Work of 3.0 × 1 0 − 4 joule is required to be done in increasing the size of a soap film from 10 cm × 6 cm to 10 cm × 11 cm . The surface tension of the film is

Surface tension, S = increase in area work done As the soap film has two surfaces. S = 2 × ( 10 × 11 − 10 × 6 ) × 1 0 − 4 m 2 3.0 × 1 0 − 4 J = 3 × 1 0 − 2 N / m .

Start Free Mock LoginAlready registered? Login here

Physics - Fluid Question with Solution | TestHub

PhysicsFluidSurface TensionMedium2 minPYQ_2007

PhysicsMediumsingle choice

Work of $3.0 \times 1 0^{- 4}$ joule is required to be done in increasing the size of a soap film from $10 cm \times 6 cm$ to $10 cm \times 11 cm$ . The surface tension of the film is

Options:

Answer:

Solution:

Surface tension, $S = \frac{work done}{increase in area}$ As the soap film has two surfaces. $S = \frac{3.0 \times 1 0 ^{- 4} J}{2 \times ( 10 \times 11 - 10 \times 6 ) \times 1 0 ^{- 4} m ^{2}} = 3 \times 1 0^{- 2} N / m .$

Stream:AIIMSSubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: PYQ_2007

Doubts & Discussion

Loading discussions...