question_answer

A balloon with mass m is descending down with an acceleration a (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration a? [AIPMT 2014]

A)  \[\frac{2ma}{g+a}\]

B)  \[\frac{2ma}{g-a}\]

C)  \[\frac{ma}{g+a}\]

D)  \[\frac{ma}{g-a}\]

Correct Answer: A

Solution :

When the balloon is descending down with acceleration a So,        \[mg-B=m\times a\,\] ?(i) [B ® 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' \[=m\left[ \frac{1-(g-a)}{(g+a)} \right]=m\left[ \frac{(g+a)-(g-a)}{(g+a)} \right]\] \[=m\left[ \frac{g+a-g+a}{g+a} \right]\Rightarrow \Delta m=\frac{2ma}{g+a}\]