In Bernoulli’s theorem only law of conservation of energy is obeyed.

When radius is decrease by $$\Delta R,$$
$$\eqalign{ & 4\pi {R^2}\Delta R\rho L = 4\pi T\left[ {{R^2} - {{\left( {R - \Delta R} \right)}^2}} \right] \cr & \Rightarrow \rho {R^2}\Delta RL = T\left[ {{R^2} - {R^2} + 2R\Delta R - \Delta {R^2}} \right] \cr & \Rightarrow \rho {R^2}\Delta RL = T2R\Delta R\left[ {\Delta R\,\,{\text{is very small}}} \right] \cr & \Rightarrow R = \frac{{2T}}{{\rho L}} \cr} $$

Pressure difference between lungs and atmosphere
$$\eqalign{ & = \left( {760 - 750} \right)mm\,{\text{of}}\,Hg \cr & = 10\,mm\,{\text{of}}\,Hg = 1\,cm\,{\text{of}}\,Hg \cr} $$
Let the boy can suck water from depth $$h.$$
Then
Pressure difference $$ = h{\rho _{{\text{water}}}}g = 1\,cm\,{\text{of}}\,Hg$$
$$\eqalign{ & {\text{or,}}\,h \times 1g\,c{m^{ - 3}} \times 980\,cm\,{s^{ - 2}} \cr & = 1\,cm \times 13.6g\,c{m^{ - 3}} \times 980\,cm\,{s^{ - 2}} \cr & \therefore h = 13.6\,cm \cr} $$
The boy can suck water from the depth of $$13.6\,cm$$

Force = weight suspended + weight of $${\frac{{3L}}{4}}$$ of wire
$$\eqalign{ & = {W_1} + \frac{{3W}}{4} \cr & {\text{stress}} = \frac{{{\text{force}}}}{{{\text{area}}}} \cr} $$

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

The given graph does not obey Hooke's law. and there is no well defined plastic region. So the graph represents elastomers.

Mass of liquid in horizontal portion of $$U$$-tube $$ = Ad\rho $$
Pseudo force on this mass $$ = Ad\rho a$$
Force due to pressure difference in the two limbs $$ = \left( {{h_1}\rho g - {h_2}\rho g} \right)A$$
Equating both the forces
$$\eqalign{ & \left( {{h_1} - {h_2}} \right)\rho gA = Ad\rho a \cr & \Rightarrow \left( {{h_1} - {h_2}} \right) = \frac{{Ad\rho a}}{{\rho gA}} = \frac{{ad}}{g} \cr} $$

When the bubble gets detached,
Buoyant force $$=$$ force due to surface tension

Force due to excess pressure $$=$$ upthrust
Access pressure in air bubble $$ = \frac{{2T}}{R}$$
$$\eqalign{ & \frac{{2T}}{R}\left( {\pi {r^2}} \right) = \frac{{4\pi {R^3}}}{{3T}}{\rho _w}g \cr & \Rightarrow {r^2} = \frac{{2{R^4}{\rho _w}g}}{{3T}} \cr & \Rightarrow r = {R^2}\sqrt {\frac{{2{\rho _w}g}}{{3T}}} \cr} $$

$$VAB$$  is the given cone. Let its height be $$h$$ and semi-vertical angle $$\alpha .$$ Let the base $$AB$$  of the cone be in the surface. $$CD$$  is the surface of separation of two liquids, $$O$$ and $$O'$$ are the centres of the base $$AB$$  and surface of separation $$CD.$$

$$\therefore $$ For equilibrium, weight of the cone = (weight of liquid of density $${{\sigma _1}}$$ displaced) + (weight of liquid of density $${{\sigma _2}}$$ displaced)
$$\eqalign{ & {\text{or}}\,\,\frac{1}{3}\pi {h^3}{\tan ^2}\alpha \rho g = \frac{1}{3}\pi {z^3}{\tan ^2}\alpha {\sigma _1}g + \frac{1}{3}\pi \left( {{h^3} - {z^3}} \right){\tan ^2}\alpha .{\sigma _2}g \cr & {\text{or}}\,\,{h^3}\rho = {z^3}{\sigma _1} + \left( {{h^3} - {z^3}} \right){\sigma _2} \cr & {\text{or}}\,\,{h^3}\left( {\rho - {\sigma _2}} \right) = {z^3}\left( {{\sigma _1} - {\sigma _2}} \right) \cr & {\text{or}}\,\,z = h{\left( {\frac{{\rho - {\sigma _2}}}{{{\sigma _1} - {\sigma _2}}}} \right)^{\frac{1}{3}}}. \cr} $$

If $$\ell $$ is the original length of wire, then change in length of first wire,
$$\Delta {\ell _1} = \left( {{\ell _1} - \ell } \right)$$
Change in length of second wire,
$$\eqalign{ & \Delta {\ell _2} = \left( {{\ell _2} - \ell } \right) \cr & {\text{Now,}}\,Y = \frac{{{T_1}}}{A} \times \frac{\ell }{{\Delta {\ell _1}}} = \frac{{{T_2}}}{A} \times \frac{\ell }{{\Delta {\ell _2}}} \cr & {\text{or}}\,\,\frac{{{T_1}}}{{\Delta {\ell _1}}} = \frac{{{T_2}}}{{\Delta {\ell _2}}}\,\,{\text{or}}\,\,\frac{{{T_1}}}{{{\ell _1} - \ell }} = \frac{{{T_2}}}{{{\ell _2} - \ell }} \cr & {\text{or}}\,\,{T_1}{\ell _2} - {T_1}\ell = {T_2}{\ell _1} - \ell {T_2} \cr & {\text{or}}\,\,\ell = \frac{{{T_2}{\ell _1} - {T_1}{\ell _2}}}{{{T_2} - {T_1}}} \cr} $$