Surface Tension: Excess of Pressure Numericals Class-11 Nootan ISC Physics

Surface Tension: Excess of Pressure Numericals Class-11 Nootan ISC Physics Solutions Ch-16. Step by step Solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Surface Tension Excess of Pressure Numericals Class-11 Nootan ISC Physics

Surface Tension: Excess of Pressure Numericals Class-11 Nootan ISC Physics Solutions

Board ISC
Class 11
Subject Physics
Writer Kumar and Mittal
Publication  Nageen Prakashan
Chapter-16 Surface Tension
Topics Numericals on Excess of Pressure
Academic Session 2024-2025

Numericals on Excess of Pressure

Surface Tension:  Class-11 Nootan ISC Solutions Ch-16 of Kumar and Mittal Physics , Nageen Prakashan

Que-13: The air pressure inside a soap bubble of diameter 3.5 mm is 8 mm of water column above the atmospheric pressure. Calculate the surface tension of soap solution.

Ans- Δp = 4 T / R

=> T = Δp x R / 4

=> (8 x 10^-3 x 1 x 10³ x 9.8 x 1.75 x 10^-3) / 4      {p = h ρ g}

=> 3.43 x 10^-2 N/m

Que-14: What would be the excess pressure above atmosphere inside an air bubble of 0.2 mm radius situated just below the surface of water? The surface tension of water is 0.07 N/m. Express this excess pressure in terms of the height of mercury column.

Ans- Δp = 2 T / R

=> (2 x 0.07) / (0.2 x 10^-3)

=> 700 N/m²

again  p = h ρ g

=> h = p / ρ g

=> 700 / (13.6 x 10^-3 x 9.8)

=> 5.25 x 10^-3

=> 5.25 mm of hg

Que-15: Find the depth at which an air-bubble of radius 0.7 mm will remain in equilibrium in water. Given surface tension of water = 7.0 × 10^-2 N/m (g = 10 m/s²)

Ans- Δp = 2 T / R

{inside water there is only one free surface}

=> Δp = (2 x 7.0 x 10^-2) / (0.7 x 10^-3)

=> 200 N/m²

now h = Δp / ρ g

=> 200 / (1 x 10³ x 10)

=> 0.02 m = 2 cm

Que-16: There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 N/m and density 10³ kg/m³. The bubble is at a depth of 10.0 cm below the free surface of liquid. By what amount is the pressure inside the bubble greater than the atmospheric pressure?

Ans- excess pressure = Δp inside + pressure of liquid

=> (2T / R) +  h ρ g

=> [(2 x 0.075) / (1 x 10^-3)] + 0.1 x 9.8 x 1 x 10³

=> 150 + 980

=> 1130 N/m²

Que-17: What will be the total pressure inside a spherical air bubble of radius 0.2 mm at a depth of 2 m below the surface of a liquid of density 1.1 × 10³ kg/m³ and surface tension 5 x 10^-2 N/m. Atmospheric pressure is 1.1 x 10^5 N/m².

Ans- Total pressure inside the bubble

=> 2 T / R + h ρ g + Atmospheric pressure

=> [(2 x 5 x 10^-2) / (0.2 x 10^-3)] + (1.1 x 10³ x 9.8 x 2) + (1.1 x 10^5)

=> 500 + 1.1 x 10³ (19.6 + 100)

=> 500 + 1.1  x 119 x 10³

=> 0.005 x 10^5 + 1.1 x 1.19 x 10^5

=> (0.005 + 1.1 x 1.19) x 10^5

=> 1.22 x 10^5 N/m²

—:  end of Surface Tension: Excess of Pressure Numericals Class-11 Nootan ISC Ch-16 Solutions :—

Return to :  Nootan Solutions for ISC Physics Class-11 Nageen Prakashan

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