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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.
Solution :
Step 1: Identifying Physical Principles and Constraints
To derive the atmospheric pressure and density distribution under isothermal conditions, we consider a small element of air at a height $y$ with thickness $dy$ and cross-sectional area $A$.
- The weight of this air element acts downwards: $dW = (dm)g = (\rho A dy)g$ [Per Newton’s Law of Gravitation].
- The pressure difference $dP$ between the top and bottom of the element must support its weight for hydrostatic equilibrium: $dP = -(\rho g) dy$ [Per the Hydrostatic Law for fluids in equilibrium].
Step 2: Incorporating the Equation of State
For an ideal gas at constant temperature $T$, the pressure $P$ and density $\rho$ are related by the ideal gas law:
$PV = nRT \implies P = \left(\frac{n}{V}\right)RT \implies P = \frac{\rho}{M}RT$
Where $M$ is the molar mass of air and $R$ is the universal gas constant. Thus, we express density as:
$\rho = \frac{PM}{RT}$
Step 3: Setting up the Differential Equation
Substitute the expression for $\rho$ from Step 2 into the hydrostatic equilibrium equation from Step 1:
$dP = -\left(\frac{PM}{RT}\right)g dy$
Rearranging the variables to isolate $P$ and $y$:
$\frac{dP}{P} = -\left(\frac{Mg}{RT}\right) dy$
Step 4: Integration
Integrate both sides with boundary conditions: at $y = 0$, $P = P_0$, and at height $y$, pressure is $P$.
$\int_{P_0}^{P} \frac{dP}{P} = -\int_{0}^{y} \left(\frac{Mg}{RT}\right) dy$
$\ln\left(\frac{P}{P_0}\right) = -\left(\frac{Mg}{RT}\right)y$
$P = P_0 e^{-(Mg/RT)y}$
Step 5: Deriving the Density Law
Since $\rho = \frac{PM}{RT}$, the density at any height $y$ is:
$\rho = \frac{M}{RT} \left[ P_0 e^{-(Mg/RT)y} \right]$
Given that at sea level ($y=0$), the density $\rho_0 = \frac{P_0 M}{RT}$, we substitute this into the equation:
$\rho = \rho_0 e^{-(Mg/RT)y}$
By defining the constant $y_0 = \frac{RT}{Mg}$, the expression becomes the requested form:
$\rho = \rho_0 e^{-y/y_0}$
Visual Representation of Density Decay
Final Solution: The law of atmospheres is obtained by solving the hydrostatic equation $dP = -\rho g dy$ coupled with the ideal gas law $\rho = PM/RT$, yielding $\rho = \rho_0 e^{-y/y_0}$ where $y_0 = \frac{RT}{Mg}$.
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.
- Q10.22(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).
- 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(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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