A ballon filled with helium rises against gravity increasing its potential energy . The speed ofthee ballonalso increases as it rises . How do you reconcile this with the law of conservation ofmechanical energy ? You can neglect viscous drag of air and assume that density of air is constant .

A ballon filled with helium rises against gravity increasing its poten

1.	A ballon filled with helium rises against gravity increasing its potential energy . The speed ofthee ballonalso increases as it rises . How do you reconcile this with the law of conservation ofmechanical energy ? You can neglect viscous drag of air and assume that density of air is constant .
Answer» Solution :m = Mass of ballon V = Volumeof ballon `rho_("He") ` = Density of helium `rho_("air")` = Denity of air Volume V of ballon displaces volume V of air . So , `V(rho_(air)-rho_(He)) g = ma = m (dv)/(dt)""…(i)` Integrating with respect to t , ` rArr V(rho_("air") -rho_("He")) "gt" = mv ""...(ii)` `1/2 mv^(2) =1/2 m ((V^(2))/(m^(2))) (rho_(air)-rho_("He"))^(2) g^(2)t^(2)` `= 1/(2m) V^(2)(rho_(air)-rho_(He))^(2) g^(2)t^(2)` If the ballon rises to a HEIGHT h , from s = ut + `1/2 at^(2)` ` :. h = 1/2 "at"^(2)` `=1/2(V(rho_(air)-rho_(He)))/m "gt"^(2) ( :. U = 0 ) ....(iii)` From (iii) and (ii) `1/2 mv^(2) = [V (rho_(a)-rho_(He))"gt"^(2)][1/(2m)V(rho_(air)-rho_(He))"gt"^(2)]` ` =V(rho_(a)-rho_(He))gh` Therefore , `1/2 mv^(2) +V_(rho_(He))=gh =V_("pair") hg ` `KE_("ballon")+PE_("ballon") ` = Change in PE of air . So, as the ballon goes up , an equalvolume of air comes down , INCREASE in PE and KE of the ballon is EQUAL to decrease inPE of air .