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 \[=m-m'\] |
\[=m\left[ \frac{1-(g-a)}{(g+a)} \right]=m\left[ \frac{(g+a)-(g-a)}{(g+a)} \right]\] |
\[=m\left[ \frac{g+a-g+a}{g+a} \right]\Rightarrow \Delta m=\frac{2ma}{g+a}\] |
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