InterviewSolution
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InterviewSolution
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(a) It is known that density `rho` of air decrease with height y(in metres ) as `rho =rho_(0)e^(-y//y_(0))` where `rho_(0)=1.25kgm^(-3)` is the density at see level , and `y_(0)` is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmospere remians 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 a payload of `400g`. Assume that the balloon maintains constant radius as it rises. How high does it rise? [take `y_(0)=800m` and `rho_(He) = 0.18kg//m^(3)`]. |
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Answer» We know that the rate of decrease of density `rho` of air with height y is directly proportional to density `rho` i.e., `-(d rho)/(dy) prop rho or (dp)/(dy) =-krho` where `K` is a constant of proportional . Here -ve sign shows that `rho` decreases as y increase. `:. (d rho)/(rho)=-K dy` Intergrating it within constions , as `y` changes from `0` to `y` , density change from `rho_(0)` to `rho`, we have `int_(rho_0)^(rho) (d rho)/(rho) = -int_(0)^(y) K dy` `[ log_(e)rho]_(rho_0)^(rho) = -Ky` or `log_(e)rho-log_(e)rho_(0)=-Ky` or `log(rho)/(rho_0) = -Ky` or `(rho)/(rho_0) = e^(-Ky)` or `rho=rho_(0)e^(-Ky)` Here `K` is a constant . suppose `y_(0)` is a constant such that `K=1//y_(0)`, then `rho=rho_(0)e^(-y//y_(0))`. (b) The balloon will rise to a height, where its density becomes equal to the air at that height. Density of balloon, `rho = (mass)/(volume) = ("pay load" + "mass of He")/("volume") = ((400+1425xx0.18)kg)/(1425m^(3)) = (656.5)/(1425) kg//m^(3)` As, `rho = rho_(0)e^(-y//y_(0)) :. (656.5)/(1425) = 1.25e^(-y//8000)` or,`e^(-y//8000) = (656.5)/(1425xx1.25) or e^(y//8000) = (1425xx1.25)/(656.5) = 2.7132` Taking log on both the sides `(y)/(8000) = log_(e)2.7132=2.3026 log_(10)2.7132 = 2.3026xx0.4335 ~~1` `y=8000xx1 = 8000m = 8km`. |
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