Answer:

Correct Answer: A

Solution:

When the balloon is descending down with acceleration a So,

$ mg-B=m\times a, $ ..(i) [B $\rightarrow$ Buoyant force] Here,

we should assume that while removing same mass the volume of balloon and hence buoyant force will not change.

Let the new mass of the balloon is m'

$ \Rightarrow $ So, mass removed $ (m-m’) $

$ \Rightarrow $ So, $ B=m’g=m’\times a, $ ..(ii)

$ \Rightarrow $ Solving Eqs. (i) and (ii), $ mg-B=m\times a $ $ B-m’g=m’\times a $ $ mg-m’g=ma+m’a $ $ (mg-ma)=m’(g+a)=m(g-a)=m’(g+a) $ $ m’=\frac{m(g-a)}{g+a} $

$ \Rightarrow $ So mass removed = m - m’ $ =m[ \frac{1-(g-a)}{(g+a)} ]=m[ \frac{(g+a)-(g-a)}{(g+a)} ] $ $ =m[ \frac{g+a-g+a}{g+a} ]\Rightarrow \Delta m=\frac{2ma}{g+a} $