The excess pressure inside an air bubble of radius $R$ inside the liquid:

\frac{2 T}{R}

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\frac{4 T}{R}

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\frac{T}{R}

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\frac{8 T}{R}

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Solution

The correct option is C $\frac{2 T}{R}$
When the air bubble is inside the liquid, there is one surface- the water to air surface where the pressure increases.
Hence the excess pressure inside the bubble is the increases in pressure across one surface
$Δ P = \frac{2 T}{R}$

Option A is correct.

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List-I	List-II
a) Radius of the common surface when two soap bubbles of radii $R_{1} a n d R_{2}$ coalesce.	e) $(R_{1}^{3} + R_{2}^{3})^{1 / 3}$
b) Radius of the large drop when two liquid drops of radii $R_{1} a n d R_{2}$ combine at constant temperature.	f) $\frac{R_{1} R_{2}}{R_{2} - R_{1}}$
c) Excess of pressure inside air bubble.	g) $\frac{2 T}{r}$
d) Excess of pressure in the air bubble in water.	h) $\frac{4 T}{r}$

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