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PhysicsFluidSurface TensionMedium2 minPYQ_2023

PhysicsMediumsingle choice

Surface tension of a soap bubble is $2.0 \times 10^{- 2} N m^{- 1}$ . Work done to increase the radius of soap bubble from $3.5 cm$ to $7 cm$ will be : [Take $π = \frac{22}{7}$ ]

Options:

Answer:

Solution:

As we know,

Work done $=$ change in surface energy of soap bubble

Therefore,

Work done $= 2 T (4 π r_{2}^{2} - 4 π r_{1}^{2})$

$= 8 π T (r_{2}^{2} - r_{1}^{2})$

$= 8 \times \frac{22}{7} \times 0.02 (7^{2} - 3 . 5^{2}) \times 10^{- 4} J$

$= 18.48 \times 10^{- 4} J$

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: PYQ_2023

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