First slide

Surface tension

Question

Pressure inside a soap bubble, formed in vacuum is P. When this bubble in taken in a closed chamber filled with gas at pressure P/2, its radius is reduced to half of its initial value. Now the pressure inside the bubble will be

Easy

A

3.5 P
B

2.5 P
C

1.5 P
D

4.5 P

Solution

Initially, $P = \frac{4 S}{r}$
Finally, $P^{1} - \frac{P}{2} = Δ P = \frac{4 S}{r / 2}$
$\Rightarrow P^{1} = \frac{P}{2} + 2 (\frac{4 S}{r}) = \frac{P}{2} + 2 P = \frac{5}{2} P$

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