A balloon with mass M is descending down with an acceleration `a (a lt g)`. What mass of its contents must be removed so that it starts moving up with same acceleration `a`?

Options

A. `(2ma)/(g+a)`
B. `(2ma)/(g-a)`
C. `(ma)/(g+a)`
D. `(ma)/(g-a)`

Answer

A

Explanation

(a) When the balloon is descending down with acceleration a
So, `mg-B=mxxxa`…..(i) (B`rarr` Buoyant force)
Here, we should assume that while removing some mass the volume of balloon and hence buoyant force will not change.
Let the new mass of the balloon is m'.
So, mass removed =m-m'
So, `B-m'g=m'zza`....(ii)
On solving Eqs. (i) nad (ii), we get
` mg-B=mxxa`
`B-m'g=m'xxa`
`implies mg-m'g =ma+m'a`
`implies (mg-ma)=m'(g+a)`
` implies m(g-a)=m'(g+a)`
`therefore m'(m(g-a))/(g+a)`
So, mass removed, `Delta m=m-m'`
`=m[1-((g-a))/((g+a))]=m[((g+a)-(g-a))/(g+a)]`
`=m[(g+a-g+a)/(g+a)]=(2ma)/(g+a)`