Mechanical Properties of Fluids NCERT Solutions Class 11 Physics - Solved Exercise Question 10.31

Question 10.31:
(a) It is known that density ρ of air decreases with height y as ρ0e-y/y0
Where ρ0 = 1.25 kg m-3
is the density at sea level, and y0 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 gremains constant.
(b) A large He balloon of volume 1425 m3 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 y0= 8000 m and ρHe = 0.18 kg m-3 ]

(a) Volume of the balloon, V = 1425 m3
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s2
y0 = 8000 m
ρHe = 0.18 kg m-3
ρ0 = 1.25 kg m-3
Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as:
ρ = ρ0e-y/y0
ρ / ρ0 = e-y/y0     ....(i)

This density variation is called the law of atmospherics.
It can be inferred from equation (i) that the rate of decrease of density with height is directly proportional to ρ, i.e.,
-(dρ / dy) ρ
(dρ / dy) = -kρ
(dρ / ρ) = -k dy
Where, k is the constant of proportionality
Height changes from 0 to y, while density changes from ρ0 to ρ
Integrating the sides between these limits, we get:
Comparing equations (i) and (ii), we get:
y0 = 1/k
k = 1/y0   ....(iii)
From equations (i) and (iii), we get
ρ = ρ0e-y/y0
(b) Density ρ = Mass / Volume
= (Mass of the payload + Mass of helium) / Volume
( m + VρHe) / V
= (400 + 1425 X 0.18) / 1425
= 0.46 kg m-3
From equations (ii) and (iii), we can obtain y as:
ρ = ρ0e-y/y0
loge(ρ / ρ0)  =  -y / y0
∴ y = - 8000 X loge(0.46 / 1.25)
= -8000 X (-1)
= 8000 m  =  8 km
Hence, the balloon will rise to a height of 8 km.

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