$$\eqalign{ & \left( {{m_{{\text{concrete}}}} + {m_{{\text{sawdust}}}}} \right)g = V{\rho _w}g \cr & 2.4 \times \frac{4}{3}\pi \left( {{R^3} - {r^3}} \right) + 0.3 \times \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {R^3} \times 1 \cr & {\text{or}}\,\,{R^3} = \frac{3}{2}{r^3} \cr & \therefore \frac{{{m_{{\text{concrete}}}}}}{{{m_{{\text{sawdust}}}}}} = \frac{{{\rho _{{\text{concrete}}}}\frac{4}{3}\pi \left( {{R^3} - {r^3}} \right)}}{{{\rho _{{\text{sawdust}}}}\frac{4}{3}\pi {r^3}}} \cr & = \frac{{2.4}}{{0.3}}\frac{{\left( {\frac{3}{2}{r^3} - {r^3}} \right)}}{{{r^3}}} = 4 \cr} $$

$$\eqalign{ & l = \frac{{FL}}{{AY}} \Rightarrow l \propto \frac{1}{{{r^2}}}\,\,\left( {F,\;L\,{\text{and}}\,Y\,{\text{are}}\,{\text{same}}} \right) \cr & \frac{{{l_A}}}{{{l_B}}} = {\left( {\frac{{{r_B}}}{{{r_A}}}} \right)^2} = {\left( {\frac{{{r_B}}}{{2{r_B}}}} \right)^2} = \frac{1}{4} \Rightarrow {l_A} = 4{l_B} \cr & {\text{or}}\,\,{l_B} = \frac{{{l_A}}}{4} \cr} $$

Given that $$\frac{F}{A} = 4.8 \times {10^7}N{m^{ - 2}}$$
$$\eqalign{ & \therefore F = 4.8 \times {10^7} \times A \cr & {\text{or}}\,\,\frac{{m{v^2}}}{r} = 4.8 \times {10^7} \times {10^{ - 6}} = 48 \cr & {\text{or}}\,\frac{{m{r^2}{\omega ^2}}}{r} = 48\,\,{\text{or}}\,\,{\omega ^2} = \frac{{48}}{{mr}} \cr & \omega = \sqrt {\left( {\frac{{48}}{{10 \times 0.3}}} \right)} = \sqrt {16} = 4\,rad/\sec \cr} $$

When terminal velocity is reached then body moves with constant velocity hence, acceleration is zero.

Case (i) At equilibrium, $$T= W$$
$$Y = \frac{{\frac{W}{A}}}{{\frac{\ell }{L}}}\,.....\,(1)$$
Case (ii) At equilibrium, $$T= W$$
$$Y = \frac{{\frac{W}{A}}}{{\frac{{\frac{\ell }{2}}}{{\frac{L}{2}}}}}\,\,\,\, \Rightarrow Y = \frac{{\frac{W}{A}}}{{\frac{\ell }{L}}}$$
$$ \Rightarrow $$ Elongation is the same.

Let the liquid rise to a height $$h.$$
$$\eqalign{ & T = \frac{{h\rho rg}}{2} \cr & \therefore hr = \frac{{2T}}{{\rho g}} \cr} $$
If the tube is of height $${h_1} < h$$
$$\eqalign{ & {h_1}{r_1} = hr = \frac{{2T}}{{\rho g}} \cr & \therefore {r_1} = \frac{{2T}}{{{h_1}\rho g}} = \frac{{2 \times 0.073}}{{\frac{{25}}{{1000}} \times 1000 \times 9.8}}\left\{ {25\,mm = \frac{{25}}{{1000}}m} \right\} \cr & = \frac{{2 \times 0.073}}{{9.8 \times 25}} = 0.0006\,m = 0.6\,mm \cr} $$

$$\eqalign{ & {\text{As energy released}} = \left( {{A_f} - {A_i}} \right)T \cr & {\text{where,}}\,{A_i} = 4\pi {R^2} = \frac{3}{R} \times \left( {\frac{{4\pi }}{3} \times {R^3}} \right) = \frac{{3V}}{R} \cr & {\text{and}}\,{A_f} = 4\pi {r^2} = \frac{V}{{\frac{4}{3}\pi {r^2}}}4\pi {r^2} = \frac{{3V}}{r} \cr & \therefore {\text{Energy released}} = 3VT\left[ {\frac{1}{r} - \frac{1}{R}} \right] \cr} $$

The pressure at a point just outside the bubble
$$\eqalign{ & {p_0} = {P_{{\text{atm}}}} + h\rho g = {P_{{\text{atm}}}} + 10 \times {10^{ - 2}} \times {10^3} \times 9.8 \cr & = {P_{{\text{atm}}}} + 980\,N{m^{ - 2}} \cr} $$
Now excess pressure within the bubble compared to a point just outside
$$ = \frac{{2T}}{R} = \frac{{2 \times 72 \times {{10}^{ - 3}}}}{{2 \times {{10}^{ - 3}}}} = 72\,N{m^{ - 2}}$$
$$\therefore $$ Inside pressure = excess pressure + outside pressure
$$\eqalign{ & = {P_{{\text{atm}}}} + 980 + 72 \cr & = {P_{{\text{atm}}}} + 1052\,N{m^{ - 2}} \cr} $$
$$\therefore $$ Pressure within the bubble in excess of atmospheric pressure $$ = 1052\,N{m^{ - 2}}.$$

As the temperature rises the atoms of the liquid become more mobile and the coefficient of viscosity falls.

$$\eqalign{ & {P_a} + \frac{1}{2}{\rho _1}v_1^2 + 0 \cr & = Pa + \frac{1}{2}{\rho _2}v_2^2 + \left( {{\rho _1}g{h_1} + {\rho _2}g{h_2}} \right) \cr & {\text{As}}\,{v_2} < < {v_1}. \cr & \therefore {v_1} = \sqrt {2g\left( {{h_1} + {h_2}\frac{{{\rho _2}}}{{{\rho _1}}}} \right)} \cr} $$