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Surface Tension Physics Mechanical Properties of Fluids Physics Class 11th

If temperature of a liquid is increased, choose the correct option regarding change in its surface tension and viscosity.

Option 1 -

Surface tension decreases while viscosity increases

Option 2 -

Surface tension increases while viscosity decreases

Option 3 -

Both increases

Option 4 -

Both decreases

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 weeks ago

Correct Option - 4

Detailed Solution:

Fact. Due to decrease in intermolecular forces

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n droplets of water each of radius r coalesce to form a big drop of radius R. If the energy released during coalescence goes into heating the drop, then the rise in temperature will be (take T as surface tension of water and J as mechanical equivalent of heat)

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$R = n^{1 / 3} r$

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A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be:

Raj Pandey

Gain in surface energy, $Δ U = T Δ A$

from volume centenary, $\frac{4}{3} π R^{3} = 64 \times \frac{4}{3} π r^{3}$

$\Rightarrow r = \frac{R}{4}$

Initial surface area, Ai = 4pR²

final surface area, $A_{f} = 64 \times 4 π r^{2}$

$∴ Δ U = 12 π R^{2} . T = 0.075 \times 12 \times 3.14 \times 1 0^{- 4} = 2.8 \times 1 0^{- 4} J$

When two soap bubbles of radii a and b (b > a) coalesce, the radius of curvature of common surface is:

Vishal Baghel

P? - P? = 4T/a

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Also, P? - P? = 4T/r

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A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be

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From volume conservation

$n \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R^{3} = n r^{3} . . . . . . . . . (1)$

Decrease in surface area = $n \times 4 π r^{2} - 4 π R^{2}$

$(Δ A) = 4 π [n r^{2} - R^{2}] = 4 π [\frac{n \times r^{3}}{r} - R^{2}] = 4 π [\frac{R^{3}}{r} - R^{2}] = 4 π R^{3} [\frac{1}{r} - \frac{1}{R}]$

Energy released (W) = $T \times Δ A = 4 π R^{3} T [\frac{1}{r} - \frac{1}{R}]$

Heat produced (Q) = $\frac{W}{J} = \frac{4 π T R^{3}}{J} [\frac{1}{r} - \frac{1}{R}]$

$N o w, Q = m s Δ θ$

$\Rightarrow \frac{4 π R^{3} T}{J} [\frac{1}{r} - \frac{1}{R}] = (\frac{4}{3} π R^{3}) \times | x | \times Δ θ = \frac{3 T}{J} [\frac{1}{r} - \frac{1}{R}]$

A water drop of radius 1cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be:

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$\frac{4}{3} π R^{3} = 729 \frac{4}{3} π r^{3}$

R³= 729r³

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$\begin{array}{l} = 75.39 \times 1 0^{- 5} J \\ Δ u = 7.5 \times 1 0^{- 4} J \end{array}$

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