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A balloon with mass mm is descending down with an acceleration aa (where a<ga < g). How much mass should be removed from it so that it starts moving up with an acceleration aa?

A

2mag+a\frac{2ma}{g+a}

B

2mag−a\frac{2ma}{g-a}

C

mag+a\frac{ma}{g+a}

D

mag−a\frac{ma}{g-a}

Step-by-Step Solution

  1. Case 1 (Descending):
  • Let the upthrust (buoyant force) be BB.
  • Forces acting on the balloon: Weight mgmg (downwards) and Upthrust BB (upwards).
  • Since it is accelerating downwards with aa, the net force equation is: mg−B=ma  ⟹  B=m(g−a)mg - B = ma \implies B = m(g - a) ... (i) (Reference: NCERT Class 11, Physics Part I, Chapter 5, Newton's Second Law).
  1. Case 2 (Ascending):
  • Let mass Δm\Delta m be removed. The new mass is (m−Δm)(m - \Delta m).
  • The upthrust BB remains the same (assuming volume doesn't change significantly).
  • The balloon accelerates upwards with aa. The net force equation is: B−(m−Δm)g=(m−Δm)aB - (m - \Delta m)g = (m - \Delta m)a
  1. Calculation:
  • Substitute BB from equation (i) into the second equation: m(g−a)−(m−Δm)g=(m−Δm)am(g - a) - (m - \Delta m)g = (m - \Delta m)a mg−ma−mg+Δmg=ma−Δmamg - ma - mg + \Delta mg = ma - \Delta ma −ma+Δmg=ma−Δma-ma + \Delta mg = ma - \Delta ma Δm(g+a)=2ma\Delta m(g + a) = 2ma Δm=2mag+a\Delta m = \frac{2ma}{g + a}
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