Find the excess pressure inside a drop of mercury of radius 2 mm, a soap bubble of radius 4 mm and an air bubble of radius 4 mm formed inside a tank of water. surface tension of mercury is 0.465 n/m and soap solution and water are 0.03 N/m and 0.076 N/m respectively.

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Solution

Excessure Pressure for :
Drop of mercury $= \frac{2_{T_{s o a p}}}{R_{D r o p}} = \frac{2 \times 0.465}{2 \times 10^{- 3}} = 2 \times 232.5 N / m^{2} = 465 N / m^{2}$
Bubble of soap $= \frac{2^{2} T_{s o a p}}{R_{B u b b l e}} = \frac{2 \times 2 \times 0.03}{4 \times 10^{- 3}} = 2 \times 15 N / m^{2} = 30 N / m^{2}$
Bubble of air in water $= \frac{2^{2} T_{w a t e r}}{R_{B u b b l e}} = \frac{2 \times 2 \times 0.076}{4 \times 10^{- 3}} = 7 \times 38 N / m^{2} = 76 N / m^{2}$

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