A balloon with its contents weighing 80N is moving down with acceleration of g2ms−2. The mass to be removed from it so that the balloon moves up with acceleration of g3ms−2 is

Mg−FB=Mg2 (1)FB−(M−ΔM)g=(M−ΔM)g3 (2)(1)+(2) ΔMg=Mg2+Mg3−ΔMg3ΔM4g3=5Mg6ΔM=5Mg6x34g=5M8=5(8)8=5kg

A balloon with its contents weighing 80N is moving down with acceleration of g2ms−2. The mass to be removed from it so that the balloon moves up with acceleration of g3ms−2 is

Mg−FB=Mg2 (1)FB−(M−ΔM)g=(M−ΔM)g3 (2)(1)+(2) ΔMg=Mg2+Mg3−ΔMg3ΔM4g3=5Mg6ΔM=5Mg6x34g=5M8=5(8)8=5kg

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A balloon with its contents weighing 80N is moving down with acceleration of $\frac{g}{2} {ms}^{- 2}$ . The mass to be removed from it so that the balloon moves up with acceleration of $\frac{g}{3} {ms}^{- 2}$ is

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3kg

6kg

5kg

2kg

answer is C.

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Detailed Solution

$\begin{array}{l} Mg - F_{B} = M \frac{g}{2} (1) \\ F_{B} - (M - ΔM) g = (M - ΔM) \frac{g}{3} (2) \end{array}$
$(1) + (2) ΔMg = \frac{Mg}{2} + \frac{Mg}{3} - \frac{ΔMg}{3}$
$\begin{array}{l} ΔM (\frac{4 g}{3}) = \frac{5 Mg}{6} \\ ΔM = \frac{5 Mg}{6} x \frac{3}{4 g} = \frac{5 M}{8} = \frac{5 (8)}{8} = 5 kg \end{array}$