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PQRS is a rectangular frame of copper wi...


PQRS is a rectangular frame of copper wire shown in figure the side RS of the frame is movable. If a soap film is formed on it then what is the diameter of the wire to maintain equilibirum (given surface tension of soap solution `=0.045N//m` and density of copper `=8.96xx10^(3)kg//m^(3))`

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The correct Answer is:
1.14 mm
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A light wire AB of length 10 cm can slide on a vertical frame as shown in figure. There is a film of soap solution trapped between the frame and the wire. Find the load W that should be suspended from the wire to keep it in equilibrium. Neglect friction. Surface tension of soat solution =25 dyne cm^-1. Take g=10 ms^-2

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Knowledge Check

  • What is the potential energy of a soap film formed on a frame of area 3 xx10^(-3)m^(2) ? [ The surface tension of soap solution is 30 xx 10^(-3) N/m ]

    A
    `1.2 xx10^(-4)` J
    B
    `1.5 xx10^(-4)`J
    C
    `1.8 xx10^(-4)`J
    D
    `2.4 xx10^(-4)`J

  • What is the potential energy of a soap film formed on a frame of area 5 xx 10^(-3) m^(2) ? The surface tension of soap solution is 30 xx 10^(-3) N/m

    A
    `2xx10^(-4)`J
    B
    `2.5 xx10^(-4)`J
    C
    `3xx10^(-4)`J
    D
    `5xx10^(-4)`J

  • A soap film in formed on a frame of area 4xx10^(-3)m^(2) . If the area of the film in reduced to half, then the change in the potential energy of the film is (surface tension of soap solution =40xx10^(-3)N//m)

    A
    `32xx10^(-5)J`
    B
    `16xx10^(-5)J`
    C
    `8xx10^(-5)J`
    D
    `16xx10^(5)J`

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    U - shaped film of a liquid is formed between a wire frame and movable wire 5 cm long. Then the force required to hold the wire in equilibrium is , (Surface tension of liquid is 0.024 N/m )

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    The work done in forming a soap film of size 10 cm xx 10 cm will be , it if the surface tension of soap solution is 3xx 10^(-2) N/m

    Surface area of a soap bubble is 1.3xx10^(-4) m^2 . The work done to doble the surface area will be ( Surface tension for soap solution = 3xx10^(-3) N//m )

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