A) \[\frac{2ma}{g+a}\]
B) \[\frac{2ma}{g-a}\]
C) \[\frac{ma}{g+a}\]
D) \[\frac{ma}{g-a}\]
Correct Answer: A
Solution :
[a] Let up thrust of air be\[~{{F}_{a}}\]then For downward motion of balloon \[{{F}_{a}}=mg-ma\] \[mg-{{F}_{a}}=ma\] For upward motion \[{{F}_{a}}-\left( m-\Delta m \right)g=\left( m-\Delta m \right)a\] Therefore \[\Delta m=\frac{2ma}{g+a}\]
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