• question_answer A balloon filled with $C{{O}_{2}}$ released on earth would (neglect viscosity of air) A)  climb with an acceleration 9.8$m/{{s}^{2}}$B)  fall with an acceleration 9.8 $m/{{s}^{2}}$C)  fall with a constant acceleration 3.4$m/{{s}^{2}}$  D)  fall with acceleration and then would attain a constant velocity

If B is up thrust of air on balloon, and a is downward acceleration, then             $Mg-B=Ma$ \begin{align} & \Rightarrow a=\frac{Mg-B}{M}=g-\frac{{{V}_{pair}}g}{{{V}_{Pco}}_{_{2}}} \\ & =\left( 1-\frac{{{V}_{pair}}}{{{V}_{Pc{{o}_{2}}}}} \right)g=\left( 1-\frac{28.8}{44} \right)\times 9.8m/{{s}^{2}} \\ & =3.4m/{{s}^{2}} \\ \end{align}