**Answer the following questions:-**

**(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 ρ_{0} =1.25 kg m^{-3} is the density of air at sea level and y_{0 }is a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.

**(b)** A zeppelin of volume 1500 m^{3} 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?[ y_{0} = 8000 m and ρH_{e} =0.18 kg m^{-3}].

Asked by Pragya Singh | 1 year ago | 131

**(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 ρ_{0} = density of air at sea level ie y =0

Or, log_{e }\( \dfrac{ρ_0}{ρ}=\dfrac{-y}{y_0}\)

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 m^{3}

Mass of payload, m = 400 kg

y_{0} = 8000 m

ρ_{0} =1.25 kg m^{-3}

density of helium, ρ H_{e} =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

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Answe the following question:-

**(a) **What is the largest average velocity of blood flow in an artery of radius 2×10^{–3}m if the flow must remain lanimar?

**(b)** What is the corresponding flow rate ? (Take viscosity of blood to be 2.084 × 10^{–3} Pa s).