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' \[=m\left[ \frac{1-(g-a)}{(g+a)} \right]=m\left[ \frac{(g+a)-(g-a)}{(g+a)} \right]\] \[=m\left[ \frac{g+a-g+a}{g+a} \right]\Rightarrow \Delta m=\frac{2ma}{g+a}\]You need to login to perform this action.
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