$\eqalign{ & \eta = {10^{ - 2}}{\text{poise}} \cr & v = 18\,km/h = \frac{{1800}}{{3600}} = 5\,m/s\,\,l = 5\,cm \cr & {\text{Strain rate}} = \frac{v}{l} \cr & {\text{Coefficient of viscosity,}}\,\eta {\text{ = }}\frac{{{\text{shearing}}\,{\text{stress}}}}{{{\text{strain}}\,{\text{rate}}}} \cr & \therefore {\text{Shearing}}\,{\text{stress}} = \eta \times {\text{strain}}\,{\text{rate}} \cr & = {10^{ - 2}} \times \frac{5}{5} = {10^{ - 2}}N{m^{ - 2}} \cr} $$

Let $$\rho $$ be the density of the material. $${\rho _0}$$ be the density of water when the sphere has just started sinking, the weight of the sphere = weight of water displaced (approx).
$$\eqalign{ & \Rightarrow \frac{4}{3}\pi \left( {{R^3} - {r^3}} \right)\rho g = \frac{4}{3}\pi {R^3}{\rho _0}g \cr & \Rightarrow \left( {{R^3} - {r^3}} \right)\rho = {R^3}\rho 0 \Rightarrow \frac{{\left( {{R^3} - {r^3}} \right)}}{{{R^3}}} = \frac{{{\rho ^0}}}{\rho } \cr & \Rightarrow \frac{r}{R} = \frac{{{{\left( 7 \right)}^{\frac{1}{3}}}}}{2} \cr} $$

Initially velocity increases with respect to time & becomes constant after some time, this constant velocity is called terminal velocity.

$$\eqalign{ & {A_1}{v_1} = {A_2}{v_2} + {A_3}{v_3} \cr & A \times 3 = A \times 1.5 + 1.5\;A \times {v_3} \Rightarrow {v_3} = 1\,m/s \cr} $$

Acceleration when block $$B$$ is in the liquid,
$$\eqalign{ & {a_1} = \frac{{{m_A}g - \left( {{m_B}g - {\text{upthrust}}} \right)}}{{\left( {{m_A} + 3m} \right)}} \cr & = \frac{{{m_A}g - \left( {3mg - \frac{{3m}}{{2\rho }}\rho g} \right)}}{{\left( {{m_A} + 3m} \right)}}\left( \uparrow \right) \cr} $$
Acceleration when block $$B$$ is outside if the liquid,
$${a_2} = \frac{{3mg - {m_A}g}}{{\left( {{m_A} + 3m} \right)}}\left( \downarrow \right)$$
$$\eqalign{ & {m_A}g - \frac{3}{2}mg = 3mg - {m_A}g \cr & \Rightarrow 2{m_A}g - \frac{9}{2}mg \Rightarrow {m_A} = \left( {\frac{9}{4}} \right)m \cr} $$