$$\eqalign{ & {P_1} - {P_0} = \frac{{4S}}{{2R}}\,......\left( {\text{i}} \right) \cr & {P_2} - {P_1} = \frac{{4S}}{R}\,......\left( {{\text{ii}}} \right) \cr} $$
Add (i) and (ii),
$${P_2} - {P_0} = \frac{{6S}}{R}$$

$$\eqalign{ & {\text{Stress}} = 1\,kg\,wt/m{m^2} = 9.8\,N/m{m^2} \cr & = 9.8 \times {10^6}\,N/{m^2}. \cr & Y = 1 \times {10^{11}}N/{m^2},\frac{{\Delta \ell }}{\ell } \times 100 = ? \cr & Y = \frac{{{\text{Stress}}}}{{{\text{Strain}}}} = \frac{{{\text{Stress}}}}{{\frac{{\Delta \ell }}{\ell }}} \cr & \therefore \frac{{\Delta \ell }}{\ell } = \frac{{{\text{Stress}}}}{Y} = \frac{{9.8 \times {{10}^6}}}{{1 \times {{10}^{11}}}} \cr & \frac{{\Delta \ell }}{\ell } \times 100 = 9.8 \times {10^{ - 11}} \times 100 \times {10^6} \cr & = 9.8 \times {10^{ - 3}} \cr & = 0.0098\% \cr} $$

From Bernoulli's theorem,
$$\eqalign{ & {P_0} + \frac{1}{2}\rho v_1^2\rho gh = {P_0} + \frac{1}{2}\rho v_2^2 + 0 \cr & {v_2} = \sqrt {v_1^2 + 2gh} = \sqrt {0.16 + 2 \times 10 \times 0.2} = 2.03\,m/s \cr} $$
From equation of continuity
$$\eqalign{ & {A_2}{v_2} = {A_1}{v_1} \cr & \pi \frac{{D_2^2}}{4} \times {v_2} = \pi \frac{{D_1^2}}{4}{v_1} \cr & \Rightarrow {D_1} = {D_2}\sqrt {\frac{{{v_1}}}{{{v_2}}}} = 3.55 \times {10^{ - 3}}\,m \cr} $$

Balancing forces in horizontal direction
$$\left( {{p_0} - \rho g\frac{h}{2}} \right)\ell h + \gamma \ell = {p_0}\ell h \Rightarrow h = \sqrt {\frac{{2\gamma }}{{\rho g}}} $$

As block $$B$$ falls, $$L$$ will decrease.
Block $$B$$ will displace, volume of liquid $$\left( {{V_1}} \right)$$
Equal to its own volume when it is in the liquid.
$${V_1} = {V_B} = \frac{{{M_B}}}{{{\rho _B}}}\,......\left( {\text{i}} \right)$$
When block $$B$$ is on block $$A,$$ it will displace the volume of liquid $$\left( {{V_2}} \right)$$  whose weight is equal to the weight of the block $$B.$$
$${V_2}{\rho _L}g = {M_B}g = \left( {{V_1}{\rho _B}} \right)g\,\,{\text{or}}\,\,{V_2}{\rho _L} = {V_1}{\rho _B}.$$
Since $$H$$ increases, $${V_1} > {V_2}$$
So, $${\rho _L} > {\rho _B}.$$

Poisson's ration $$=0.5$$
Since, density is constant therefore change in volume is zero, we have
$$\eqalign{ & V = A \times 1 = {\text{constant}} \cr & \log \,V = \log A + \log 1 \cr & {\text{or}}\,\frac{{dA}}{A} + \frac{{dl}}{1} = 0 \cr & \frac{{dl}}{1} = - \frac{{dA}}{A} \cr} $$
∴ Percentage increase in length $$=4%$$

Balancing the forces acting on the drop, we get
$$\frac{4}{3}\pi {r^3}\rho g = 2\pi rT + \frac{1}{2} \cdot \frac{4}{3}\pi {r^3}\sigma \Rightarrow r = \sqrt {\frac{{3T}}{{\left( {2p - \sigma } \right)g}}} $$

Force $$F = \rho Qv = \rho a{v^2} = \rho a \times 2gh$$
Thus net force
$$\eqalign{ & = {F_2} - {F_1} \cr & = \rho av_2^2 - \rho av_1^2 \cr & = \rho a\left( {v_2^2 - v_1^2} \right) \cr & = \rho a\left[ {2g\left( {y + h} \right) - 2gy} \right] \cr & = \rho a \times 2gh \cr} $$

$$\frac{{dF}}{{dx}} = 0\,{\text{at}}\,\frac{\ell }{2}$$

Work done = Surface tension $$ \times $$ increase in area of the film
$$W = S \times \Delta A$$
Increase in area = Final area - initial area
$$\eqalign{ & = 10 \times \left( {0.5 + 0.1} \right) - 10 \times 0.5 = 1\,c{m^2} \cr & \therefore W = 72 \times 2 \times 1 = 144\,erg\,\left[ {\because {\text{There are 2 free surfaces;}}\,\therefore \Delta A = 2 \times 1} \right] \cr} $$