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Question
Find the excess pressure inside (a) a drop of mercury of radius 2 mm (b) a soap bubble of radius 4 mm and (c) an air bubble of radius 4 mm formed inside a tank of water. Surface tension of mercury, soap solution and water are 0.465 N m−1, 0.03 N m−1 and 0.076 N m−1 respectively.
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Solution
Given:
\[\text{ Radius of mercury drop r = 2 mm }= 2 \times {10}^{- 3} \text{ m }\]
\[\text{ Radius of soap bubble r = 4 mm }= 4 \times {10}^{- 3} \text{ m }\]
\[\text{ Radius of air bubble r = 4 mm } = 4 \times {10}^{- 3} \text{ m }\]
\[\text{ Surface tension of mercury T}_{Hg} = 0 . 465 \text{ N/m }\]
\[\text{ Surface tension of soap solution T}_s = 0 . 03 \text{ N/m }\]
\[\text{ Surface tension of water T}_a = 0 . 076 \text{ N/m}\]
(a) Excess pressure inside mercury drop:
\[P = \frac{2 T_{Hg}}{r}\]
\[ = \frac{0 . 465 \times 2}{2 \times {10}^{- 3}} = 465 \text{ N/ m}^2\]
(b) Excess pressure inside the soap bubble:
\[P = \frac{4 T_s}{r}\]
\[ = \frac{4 \times 0 . 03}{4 \times {10}^{- 3}} = 30 \text{ N/ m }^2 \]
(c) Excess pressure inside the air bubble:
\[P = \frac{2 T_a}{r}\]
\[ = \frac{2 \times 0 . 076}{4 \times {10}^{- 3}} = 38 \text{ N/ m}^2\]
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