What is the change in potential energy of the soap film formed on a frame of area 4 - 10-3 m2- if the area of the film is reduced to half- surface tension - 40 - 10-3 N-m

The correct option is B 16 × 10^ 5 J ...

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Surface Tension

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Question

What is the change in potential energy of the soap film formed on a frame of area $4 \times 10^{- 3} m^{2}$ , if the area of the film is reduced to half? (surface tension $= 40 \times 10^{- 3} N / m$ )

32 \times 10^{- 5} J

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16 \times 10^{- 5} J

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8 \times 10^{- 5} J

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16 \times 10^{- 3} J

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Solution

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Similar questions

4 \times 10^{- 3} m^{2}

= 40 \times 10^{- 3} N/m

4 \times 10^{- 3} m^{2} ?

40 \times 10^{- 5} H / m)

4 \times 10^{- 3} m^{2}

= 40 \times 10^{- 3} N/m

4 \times 4 c m^{2}

4 \times 5 c m^{2}

3 \times 10^{- 2} N / m

3 \times 10^{- 2} N m^{- 1}

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