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 and is found to fall
down near 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

Answer:

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Answer 2

Answer:

Explanation:

Let the force of air resistance = -kv (k= constant)

when balloon with mass M moves downward with velocity v :

Net force on balloon = 0 (As it moves with constant velocity)

⇒Mg=kv+B→(1)

Let mass m be removed from mass M of the balloon such that it moves upward with velocity v

Then, (M−m)g+kv=B→(2)

Adding equation 1 and 2 we get :

2Mg−mg=2B

⇒mg=2(Mg−B)

⇒m=  

g

2(Mg−B)

​  

 

So mass to be removed = 2(M−  

g

B

​  

)