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प्रश्न
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].
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उत्तर १
`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.
उत्तर २
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
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