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A liquid is kept in a beaker of radius 4...

A liquid is kept in a beaker of radius 4 cm. Consider a diameter of the beaker on the surface of the water. Find the force by which the surface on one side of the diameter pulls the surface on the other side. Surface tension of liquid `=0.075 Nm^(-1)`

Text Solution

Verified by Experts

The length of the diameter is
l=2r=8 cm =0.08 m
The surface tension is `S=F//l`. Thus,
`F=Sl=(0.075 Nm^(-1))xx(0.08 m)=6xx10^(-3)N`
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Knowledge Check

  • If there is no difference of pressure on the two sides of the surface, then the liquid surface is

    A
    concave
    B
    convex
    C
    cylindrical
    D
    plane

  • A ring of radius 0.75 cm is floating on the surface of water . If surface tension of water is 0.07 N/m the force required to lift the ring from the surface of water will be

    A
    `66xx 10^(-1)` N
    B
    ` 66 xx 10^(-2)`N
    C
    `1.05 xx 10^(-3)` N
    D
    `66 xx 10^(-4)` N

  • If the surface tension of a liquid is 5 N/m then the surface energy of the liquid film on a ring of area 0.15 m^(2) is

    A
    `1.5`J
    B
    `3.0`J
    C
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    D
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