### Q.31:- (a)It is known that density *ρ* of air decreases with height y as ρ_{0}e^{-y/y0}

### Where ρ_{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 gremains constant.

(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 ρ_{He} = 0.18 kg m^{-3} ]

**Answer:-****
**(a) Volume of the balloon,

*V*= 1425 m

^{3}

Mass of the payload,

*m*= 400 kg

Acceleration due to gravity, g = 9.8 m/s

^{2}

y

_{0}= 8000 m

ρ

_{He}= 0.18 kg m

^{-3}

ρ

_{0}= 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 = ρ

_{0}e

^{-y/y0}

ρ / ρ

_{0 }= 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 ρ_{0} to ρ

Integrating the sides between these limits, we get:

**(i)**and

**(ii)**, we get:

y_{0} = 1/k

k = 1/y_{0} ….**(iii)**

From equations **(i)** and **(iii)**, we get

ρ = ρ_{0}e^{-y/y0}

(b) Density ρ = Mass / Volume

= (Mass of the payload + Mass of helium) / Volume

( m + Vρ_{He}) / V

= (400 + 1425 × 0.18) / 1425

= 0.46 kg m^{-3}

From equations **(ii)** and **(iii)**, we can obtain y as:

ρ = ρ_{0}e^{-y/y0}

log_{e}(ρ / ρ_{0}) = -y / y_{0}

∴ y = – 8000 × log_{e}(0.46 / 1.25)

= -8000 × (-1)

= 8000 m = 8 km

Hence, the balloon will rise to a height of 8 km.