Question

3. A balloon of volume 120 m3 is filled with hot air, of density 0.38 kgm . If the fabric of balloon
weighs 12 kg, such that an additional equipment of wt. x. is attached to it, calculate the magnitude
of x. Density of cold air is 1.30 kgm-3.
[98.4 kgf]​

Answer 1

Step-by-step explanation:

Given  A balloon of volume 120 m3 is filled with hot air, of density 0.38 kgm . If the fabric of balloon weighs 12 kg, such that an additional equipment of wt. x. is attached to it, calculate the magnitude

• Volume of balloon = V = 120 cu m
• Density of hot air = ρ = 0.38 kg / m^3
• Mass of the balloon = 12 kg
• Additional equipment of weight attached to balloon is x
• Density of cold air = ρ = 1.30 kg/m^3
• Weight of hot air = volume of balloon x density of hot air x g
•                              = 120 x 0.38 x g
•                               = 45.6 kgf
• Downward thrust = weight of fabric balloon + weight of hot air inside balloon + weight of the equipment attached
•                                 = 12 + 45.6 + x
•                                = 57.6 + x
• So Upthrust = Weight of cold air displaced by balloon.
•                     = Volume x ρ cold air x g
•                    = 120 x 1.3 x g
•                     = 156 kgf
• Now by the law of floation  
•                   Downward thrust = Upthrust
•                          57.6 + x = 156
•                           Or x = 156 – 57.6
•                         Or x = 98.4 kgf

Reference link will be

https://brainly.in/question/8281885

Answer 2

Answer:

X=98.4 Kgf is the final answer