For brittle material, there is no yield point.

$$\eqalign{ & Y = \frac{{F\ell }}{{A\Delta \ell }}; \Rightarrow \,F = \frac{{YA\Delta \ell }}{\ell } \cr & \frac{{{F_1}}}{{\;{F_2}}} = \frac{{{A_1}}}{{\;{A_2}}} \times \frac{{{l_2}}}{{{l_1}}} = \frac{{r_1^2}}{{r_2^2}} \times \frac{{{l_2}}}{{{l_1}}} = \frac{4}{1} \times \frac{1}{4} = 1:1 \cr} $$

We know that excess pressure in a soap bubble is inversely proportional to its radius. The soap bubble at end 1 has small radius as compared to the soap bubble at end 2 (given). Therefore excess pressure at 1 is more.

As the value is opened, air flows from end 1 to end 2 and the volume of soap bubble at end 1 decreases.

Due to acceleration down the plane, both body $$B$$ and liquid will received the same component of acceleration down the plane.

$$\eqalign{ & {T_1} - {T_2} = m{\omega _0}^2\left( {2l} \right) \cr & {T_2} = m{\omega _0}^2\left( {3l} \right) \cr & \therefore {T_1} = 5m{\omega _0}^2l. \cr & \frac{{{\text{Energy}}\left( A \right)}}{{{\text{Energy}}\left( B \right)}} = \frac{{{{\left( {{\text{stress}}} \right)}^2}_A{\text{volume}}\,A}}{{{{\left( {{\text{stress}}} \right)}^2}_B{\text{volume}}\,B}} \cr & = \frac{{{{\left( 5 \right)}^2}}}{{{{\left( 3 \right)}^2}}} \times \left( {\frac{{A \times 2l}}{{A \times l}}} \right) = \frac{{50}}{9} \cr} $$

$$\eqalign{ & \Delta L = \frac{{LF}}{{AY}},\,{\text{where}}\,L = 3\,m, \cr & A = \pi {\left( {1.0 \times {{10}^{ - 3}}\,m} \right)^2} = 3.14 \times {10^{ - 6}}\,{m^2}, \cr} $$
and since each wire supports one-quarter of the load,
$$\eqalign{ & F = \frac{{\left( {50\;kg} \right)\left( {9.8\,m/{s^2}} \right)}}{4} = 123\,N \cr & \Delta L = \frac{{\left( {3\;m} \right)\left( {123\,N} \right)}}{{\left( {3.14 \times {{10}^{ - 6}}\,{m^2}} \right)\left( {1.8 \times {{10}^{11}}\,N/{m^2}} \right)}} \cr & = 65 \times {10^{ - 5}}\,m\,\,{\text{or}}\,\,0.65\,mm \cr} $$

$$\eqalign{ & e = \frac{{\Delta Y}}{\ell } = \frac{{1 \times {{10}^{ - 3}}}}{4} \cr & u = \frac{{{e^2}Y}}{2} = {\left( {\frac{{{{10}^{ - 3}}}}{4}} \right)^2} \times \frac{{2 \times {{10}^{11}}}}{2} = 0.075\,J. \cr} $$

When small droplets coelesce to form a bigger drop, energy released in this process is given by,
$$4\pi {R^3}T\left[ {\frac{1}{r} - \frac{1}{R}} \right]$$
According to question
$$\eqalign{ & \frac{1}{2}m{v^2} = 4\pi {R^3}T\left[ {\frac{1}{r} - \frac{1}{R}} \right] \cr & \Rightarrow \frac{1}{2}\left[ {\frac{4}{3}\pi {R^3}\rho } \right]{v^2} = 4\pi {R^3}T\left[ {\frac{1}{r} - \frac{1}{R}} \right] \cr & \Rightarrow {V^2} = \frac{{6T}}{\rho }\left[ {\frac{1}{r} - \frac{1}{R}} \right] \Rightarrow V = {\left[ {\frac{{6T}}{\rho }\left( {\frac{1}{r} - \frac{1}{R}} \right)} \right]^{\frac{1}{2}}} \cr} $$

If $${v_2}$$ is the required speed then
$$\eqalign{ & {A_1}{v_1} = {A_2}{v_2} \cr & {\text{or}}\,\,{v_1} = \frac{{{A_2}}}{{{A_1}}}{v_2} = 0 .1{v_2}\,......\left( {\text{i}} \right) \cr} $$
From Bernoulli’s equation, we have
$${P_a} + \frac{1}{2}\rho v_1^2 + \rho g\left( {3 - 0.525} \right) = {P_a} + \frac{1}{2}\rho v_2^2 + 0\,......\left( {{\text{ii}}} \right)$$
After solving above equations, we get
$$v_2^2 = 50\,{m^2}/{s^2}$$

$$L = \frac{S}{{dg}} = \frac{{{{10}^6}}}{{3 \times {{10}^3} \times 10}} = \frac{{100}}{3} = 34\,m$$