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Class 10 Class 12

Mechanical Properties of Fluids

A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and a light slider supports a weight of 1.5x10^–2N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?

The thin soap film formed between the wire and slider has two free surfaces.

Therefore, total length of free surface that touches with slider is 2L = 60cm = 0.60 m.

Because the slider can support 1.5x10^–2N weight.

$therefore space space space space space space space space space Surface space tension space straight T equals fraction numerator 1.5 cross times 10 to the power of negative 2 end exponent over denominator 0.60 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2.5 cross times 10 to the power of negative 2 end exponent straight N divided by straight m$

151 Views

Two narrow bores of diameters 3-0 mm and 6.0 mm are joined together to form a U-tube open at both the ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? (Surface tension of water at the temperature of the experiment is 72 dyne/cm. Take angle of contact to be zero and density of water to be 1.0 g/cc.)

We know water rises in the capillary tube.

Height to which water rises in the capillary tube is,

Therefore the height h₁of the water that rises in the limb of diameter 3 mm is given by,

The height h₂of the water that will rise in the limb of diameter 6 mm is given by,

Hence difference in the levels of two limbs is,

133 Views

Mercury has an angle of contact equal to 140° with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? (Surface tension of mercury at the temperature of the experiment is 0.465 N/m. Density of mercury = 13.6x10^–3kgm^–3.)

Here,

Angle of contact, $θ = 140 °$

Radius of the narrow tube, r = 1.0mm=10^-3m
Surface tension, T = 0.465 N/m
Density of mercury, $ρ$ = 13600 kg/m³
We know the height by which liquid rises in the tube is,

$h = \frac{2 T cosθ}{rpg} = \frac{2 \times 0.465 \cos 140}{10^{- 3} \times 13600 \times 9.8} = - 0.00535 m = - 5.35 m$

Negative sign means that mercury will dip down.

$∴ Mercury will dip down by 5.35 mm .$

135 Views

What is the pressure inside a drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 0.465N/m. The atmospheric pressure is 1.01 x10⁵ Pa. Also give the excess pressure inside the drop.

Given,

Radius of drop of mercury, r = 3.0 mm

Surface tension of mercury = 0.465 N/m

Atmospheric pressure, P_A = 1.01

\times

10⁵ Pa

Excess of pressure inside the liquid drop is given by,

p = \frac{2 T}{R} Here, T = 0.465 N / m R = 3 mm = 3 \times 10^{- 3} m ∴ E x c e s s p r e s s u r e, p = \frac{2 \times 4.65 \times 10^{- 1}}{3 \times 10^{- 3}} = 310 Pa = 0.0031 \times 10^{5} Pa ∴ Pressure inside the drop = P_{0} + p = 1.01 x 10^{5} + 0.0031 \times 10^{5} Pa = 1.0131 \times 10^{5} Pa

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If an air bubble of 5.0 mm is formed at depth of 40.0 cm inside a container containing soap solution (of relative density 1.20), what would be the pressure inside the bubble?(1AP = 1.013 x10⁵ Pa and surface tension of soap solution is 0-025 N/m)

Total pressure inside the air bubble in the soap solution is the sum of the atmospheric pressure, pressure due to height column of soap solution and excess of pressure due to surface tension.

That is,