NEET Solved Paper 2014 Question 6
Question: 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]
Options:
A) $ \frac{2ma}{g+a} $
B) $ \frac{2ma}{g-a} $
C) $ \frac{ma}{g+a} $
D) $ \frac{ma}{g-a} $
Show Answer
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} $
