Q: The pressure of air inside a soap bubble of diameter 0.7cm is 8mm of water above atmospheric pressure, the surface tension of soap solution is :

Excess pressure inside a soap bubble is given by-ΔP=r4T⟹ρgh=r4T⟹103×9.8×0.008=0.00354T⟹T=6.86×10−2 N/m

Q: What should be the excess pressure inside a water drop whose radius is 2 mm. Surface tension of water is 0.075 N/m.

Given, R=2 mm=2×10−3 m T=0.075 N/m;Pex= ?∴ Since the pressure inside the water drop is more∴Pex=R2T=2×10−32×0.075=0.075×103 or Pex=75 N/m2

Question 1

Excess pressure can be (2T/R) for

Accepted Answer

i. Excess pressure inside a spherical liquid drop =2T/R, because there is only one free surface hereii. In case of spherical meniscus of radius of curvature R, also excess pressure =2T/R because again there is only one free surface.iii. Excess pressure inside a cylindrical drop of radius R=T/R. Hence, for a cylindrical bubble in air, excess pressure is 2T/R, because there exist two free surfaces in this case.iv. For a spherical bubble in water, excess pressure =T/R, as there is only one free surface.Hence, all the four options are correct.

Question 2

Derive an expression for the excess pressure inside a liquid drop.

Accepted Answer

The net upward force on the top hemisphere of the liquid is just the pressure difference times the area of the equatorial circle :
Fupward ​=(P1′​−P0​)πr2 where Fupward ​= upward force on the liquid due to aircurrents;
P1′​= Final pressure; P0​= Initial pressure; r= radius of the drop, assuming it to be spherical.
The surface tension force downward around circle is equal to the surface tension of the circumference, since only one surface contributed to the force :
Thus, Fdownward ​=T(2πr) where, T= surface tension.
This gives, P1​−P0​=r2 T​ which represents the excess pressure inside the liquid drop.

Question 3

Assertion :A large soap bubble expands while a small bubble shrinks, when they are connected to each other by a capillary tube. Reason :The excess pressure inside bubble (or drop) is inversely proportional to the radius.

Accepted Answer

Since the excess pressure due to surface tension is inversely proportional to its radius, it follows that smaller the bubble, greater is the excess pressure. Thus, when the larger and the smaller bubbles are put in communication, air starts passing from the smaller into the large bubble because excess pressure inside the former is greater than inside the latter. As a result, the smaller bubble shrinks and the larger one swells.

Question 4

There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 N/m and density 1000 kg/m3 The bubble is at a depth of 10cm below the free surface. By what amount is the pressure inside the bubble greater than the atmospheric pressure? (Take g=9.8m/s2)

Accepted Answer

Given : Depth of air bubble below the water surface h=10 cm=0.1 m Radius of air bubble R=1 mm=0.001 m Surface tension of water S=0.075 N/mExcess pressure inside the air bubble ΔP=ρgh+R2S​∴ΔP=1000×9.8(0.1)+0.0012(0.075)​=1130 N/m2

Question 5

What is the excess pressure inside a soap bubble with a radius of 1.5cm and the surface tension of 0.03N/m?

Accepted Answer

Under equilibrium, a static soap bubble has an internal pressure,P=R4T​Where P is the excess pressure, T is the tension and R is the radius of the sphere. So for a bubble of radius 1.5cm or 0.015m and tension 0.03Nm−−1 we get, ΔP=0.0154×0.03​=8Pa

Question 6

The pressure of air inside a soap bubble of diameter 0.7cm is 8mm of water above atmospheric pressure, the surface tension of soap solution is :

Accepted Answer

Excess pressure inside a soap bubble is given by-​ΔP=r4T​⟹ρgh=r4T​⟹103×9.8×0.008=0.00354T​⟹T=6.86×10−2 N/m​

Question 7

What should be the excess pressure inside a water drop whose radius is 2 mm. Surface tension of water is 0.075 N/m.

Accepted Answer

Given, R=2 mm=2×10−3 m T=0.075 N/m;Pex​= ?∴ Since the pressure inside the water drop is more​∴Pex​=R2T​=2×10−32×0.075​=0.075×103 or Pex​=75 N/m2​

Question 8

There are two liquid drops of different radii. The excess pressure inside over the outside is

Accepted Answer

The excess pressure inside a drop is always more than that on the outside and this excess pressure is given by, ΔP=R2T​Therefore, ΔP∝R−1Hence, the drop with a smaller radius will have a higher excess pressure inside the drop.

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Solution for Class 11, Physics, Mechanical Properties of Fluids

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