Correct Answer - a
Let `F` be the upthrust on balloon due to air
Case (i) When balloon is descending down with
acceleration a then as shown in
`mg - F = ma`
Case (ii) Let `m_(0)` be the mass removed from the ballon then balloon moves upwards with acceleration
Now `F - (m - m_(0)) g = (m - m_(0))` a ..(ii)
Adding (i) and (ii) we get
`mg - (m - m_(0)) g = ma + (m -m_(0))a`
Solving it we get `m_(0) = (2ma)/(g + a)`
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