1. Pressure inside an air bubble in a liquid

For an air bubble at depth $h$ inside a liquid, the excess pressure inside the bubble over atmospheric pressure is:

$P_{\text{inside}}-P_{\text{atm}}=\rho g h + \frac{2T}{r}$

Here:

• $\rho = 1000\ \mathrm{kg/m^3}$
• $g = 10\ \mathrm{m/s^2}$
• $h = 20\ \mathrm{cm} = 0.2\ \mathrm{m}$
• $r = 0.1\ \mathrm{cm} = 0.001\ \mathrm{m}$
• Given excess pressure:

$P_{\text{inside}}-P_{\text{atm}}=2100\ \mathrm{N/m^2}$

2. Calculate hydrostatic pressure at the bubble depth

$\rho g h = 1000 \times 10 \times 0.2 = 2000\ \mathrm{N/m^2}$

3. Use the pressure relation

$2100 = 2000 + \frac{2T}{r}$

So,

$\frac{2T}{r} = 100$

Substitute $r=0.001\ \mathrm{m}$:

$\frac{2T}{0.001}=100$

$2T = 0.1$

$T = 0.05\ \mathrm{N/m}$

4. Check options

• A: $0.02$ ❌
• B: $0.05$ ✅
• C: $0.25$ ❌
• D: $0.1$ ❌

Therefore, the correct answer is Option B.