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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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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Excessure Pressure for :

Drop of mercury

= \frac{2 _{T_{so a p}}}{R _{Dro p}} = \frac{2 \times 0.465}{2 \times 1 0 ^{- 3}} = 2 \times 232.5 N / m^{2} = 465 N / m^{2}

Bubble of soap

= \frac{2 ^{2} T _{so a p}}{R _{B u bb l e}} = \frac{2 \times 2 \times 0.03}{4 \times 1 0 ^{- 3}} = 2 \times 15 N / m^{2} = 30 N / m^{2}

Bubble of air in water

= \frac{2 ^{2} T _{w a t er}}{R _{B u bb l e}} = \frac{2 \times 2 \times 0.076}{4 \times 1 0 ^{- 3}} = 7 \times 38 N / m^{2} = 76 N / m^{2}

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Question Text	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.
Answer Type	Text solution:1
Upvotes	151

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.

Text solutionVerified

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