Question

If the surface tension at the water-air interface is 0.0718 n/m, how much does the pressure in a cylindrical jet of water 4 mm in diameter exceed the pressure of the surrounding atmosphere

Answer 1

Put the value in the formula
2s/r.

Answer 2

Answer:

The pressure in a cylindrical jet of water 4 mm in diameter exceeds the pressure of the surrounding atmosphere by 71.8 Nm⁻².

Concept:

Surface tension: The surface tension of a liquid is defined as the force per unit length in the plane of the liquid surface, acting at right angles on either side of an imaginary line drawn on that surface. The SI unit of surface tension is 'newton/meter' (Nm⁻¹).

Excess pressure inside a liquid is given by,

P = 2T/R
where P = Excess pressure

T = Tension at the surface

R = Radius of the bubble

Solution:

Here, T = 0.0718 Nm⁻¹

Diameter of cylindrical jet, D = 4 mm = 4 × 10⁻³ m

                                                = 0.004 m

Radius of cylindrical jet, R = D/2

                                            = 0.004/2

                                             = 0.002 m

∴ Excess pressure, P = 2T/R

                                   = 2(0.0718)/0.002

                                   = 0.1436/0.002

                                   = 71.8 Nm⁻²

Hence,  the pressure in a cylindrical jet of water 4 mm in diameter exceeds the pressure of the surrounding atmosphere by 71.8 Nm⁻².

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