Study Notes

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 $v$? 

(A) $2\left\{ {M + \frac{B}{g}} \right\}$
(B) $2\left\{ {M – \frac{B}{g}} \right\}$
(C) $\left\{ {M + \frac{B}{g}} \right\}$
(D) $\left\{ {M – \frac{B}{g}} \right\}$

The correct answer is B. $2\left\{ {M – \frac{B}{g}} \right\}$

Solution:

Let, the air resistance force is $F$ and Buoyant force is $B$. Given that

$${F}_{a} \propto {v},$$ where ${v} \rightarrow$ velocity

$$\Rightarrow F_{a}=k v,$$ where $k \rightarrow$ proportionality constant When the balloon is moving downward,

$${B}+{kv}={Mg} \quad …(i)$$

$$\Rightarrow M=\frac{B+k v}{g}$$

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

Here,

$${B}-({mg}+{kv})=0$$

$$\Rightarrow {B}={mg}+{kv}$$

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

So, amount of mass that should be removed, 

$$M-m=\frac{B+k v}{g}-\frac{B-k v}{g}$$$$M-m=\frac{B+k v-B+k v}{g}$$$$M-m=\frac{2 k v}{g}$$From equation $(i)$ we get, $$M-m=\frac{2(M g-B)}{g}$$$$M – m = 2\left\{ {M – \frac{B}{g}} \right\}$$

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