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 to the direction of velocity and is proportional to it. The balloon carries a mass M ans id found to fall down near the earth's surface with a constatnt velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v ?

Answer 1

Let the air resistane force is $F_a$ and Buoyant force is B.

Given that , $F_a=av$

where v = velocity

$\Rightarrow F_m=kv$

where k = proportionality constant

When the balloon is moving downward

B+ kv = Mg ...(i)

$\Rightarrow M=\frac{B+kv}{g}$

For the balloon to rise with a constant velocity v (upward), let the mass be m

Here, B - (mg +kv) = 0 ...(ii)

$\Rightarrow m=\frac{B+kv}{g}$

$\therefore$ Amount of mass that should be removed = M -m

$=\frac{B+kv}{g}-\frac{B-kv}{g}$

$=\frac{B+kv-B+kv}{g}$

= $\frac{2kv}{g}=\frac{2(Mg-B)}{g}$

$=2\{M-\frac{B}{g}\}$