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Mechanical Properties Of Fluids

Question

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The work done in increasing the size of a soap film for 10cm × 6cm to 10cm × 11cm is 3 × 10^-4 J. The surface tension of the film is

1.0 × 10^-2 N/m

6.0 × 10^-2 N/m

3.0 × 10^-2 N/m

1.5 × 10^-2N/m

Solution

C.

3.0 × 10^-2 N/m

Given:-

Increasing size ΔA = (10 × 10^-2 × 11 × 10^-2 - 10 × 10^-2 × 6 × 10^-2 )

ΔA = ( 10 × 11 - 10 × 6 )

Work done = area increased × surface tension

surface tension

S_T = $\frac{3 \times 10^{- 4}}{2 \times (10 \times 11 - 10 \times 6) \times 10^{- 4}}$

(since soap film has two surfaces)

S_T = 3 × 10^-2 N/m

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