1. Forces on the bubble just before detachment

A bubble at the bottom experiences:

• Upward buoyant force due to displaced water
• Downward force due to surface tension along the circular contact line

Just before detachment, these balance.

2. Buoyant force

The bubble is approximately spherical with radius $R$, so its volume is

$V = \frac{4}{3}\pi R^3$

Hence the buoyant force is

$F_b = \rho_w g V = \rho_w g \cdot \frac{4}{3}\pi R^3$

3. Surface tension force holding the bubble

The bubble touches the vessel over a circular rim of radius $r$.

Surface tension force acts along the circumference:

$\text{circumference} = 2\pi r$

Since the bubble has two liquid surfaces (inner and outer surface of the thin film), the total downward force due to surface tension is

$F_T = 2 \times (2\pi r T) \sin\theta$

For $r \ll R$, just before detachment the bubble is nearly spherical and at the contact rim,

$\sin\theta \approx \frac{r}{R}$

So,

$F_T = 4\pi r T \cdot \frac{r}{R} = \frac{4\pi T r^2}{R}$

4. Condition for detachment

At the point of detachment,

$F_b = F_T$

So,

$\rho_w g \cdot \frac{4}{3}\pi R^3 = \frac{4\pi T r^2}{R}$

Cancel $4\pi$:

$\frac{1}{3}\rho_w g R^3 = \frac{T r^2}{R}$

$r^2 = \frac{\rho_w g R^4}{3T}$

$r = R^2\sqrt{\frac{\rho_w g}{3T}}$

This would match option A if bubble had one surface only.

5. Correct treatment for a vapour bubble in water

Here the bubble is a vapour/gas bubble inside water, so there is only one liquid-gas interface, not a soap bubble film. Therefore surface tension force is

$F_T = (2\pi r T)\sin\theta$

Again using

$\sin\theta \approx \frac{r}{R}$

we get

$F_T = 2\pi r T \cdot \frac{r}{R} = \frac{2\pi T r^2}{R}$

Now balance with buoyancy:

$\rho_w g \cdot \frac{4}{3}\pi R^3 = \frac{2\pi T r^2}{R}$

Cancel $2\pi$:

$\frac{2}{3}\rho_w g R^3 = \frac{T r^2}{R}$

$r^2 = \frac{2\rho_w g R^4}{3T}$

Therefore,

$\boxed{r = R^2\sqrt{\frac{2\rho_w g}{3T}}}$

6. Option check

• A: $R^2\sqrt{\dfrac{\rho_w g}{3T}}$
• B: $R^2\sqrt{\dfrac{\rho_w g}{6T}}$
• C: $R^2\sqrt{\dfrac{\rho_w g}{T}}$
• D: $R^2\sqrt{\dfrac{2\rho_w g}{3T}}$

Hence the correct option is

$\boxed{\text{D}}$