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It is known that density rho of air decreases with heighty as -y//y_(0)rho=rho_(o)e where rho_(o)=1.25kgm^(-3) is the density at sea level and y_(o) is a constant . Thisdensity 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 1425m^(3) is used to lift apayload of 400 kg . Assume that the balloon maintains constant radius as it rises . How high does it rise ? (y_(o)=8000mandrho_(He)=0.18kgm^(-3)). |
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Answer» Solution :The rate of decrease of DENSITY `rho` of air is directly proportional to the height y. It is given as , `-(drho)/(dy)proprhoor-(drho)/(dy)=-krho` where K= CONSTANT and negative sign indicates that density decreases as height increases. From height 0 to y density becomes `rho_(o)` to `rho` . By integrating above equation, `int_(rho_(o))^(rho)(1)/(rho)=-kint_(o)^(y)dy` `[Inrho]_(rho_(o))^(rho)=-k[y]_(o)^(y)` `[In rho-Inrho_(o)]=-k[y=0]` ln `((rho)/(rho_(o)))=-ky` `log_(e)((rho)/(rho_(o)))=-ky` `therefore(rho)/(rho_(o))=e^(-ky)` `thereforerho=rho_(o)^(e-ky)` TAKING constant `k=(1)/(y_(o))` `rho=rho_(o)e^((y)/(y_(0)))` The balloon will rise upto a height where density of air equal to the density of balloon Volume of ballon `V=1425m^(3)` MASS of He has in balloon, `V=1425xx0.18=256.5kg` Total mass of balloon (with payload), `M=400+256.5` `=656.5kg` Density of balloon `rho=(M)/(V)=(656.5)/(1425)` `rho=0.46kg//m^(3)` `y_(o)=8000m` `rho_(o)=1.25kg//m^(3)` `rho=0.46kg//m^(3)` `rho=rho_(o)e^((y)/(y_(o)))` `therefore0.46=1.25_(e)^((-y)/(8000))` `thereforee^((y)/(8000))=(1.25)/(0.46)=2.7` `thereforey=8000loge^(2.7)` `=8000xx1.025` `=8200m` `=8.2km` |
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