Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 10.19:

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10^–1 N m^–1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.

Answer:

Answer: 

Radius of the mercury drop, r = 3.00 mm = 3 × 10^–3 m

Surface tension of mercury, S = 4.65 × 10^–1 N m^–1

Atmospheric pressure, P₀ = 1.01 × 10⁵ Pa

Total pressure inside the mercury drop

= Excess pressure inside mercury + Atmospheric pressure

= 1.0131 × 10⁵

= 1.01 ×10⁵ Pa

Excess pressure 

 = 310 Pa

Question 10.20:

What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10^–2 N m^–1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 10⁵ Pa).

Answer:

Excess pressure inside the soap bubble is 20 Pa;

Pressure inside the air bubble is 

Soap bubble is of radius, r = 5.00 mm = 5 × 10^–3 m

Surface tension of the soap solution, S = 2.50 × 10^–2 Nm^–1

Relative density of the soap solution = 1.20

∴Density of the soap solution, ρ = 1.2 × 10³ kg/m³

Air bubble formed at a depth, h = 40 cm = 0.4 m

Radius of the air bubble, r = 5 mm = 5 × 10^–3 m

1 atmospheric pressure = 1.01 × 10⁵ Pa

Acceleration due to gravity, g = 9.8 m/s²

Hence, the excess pressure inside the soap bubble is given by the relation:

Therefore, the excess pressure inside the soap bubble is 20 Pa.

The excess pressure inside the air bubble is given by the relation:

Therefore, the excess pressure inside the air bubble is 10 Pa.

At a depth of 0.4 m, the total pressure inside the air bubble

= Atmospheric pressure + hρg + P’

Therefore, the pressure inside the air bubble is 

Question 10.21:

A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm². The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.

Answer:

Base area of the given tank, A = 1.0 m²

Area of the hinged door, a = 20 cm² = 20 × 10^–4 m²

Density of water, ρ₁ = 10³ kg/m³

Density of acid, ρ₂ = 1.7 × 10³ kg/m³

Height of the water column, h₁ = 4 m

Height of the acid column, h₂ = 4 m

Acceleration due to gravity, g = 9.8

Pressure due to water is given as:

Pressure due to acid is given as:

Pressure difference between the water and acid columns:

Hence, the force exerted on the door = ΔP × a

= 2.744 × 10⁴ × 20 × 10^–4

= 54.88 N

Therefore, the force necessary to keep the door closed is 54.88 N.

Question 10.22:

A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.

(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.

(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

Answer:

Answer: (a) 96 cm of Hg & 20 cm of Hg; 58 cm of Hg & –18 cm of Hg

(b) 19 cm

(a) For figure (a)

Atmospheric pressure, P0 = 76 cm of Hg

Difference between the levels of mercury in the two limbs gives gauge pressure

Hence, gauge pressure is 20 cm of Hg.

Absolute pressure = Atmospheric pressure + Gauge pressure

= 76 + 20 = 96 cm of Hg

For figure (b)

Difference between the levels of mercury in the two limbs = –18 cm

Hence, gauge pressure is –18 cm of Hg.

Absolute pressure = Atmospheric pressure + Gauge pressure

= 76 cm – 18 cm = 58 cm

(b) 13.6 cm of water is poured into the right limb of figure (b).

Relative density of mercury = 13.6

Hence, a column of 13.6 cm of water is equivalent to 1 cm of mercury.

Let h be the difference between the levels of mercury in the two limbs.

The pressure in the right limb is given as:

Atmospheric pressure + 1 cm of Hg

= 76 + 1 = 77 cm of Hg … (i)

The mercury column will rise in the left limb.

Hence, pressure in the left limb, 

Equating equations (i) and (ii), we get:

77 = 58 + h

∴h = 19 cm

Hence, the difference between the levels of mercury in the two limbs will be 19 cm.

Question 10.23:

Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?

Answer:

Answer: Yes

Two vessels having the same base area have identical force and equal pressure acting on their common base area. Since the shapes of the two vessels are different, the force exerted on the sides of the vessels has non-zero vertical components. When these vertical components are added, the total force on one vessel comes out to be greater than that on the other vessel. Hence, when these vessels are filled with water to the same height, they give different readings on a weighing scale.