QJEEAdv20Phfa86: JEE Advanced 2020 Physics - Class 11 JEEAdv20Phfa86 Solution: A spherical bubble inside water has radius R. Take the pressure …

As change in the kinetic energy of the surface of the bubble is zero.

$\therefore$ W_external + W_water + W_gas = 0 ....(i)

where W_external = work done by external agent

W_water = work done by water

W_gas = work done by gas

For adiabatic process, ${P_1}V_1^\gamma = {P_2}V_2^\gamma$

Here, ${P_0}{\left[ {{4 \over 3}\pi {R^3}} \right]^{{{41} \over {30}}}} = P{\left[ {{4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]^{{{41} \over {30}}}}$

$\Rightarrow$ P = ${P_0}{\left[ {{R \over {R - a}}} \right]^{{{41} \over {10}}}}$

Now, W_gas = ${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$

= ${{{P_0} \times {4 \over 3}\pi {R^3} - P \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$

= ${{{P_0} \times {4 \over 3}\pi {R^3} - {P_0}{{\left[ {{R \over {R - a}}} \right]}^{{{41} \over {10}}}} \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$

= ${{{P_0} \times {4 \over 3}\pi {R^3}\left[ {1 - {{\left( {{R \over {R - a}}} \right)}^{{{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]} \over {{{11} \over {30}}}}$

= ${{{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {{{R - a} \over R}} \right)}^{ - {{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]}$

= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {1 - {a \over R}} \right)}^{ - {{11} \over {10}}}}} \right]$

= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - 1 - {{11a} \over {10R}} - {{\left( { - {{11} \over {10}}} \right)\left( { - {{11} \over {10}} - 1} \right)} \over 2}{{{a^2}} \over {{R^2}}}} \right]$

= -4P₀ $\pi$ R² $a$ - ${{40 \times 21} \over {100 \times 2}}{P_0}\pi R{a^2}$

= -4P₀ $\pi$ R² $a$ - $4.2{P_0}\pi R{a^2}$

W_water = P₀dV

= P₀ ${\left[ {{4 \over 3}\pi {R^3} - {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]}$

= ${{{4{P_0}} \over 3}\pi \left[ {{R^3} - {{\left( {R - a} \right)}^3}} \right]}$

= ${{{4{P_0}} \over 3}\pi \left[ {\left[ {R - \left( {R - a} \right)} \right]\left[ {{R^2} + R\left( {R - a} \right) + {{\left( {R - a} \right)}^2}} \right]} \right]}$

= ${{{4{P_0}} \over 3}\pi \left[ {\left[ a \right]\left[ {3{R^2} - 3Ra + {a^2}} \right]} \right]}$

= ${{{4{P_0}} \over 3}\pi \left[ {\left( {3{R^2}a - 3R{a^2} + {a^3}} \right)} \right]}$

= ${4{P_0}\pi \left[ {{R^2}a - R{a^2}} \right]}$ [ignore ${{a^3}}$ term as no ${{a^3}}$ term in the question]

$\therefore$ W_gas + W_water = - $4{P_0}\pi R{a^2}$ - $4.2{P_0}\pi R{a^2}$

= $- 4{P_0}\pi R{a^2}\left[ {1 + 1.05} \right]$

= $- 4{P_0}\pi R{a^2}\left[ {2.05} \right]$

From (i), W_external = - (W_gas + W_water)

= $4{P_0}\pi R{a^2}\left[ {2.05} \right]$

$\therefore$ X = 2.05