(a) According to the law of atmospheres density of air decreases with increase in height y as $$ρ=\rho _{0}e^{\dfrac{-y}{y_{0}}}$$. Where ρ0 =1.25 kg m-3 is the density of air at sea level and yis a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.

(b) A zeppelin of volume 1500 m3 is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y0 = 8000 m and ρHe =0.18 kg m-3].

Asked by Pragya Singh | 1 year ago |  131

##### Solution :-

(a) We know that rate of decrease of density ρ of air is directly proportional  to the height y.

i.e., $$\dfrac{ d \rho }{ d y } = \;- \dfrac{ \rho }{ y_{ 0 }}$$. . . . . . . . . . . . . . ( 1 )

Where y is the constant of proportionality and the –ve sign indicates the decrease in density with increase in height.

Integrating equation (1) , we get :

$$\int_{\rho _{0}}^{\rho }\dfrac{d}{\rho } =\; – \int_{0}^{y } \dfrac{1}{y_{0}}dy​$$

$$[ log \rho ]_{\rho _{0}}^{\rho } = -[\dfrac{y}{y_{0}}]_{0}^{y}$$

Where ρ0 = density of air at sea level ie y =0

Or, log$$\dfrac{ρ_0}{ρ}=\dfrac{-y}{y_0}$$

Therefore, $$\rho =\rho _{0}\;e^{\dfrac{y}{y_{0}}}$$

(b).  Given:

Density ρ = $$\dfrac{ Mass }{ Volume }​$$

$$= \dfrac{ Mass \;of \;payload + Mass \;of \;helium }{ Volume }$$

Volume of zeppelin  = 1500 m3

Mass of payload, m = 400 kg

y0 = 8000 m

ρ0 =1.25 kg m-3

density of helium, ρ He =0.18 kg m-3

= $$\dfrac{ 400 + 1500 \times 0.18 }{ 1500 }$$​  [ Mass = volume × density]

=  0.45 kg m-3

Using equation (1), we will get:

$$\rho =\rho _{0}e^{\frac{y}{y_{0}}}​$$

$$\\\Rightarrow log_{e}(\dfrac{\rho _{0}}{\rho })= \dfrac{y_{0}}{y}$$

$$\\y= \dfrac{y_{0}}{log_{e}(\dfrac{\rho _{0}}{\rho })}\\$$

$$\\\Rightarrow y= \dfrac{8000}{log_{e}(\dfrac{1.25}{0.45 })}$$

Therefore,  y ≈ 8km

Answered by Pragya Singh | 1 year ago

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