First slide

Surface tension

Question

Excess pressure in a soap bubble of radius R is P. What will be the excess pressure in a drop of same soap solution of radius $\frac{R}{8} ?$

Easy

A

$\frac{P}{4}$
B

8P
C

4P
D

2P

Solution

(3)
$\begin{array}{l} F o r t h e s o a p b u b b l e, P = \frac{4 T}{R} \\ F o r t h e d r o p o f s o a p s o l u t i o n, P^{'} = \frac{2 T}{r} = \frac{2 T}{R / 8} \\ ∴ \frac{P^{'}}{P} = \frac{8 \times 2 T}{R} \times \frac{R}{4 T} = 4 \Rightarrow P^{'} = 4 P \end{array}$

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