AP EAMCET Engineering 18 Sep 2020 Shift 2 Paper

The given situation is shown in the figure In above figures, M is the mass of balloon and M′ is the mass remove from the balloon.
According to Newton's law of motion in first case,
According to Newton's second law of motion, for second case,
F−(M−M′)g=(M−M′)a
⇒‌‌F−Mg+M′g=Ma−M′a
[ from Eq. (i)]
⇒‌‌M′g+M′a=Ma+Ma
⇒‌‌M′(g+a)=2Ma
⇒‌‌M′=‌

2Ma

g+a