The pressure inside a small air bubble of radius 0.1mm situated just below the surface of water will be equal to (Take surface tension of water 70 x10⁻³Nm⁻¹ and atmospheric pressure = 1.013 x 10⁵ Nm⁻² )

The correct option is 1.027 x 10⁵ Pa.

Step-by-step Derivation and Explanation:

1. Identify the given parameters:
Radius of the air bubble, $r=0.1\text{mm}=0.1\times {10}^{-3}\text{m}={10}^{-4}\text{m}$
Surface tension of water, $T=70\times {10}^{-3}{\text{N m}}^{-1}$
Atmospheric pressure, $P_{\text{atm}}=1.013\times {10}^5{\text{N m}}^{-2}$ (or Pa)

2. Formula for excess pressure:
An air bubble inside water has only one liquid-gas interface. Therefore, the excess pressure $\mathrm{\Delta P}$ inside the bubble is given by:
$\mathrm{\Delta P}=\frac{2T}{r}$

3. Calculate the excess pressure:
Substituting the given values into the formula:
$\mathrm{\Delta P}=\frac{2\times 70\times {10}^{-3}}{{10}^{-4}}$
$\mathrm{\Delta P}=140\times {10}^1$
$\mathrm{\Delta P}=1400\text{Pa}=0.014\times {10}^5\text{Pa}$

4. Calculate the total pressure inside the bubble:
Since the bubble is situated just below the surface of water, the pressure outside the bubble is equal to the atmospheric pressure. The total pressure $P$ inside the bubble is the sum of the atmospheric pressure and the excess pressure:
$P=P_{\text{atm}}+\mathrm{\Delta P}$
$P=1.013\times {10}^5\text{Pa}+0.014\times {10}^5\text{Pa}$
$P=(1.013+0.014)\times {10}^5\text{Pa}$
$P=1.027\times {10}^5\text{Pa}$

Thus, the pressure inside the air bubble is $1.027\times {10}^5\text{Pa}$.