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Q10.22(b):
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. (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).
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. (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).
Solution :
Step 1: Establishing Initial Parameters and Hydrostatic Equilibrium
From the problem statement and Figure 10.25 (b), let the initial pressure of the gas in the enclosure be $P_g$. In the right limb, the column of mercury is exposed to the atmosphere ($P_{atm} = 76 \text{ cm of Hg}$). In the left limb, the mercury column is connected to the gas enclosure. Let $h$ be the height difference between the mercury levels in the two limbs. According to the law of hydrostatics [Per the Law of Communicating Vessels], the pressure at a horizontal level in a continuous static fluid is equal.
The equilibrium condition for the manometer in state (b) is given by:
$P_g + h_{left} \cdot \rho_{Hg} \cdot g = P_{atm} + h_{right} \cdot \rho_{Hg} \cdot g$
$P_g = P_{atm} - h \cdot \rho_{Hg} \cdot g$ (where $h$ is the difference in mercury levels).
Step 2: Analysis of Adding Immiscible Water
We add $h_w = 13.6 \text{ cm}$ of water to the right limb. Since water is immiscible with mercury and $\rho_w < \rho_{Hg}$, the water floats on top of the mercury in the right limb. This increases the total pressure on the mercury surface in the right limb. The density of water is $\rho_w = 1 \text{ g/cm}^3$ and the density of mercury is $\rho_{Hg} = 13.6 \text{ g/cm}^3$.
The equivalent pressure head of water in terms of mercury is calculated as:
$h_{eq} = h_w \times \frac{\rho_w}{\rho_{Hg}}$
$h_{eq} = 13.6 \text{ cm} \times \frac{1 \text{ g/cm}^3}{13.6 \text{ g/cm}^3} = 1 \text{ cm of Hg}$
Step 3: Calculating the Shift in Mercury Levels
Let the new difference in mercury levels be $h'$. The new pressure equilibrium equation at the common interface level in the left limb becomes:
$P_g + h_1 \rho_{Hg} g = P_{atm} + h_2 \rho_{Hg} g + h_w \rho_w g$
Because the gas volume is assumed constant and the tube cross-section is uniform, if the mercury level in the right limb drops by $x$, the level in the left limb rises by $x$. The new difference $h'$ relates to the previous difference $h$ as $h' = h + 2x$. However, the hydrostatic balance dictates that the added 1 cm of mercury pressure equivalent must be distributed across the two columns.
When 13.6 cm of water is added, it exerts a pressure equivalent to 1 cm of Hg. This pressure pushes the mercury in the right limb down. Consequently, the mercury level in the left limb rises. By the principle of displacement in U-tubes:
- Pressure increment on right side = $1 \text{ cm Hg}$.
- Since the pressure must balance, the mercury column height difference $h'$ must adjust to compensate for the additional $1 \text{ cm}$ of head.
- The mercury level in the right limb will descend by $0.5 \text{ cm}$, and the mercury level in the left limb will rise by $0.5 \text{ cm}$.
Summary of Level Changes
| Limb | Direction of Change | Magnitude |
|---|---|---|
| Right Limb (Water side) | Downward | $0.5 \text{ cm}$ |
| Left Limb (Gas side) | Upward | $0.5 \text{ cm}$ |
Final Solution: Upon pouring 13.6 cm of water into the right limb, the mercury level in the right limb will fall by 0.5 cm, and the mercury level in the left limb will rise by 0.5 cm, resulting in a net increase of 1 cm in the mercury level difference between the two limbs.
More Questions from Class 11 Physics Mechanical Properties of Fluids Additional Exercises
- Q10.21: A tank with a square base of area $1.0 m^2$ is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area $20 cm^2$. 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.
- Q10.22(a): 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.
- Q10.23: 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 ?
- Q10.24: During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein ? [Use the density of whole blood from Table 10.1].
- Q10.25(a): In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of blood flow in an artery of diameter $2 \times 10^{-3} m$ if the flow must remain laminar ?
- Q10.25(b): In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (b) Do the dissipative forces become more important as the fluid velocity increases ? Discuss qualitatively.
- Q10.26(a): (a) What is the largest average velocity of blood flow in an artery of radius $2 \times 10^{-3}m$ if the flow must remain lanimar?
- Q10.26(b): (b) What is the corresponding flow rate ? (Take viscosity of blood to be $2.084 \times 10^{-3} Pa s$).
- Q10.27: A plane is in level flight at constant speed and each of its two wings has an area of $25 m^2$. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be $1 kg m^{-3}$).
- Q10.28: In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius $2.0 \times 10^{-5} m$ and density $1.2 \times 10^3 kg m^{-3}$. Take the viscosity of air at the temperature of the experiment to be $1.8 \times 10^{-5} Pa s$. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air.
- Q10.29: Mercury has an angle of contact equal to $140^{\circ}$ 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 $0.465 N m^{-1}$. Density of mercury = $13.6 \times 10^3 kg m^{-3}$.
- Q10.30: Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both 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 $7.3 \times 10^{-2} N m^{-1}$. Take the angle of contact to be zero and density of water to be $1.0 \times 10^3 kg m^{-3}$ ($g = 9.8 m s^{-2}$) .
- Q10.31(a): (a) It is known that density $\rho$ of air decreases with height $y$ as $\rho = \rho_0 e^{-y/y_0}$ where $\rho_0 = 1.25 kg m^{-3}$ is the density at sea level, and $y_0$ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of $g$ remains constant.
- Q10.31(b): (b) A large He balloon of volume $1425 m^3$ is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise ? [Take $y_0 = 8000$ m and $\rho_{He} = 0.18 kg m^{-3}$].
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