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?

Answer 1

Given :

Let M be the 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)

M=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

⇒M’=B+kv /g

∴ Amount of mass that should be removed = M − M’.

∆M=B+kv/g-B-kv/g      

=(B+kv-B+kv/)g    

=2kvg

=2(Mg-B)/g      

=2(M-B)/g

Answer 2

Answer:

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