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Question
The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall to the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?
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Solution
Let M be mass of the balloon.
Let the air resistance force on balloon be F .
Given that F ∝ v.
⇒ F = kv,
where k = proportionality constant.
When the balloon is moving downward with constant velocity,
B + kv = Mg ...(i)
\[\Rightarrow M = \frac{B + kv}{g}\]
Let the mass of the balloon be M' so that it can rise with a constant velocity v in the upward direction.
B = Mg + kv
\[\Rightarrow M' = \frac{B + kv}{g}\]
∴ Amount of mass that should be removed = M − M'.
\[∆ M = \frac{B + kv}{g} - \frac{B - kv}{g}\]
\[ = \frac{B + kv - B + kv}{g}\]
\[ = \frac{2kv}{g} = \frac{2\left( Mg - B \right)}{g}\]
\[ = 2\left\{ M - \frac{B}{g} \right\}\]
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