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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 - CBSE (Science) Class 11 - Physics

#### 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 (with solutions)
Chapter 10: Mechanical Properties of Fluids
Q: 31.2 | Page no. 271

#### 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: null - Concept of Pressure. For the course CBSE (Science)
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