Pressure on the bubble at height $y$ from the bottom

If the liquid height is $H$, then the depth of the bubble below the free surface is $H-y$. Hence pressure on the bubble is $P(y)=P_0+\rho_l g(H-y).$

At the bottom $(y=0)$, $P_b=P_0+\rho_l gH.$

Adiabatic relation for the gas inside the bubble

The bubble does not exchange heat with liquid, so the process is adiabatic: $PV^\gamma=\text{constant}, \qquad \gamma=\frac53.$

Also, for an ideal gas, $PV=nRT.$

From adiabatic relation and ideal gas law, we can use $TV^{\gamma-1}=\text{constant}$ or equivalently $T^\gamma P^{1-\gamma}=\text{constant}.$

A more direct useful result is: $T\propto P^{\frac{\gamma-1}{\gamma}}.$

Since $\gamma=\frac53$, $\frac{\gamma-1}{\gamma}=\frac{2/3}{5/3}=\frac25.$

Therefore, $T=T_0\left(\frac{P(y)}{P_b}\right)^{2/5}.$

So, $T=T_0\left(\frac{P_0+\rho_l g(H-y)}{P_0+\rho_l gH}\right)^{2/5}.$

Volume of the bubble at height $y$

Using ideal gas equation, $V=\frac{nRT}{P(y)}.$

Substitute $T$: $V=\frac{nR T_0}{P(y)}\left(\frac{P(y)}{P_b}\right)^{2/5}.$

Rearranging, $V=\frac{nR T_0}{P_b^{2/5} P(y)^{3/5}}.$

Since $P_b=P_0+\rho_l gH, \qquad P(y)=P_0+\rho_l g(H-y),$ we get $V=\frac{nR T_0}{\left(P_0+\rho_l gH\right)^{2/5}\left(P_0+\rho_l g(H-y)\right)^{3/5}}.$

Buoyancy force

Buoyant force is $F_B=\rho_l gV.$

Therefore, $F_B=\frac{\rho_l nR g T_0}{\left(P_0+\rho_l gH\right)^{2/5}\left(P_0+\rho_l g(H-y)\right)^{3/5}}.$

Compare with options

This exactly matches Option B: $\boxed{\frac{\rho_l nRgT_0}{\left(P_0+\rho_l gH\right)^{2/5}\left[P_0+\rho_l g(H-y)\right]^{3/5}}}$

Comparison with stored correct answer

Stored correct answer is B, which matches the derived result.