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physics ncert solutions class 11th Physics Class 11th Mechanical Properties of Fluids

10.31 (a) It is known that density $ρ$ of air decreases with height y as

where $ρ = ρ_{o} e^{\frac{- y}{y_{o}}}$ = 1.25 kg m^–3 is the density at sea level, and $ρ_{o}$ 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 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 $y_{o}$ = 8000 m and $y_{o}$ = 0.18 kg m–3].

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Vishal Baghel | Contributor-Level 10

5 months ago

Volume of the balloon, V = 1425 $ρ_{H e}$

Mass of the payload, m = 400 kg

Acceleration due to gravity, g = 9.8 m/ $m^{3}$

$s^{2}$ = 8000 m

$y_{o}$ = 0.18 kg m^–3

$ρ_{H e}$ = 1.25 kg m^–3

Density of the balloon = $ρ_{o}$

Height to which the balloon will rise = y

Density of air decreases with height and the relationship is given by:

$ρ$ = $ρ = ρ_{o} e^{\frac{- y}{y_{o}}}$ ……(i)

Differentiating equation (i), we get

$\frac{ρ}{ρ_{o}}$ $e^{\frac{- y}{y_{o}}}$

$- \frac{d ρ}{d y}$ , where k is the constant of proportionality

$α ρ$ , height changes from 0 to y, while density changes from $\frac{d ρ}{d y} = - k ρ$ to $\frac{d ρ}{ρ} = - k d y$ . Integrating both sides between the limits, we get:

$ρ_{o}$

$ρ$ = -ky

$\int_{ρ_{o}}^{ρ} \frac{d ρ}{ρ} = - \int_{0}^{y} k d y$ = ${[\log_{e} ? ρ]}_{ρ_{o}}^{ρ}$ ….(ii)

From equation (i) and

...more

Volume of the balloon, V = 1425 $ρ_{H e}$

Mass of the payload, m = 400 kg

Acceleration due to gravity, g = 9.8 m/ $m^{3}$

$s^{2}$ = 8000 m

$y_{o}$ = 0.18 kg m^–3

$ρ_{H e}$ = 1.25 kg m^–3

Density of the balloon = $ρ_{o}$

Height to which the balloon will rise = y

Density of air decreases with height and the relationship is given by:

$ρ$ = $ρ = ρ_{o} e^{\frac{- y}{y_{o}}}$ ……(i)

Differentiating equation (i), we get

$\frac{ρ}{ρ_{o}}$ $e^{\frac{- y}{y_{o}}}$

$- \frac{d ρ}{d y}$ , where k is the constant of proportionality

$α ρ$ , height changes from 0 to y, while density changes from $\frac{d ρ}{d y} = - k ρ$ to $\frac{d ρ}{ρ} = - k d y$ . Integrating both sides between the limits, we get:

$ρ_{o}$

$ρ$ = -ky

$\int_{ρ_{o}}^{ρ} \frac{d ρ}{ρ} = - \int_{0}^{y} k d y$ = ${[\log_{e} ? ρ]}_{ρ_{o}}^{ρ}$ ….(ii)

From equation (i) and (ii), we get

$\frac{ρ}{ρ_{o}}$ = $e^{- k y}$ , k = $y_{0}$

Substituting the value of k in equation (ii), we get

$\frac{1}{k}$

Density $\frac{1}{y_{o}}$ = $ρ = ρ_{o} e^{\frac{- y}{y_{o}}} \dots \dots (i i i)$

= $ρ = \frac{M a s s}{V o l u m e} = \frac{M a s s o f t h e p a y l o a d + M a s s o f h e l i u m}{V o l u m e}$ = 0.461 kg/ $\frac{m + V ρ_{H e}}{V}$

From equation (iii), taking log on both sides

$\frac{400 + 1425 \times 0.18}{1425}$ = - $m^{3}$

y = -8000 $\log_{e} ? \frac{ρ}{ρ_{o}}$ = 8000 m = 8 km

The balloon will rise to 8 km

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