Search any question & find its solution
Question:
Answered & Verified by Expert
(a) It is known that density $\rho$ of air decreases with heighty as $\rho=\rho_0 \mathrm{e}^{-\mathrm{y} / \mathrm{y}_0}$
Where $\rho_0=1.25 \mathrm{~kg} \mathrm{~m}^{-3}$ 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.
(b) A large He balloon of volume $1425 \mathrm{~m}^3$ is used to lift a payload of $400 \mathrm{~kg}$. Assume that the balloon maintains constant radius as it rises. How high does it rise? [Take $y_0=8000 \mathrm{~m}$ and $\rho_{H e}=0.18 \mathrm{~kg} \mathrm{~m}^{-3}$ ]
Where $\rho_0=1.25 \mathrm{~kg} \mathrm{~m}^{-3}$ 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.
(b) A large He balloon of volume $1425 \mathrm{~m}^3$ is used to lift a payload of $400 \mathrm{~kg}$. Assume that the balloon maintains constant radius as it rises. How high does it rise? [Take $y_0=8000 \mathrm{~m}$ and $\rho_{H e}=0.18 \mathrm{~kg} \mathrm{~m}^{-3}$ ]
Solution:
2663 Upvotes
Verified Answer
(a) We know that rate of decrease of density $\rho$ of air is directly proportional to the height $y$. It is given as $\frac{\mathrm{d} \rho}{\mathrm{dy}}=-\frac{\rho}{\mathrm{y}_0}$,
where $y$ is a constant of proportionality and $-v e$ sign signfies that density is decreasing with increase in height. on integration, we get $\int_{\rho_0}^\rho \frac{d \rho}{\rho}=-\int_0^y \frac{1}{y_0} d y$ $\Rightarrow[\log \rho]_{\rho_0}^\rho=-\left[\frac{\mathrm{y}}{\mathrm{y}_0}\right]_0^{\mathrm{y}}$, where, $\rho_0=$ density of air at sea level i.e., $y=0$
or $\log _e \frac{\rho}{\rho_0}=\frac{y}{y_0}$ or $\rho=\rho_0 e^{-\frac{y}{y_0}}$
Hence dimensions and units of constant $y_0$ are same as of $\mathrm{y}$.
(b) Here volume of $H_{e^{\prime}}$ ballon, $V=1425 \mathrm{~m}^3$, mass of payload, $m=400 \mathrm{~kg}$
$y_0=8000 \mathrm{~m}$, density of $\mathrm{He} \rho_{\mathrm{He}}=0.18 \mathrm{kgm}^{-3}$
Mean density of balloon,
$\rho=\frac{\text { Total mass of balloon }}{\text { Volume }}=\frac{\mathrm{m}+\mathrm{V} \cdot \rho_{\mu \mathrm{e}}}{\mathrm{V}}$
$=\frac{400+1425 \times 0.18}{1425}=0.4608=0.46 \mathrm{kgm}^{-3}$
As density of air at sea level $\rho_0=1.25 \mathrm{~kg} \mathrm{~m}^{-3}$. The balloon will rise up to a height $y$ where density of air $=$ density of balloon $\rho=0.46 \mathrm{kgm}^{-3}$
$$
\begin{aligned}
\text { As } & \rho=\rho_0 \mathrm{e}^{-\frac{\mathrm{y}}{y_0}} \text { or } \frac{\rho_0}{\rho}=\mathrm{e}^{\frac{y_0}{\mathrm{y}}} \\
\therefore \log _{\mathrm{e}}\left(\frac{\rho_0}{\rho}\right) &=\frac{\mathrm{y}_0}{\mathrm{y}} \text { or } \mathrm{y}=\frac{\mathrm{y}_0}{\log _{\mathrm{e}}\left(\frac{\rho_0}{\rho}\right)}=\frac{8000}{\log _{\mathrm{e}}\left(\frac{1.25}{0.46}\right)} \\
=& 8002 \mathrm{~m} \text { or } 8.0 \mathrm{~km} .
\end{aligned}
$$
where $y$ is a constant of proportionality and $-v e$ sign signfies that density is decreasing with increase in height. on integration, we get $\int_{\rho_0}^\rho \frac{d \rho}{\rho}=-\int_0^y \frac{1}{y_0} d y$ $\Rightarrow[\log \rho]_{\rho_0}^\rho=-\left[\frac{\mathrm{y}}{\mathrm{y}_0}\right]_0^{\mathrm{y}}$, where, $\rho_0=$ density of air at sea level i.e., $y=0$
or $\log _e \frac{\rho}{\rho_0}=\frac{y}{y_0}$ or $\rho=\rho_0 e^{-\frac{y}{y_0}}$
Hence dimensions and units of constant $y_0$ are same as of $\mathrm{y}$.
(b) Here volume of $H_{e^{\prime}}$ ballon, $V=1425 \mathrm{~m}^3$, mass of payload, $m=400 \mathrm{~kg}$
$y_0=8000 \mathrm{~m}$, density of $\mathrm{He} \rho_{\mathrm{He}}=0.18 \mathrm{kgm}^{-3}$
Mean density of balloon,
$\rho=\frac{\text { Total mass of balloon }}{\text { Volume }}=\frac{\mathrm{m}+\mathrm{V} \cdot \rho_{\mu \mathrm{e}}}{\mathrm{V}}$
$=\frac{400+1425 \times 0.18}{1425}=0.4608=0.46 \mathrm{kgm}^{-3}$
As density of air at sea level $\rho_0=1.25 \mathrm{~kg} \mathrm{~m}^{-3}$. The balloon will rise up to a height $y$ where density of air $=$ density of balloon $\rho=0.46 \mathrm{kgm}^{-3}$
$$
\begin{aligned}
\text { As } & \rho=\rho_0 \mathrm{e}^{-\frac{\mathrm{y}}{y_0}} \text { or } \frac{\rho_0}{\rho}=\mathrm{e}^{\frac{y_0}{\mathrm{y}}} \\
\therefore \log _{\mathrm{e}}\left(\frac{\rho_0}{\rho}\right) &=\frac{\mathrm{y}_0}{\mathrm{y}} \text { or } \mathrm{y}=\frac{\mathrm{y}_0}{\log _{\mathrm{e}}\left(\frac{\rho_0}{\rho}\right)}=\frac{8000}{\log _{\mathrm{e}}\left(\frac{1.25}{0.46}\right)} \\
=& 8002 \mathrm{~m} \text { or } 8.0 \mathrm{~km} .
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.