Question:
A balloon of mass 'M' is rising up with an acceleration 'a'.
In order to triple its acceleration, the fraction of mass to be removed from the balloon is
(\( g \) - acceleration due to gravity)
A balloon of mass 'M' is rising up with an acceleration 'a'.
In order to triple its acceleration, the fraction of mass to be removed from the balloon is
(\( g \) - acceleration due to gravity)
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Use net force equations and substitute new conditions to find change in mass or acceleration.
Updated On: Jun 3, 2025
- \( \dfrac{a}{g+a} \)
- \( \dfrac{a}{g+2a} \)
- \( \dfrac{2a}{g+3a} \)
- \( \dfrac{2a}{g+2a} \)
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The Correct Option is C
Solution and Explanation
Initial: \( T - Mg = Ma \Rightarrow T = M(g+a) \)
New acceleration = 3a, new mass = \( M' \)
\( T = M'(g + 3a) \Rightarrow M(g+a) = M'(g + 3a) \Rightarrow \dfrac{M'}{M} = \dfrac{g+a}{g+3a} \)
Fraction removed = \( 1 - \dfrac{g+a}{g+3a} = \dfrac{(g+3a)-(g+a)}{g+3a} = \dfrac{2a}{g+3a} \)
New acceleration = 3a, new mass = \( M' \)
\( T = M'(g + 3a) \Rightarrow M(g+a) = M'(g + 3a) \Rightarrow \dfrac{M'}{M} = \dfrac{g+a}{g+3a} \)
Fraction removed = \( 1 - \dfrac{g+a}{g+3a} = \dfrac{(g+3a)-(g+a)}{g+3a} = \dfrac{2a}{g+3a} \)
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