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?   [NEET 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 \[=m-m'\]

\[=m\left[ \frac{1-(g-a)}{(g+a)} \right]=m\left[ \frac{(g+a)-(g-a)}{(g+a)} \right]\]

\[=m\left[ \frac{g+a-g+a}{g+a} \right]\Rightarrow \Delta m=\frac{2ma}{g+a}\]