An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m3. If the pressure inside the bubble is 2100 N/m2 greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g = 10 m/s2)

T is surface tensionPin air bubble=P0+ρgh+2TR Pin−P0=ρgh+2TR=2100 2TR=2100−ρgh T=R2(2100−103×10×0.2) =120(2100−2000)×10−2 =0.05

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m3. If the pressure inside the bubble is 2100 N/m2 greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g = 10 m/s2)

T is surface tensionPin air bubble=P0+ρgh+2TR Pin−P0=ρgh+2TR=2100 2TR=2100−ρgh T=R2(2100−103×10×0.2) =120(2100−2000)×10−2 =0.05

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An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m³. If the pressure inside the bubble is 2100 N/m²greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g = 10 m/s²)

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0.02

0.1

0.25

0.05

answer is D.

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Detailed Solution

T is surface tension
$P in air bubble = P_{0} + ρ g h + \frac{2 T}{R}$
$\begin{aligned} P_{in} - P_{0} = ρ g h + \frac{2 T}{R} = 2100 \\ \frac{2 T}{R} = 2100 - ρ g h \\ T = \frac{R}{2} (2100 - 10^{3} \times 10 \times 0.2) \\ = \frac{1}{20} (2100 - 2000) \times 10^{- 2} \\ = 0.05 \end{aligned}$