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Solution - 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 - Pressure - Concept of Pressure

Question

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 y0= 8000 m and `rho_"He"`= 0.18 kg m–3].

Solution 1

`Density rho = "Mass"/"volume"`

`= "Mass of the payload + Mass of helium"/"Volume"` 

`= (m+Vrho_"He")/V`

`= (400 + 1425 xx 0.18)/1425`

`= 0.46 "kg/m"^3`

From equaiton ii and iii we can obtain y as

`rho = rho_0^(e^(-y/y_0))`

`log_e = rho/rho_0 = -y/y_0`

`:. y = -8000xx log_e 0.46/1.25`

`= -8000xx -1`

= 8000 m = 8 km

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

Solution 2

Here volume of He balloon,V= `1425 m^3`, mass of payload, m = 400 kg

y_0 = 8000 m, density of He `rho_He` = `0.18 kg^(-3)`

Mean density of ballon, `rho = "Total mass of ballon"/"Volume" = (m + V.rho_"He")/V Pa`

= `(400 + 1425 xx 0.18)/1425 = 0.4608 = 0.46  kgm^(-3)`

As density of air at sea level `rho_0 = 1.25 kg m^(-3)`. The balloon will rise up to a height y = where density of air = density of ballon `rho = 0.46 kgm^(-3)` 

As `rho = rho_0^(e^(-y/y_0))` or `rho_0/rho = e^(y_0/y)`

`:. log_e (rho_0/rho) = y_0/y or y = y_0/log_e(rho_0/rho) = 8000/log_e(1.25/0.46)`

= 8002 m or 8.0 km

Is there an error in this question or solution?

APPEARS IN

NCERT Physics Textbook for Class 11 Part 2
Chapter 10: Mechanical Properties of Fluids
Q: 31.2 | Page no. 271

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Reference Material

Solution for question: 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 concept: Pressure - Concept of Pressure. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science)
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