Question

A $2kg$ weather balloon is released and begins to rise against $6.5\,N$ of viscous drag. If its buoyancy is $32\,N$ , what is its acceleration?
A. $1\,m.{s^{ - 2}}$
B. $3.0\,m.{s^{ - 2}}$
C. $2.0\,m.{s^{ - 2}}$
D. $4.0\,m.{s^{ - 2}}$

Answer 1

Hint: In order to answer this question, to find the acceleration, first we will rewrite the given facts and then we will first find the weight of the balloon (as mass is given).Then we will calculate the net upward force to apply the formula of force, to finally calculate the acceleration.

Formula-used:
To find the weight of a balloon(mass of balloon is given), we will use $\text{weight of balloon} = mg$ and then for finding the acceleration, we will apply $F = ma$ .

Complete step by step answer:
Given that: Mass of balloon, $m = 2kg$, Viscous drag $ = 6.5N$ and Buoyancy Force $ = 32N$. Differences in pressure occurring on opposite sides of an item immersed in a static fluid cause buoyancy. (Due to the fact that pressure rises with depth). (Because force is perpendicular to the surface). The net force owing to the fluid is upward in direction.

And, as we know the gravitational acceleration, $g = 9.8\,m.{s^{ - 2}}$.Now,
$\text{Weight of balloon} = mg \\
\Rightarrow \text{Weight of balloon} = 2 \times 9.8 = 19.6N \\ $
Since, balloon is rising, the viscous drag is another downward force
$19.6 + 6.5 = 26.1N$ ……..(i)
And, the net upward force $ = 32 - 26.1 = 5.9N$ ………(ii)
Now, using $F = ma$ -
Acceleration, $a = \dfrac{F}{m} = \dfrac{{5.9}}{2} = 2.95\,m.{s^{ - 2}}$
Therefore, as we can see that, in option(B) $3.0\,m.{s^{ - 2}}$ is approximately equal to the value we found, so acceleration is approximately equal to $3.0\,m.{s^{ - 2}}$ .

Hence, the correct option is B.

Note: Buoyancy is the most prevalent driving factor of convection currents, and it also applies to fluid mixes. The mathematical modelling is changed to apply to continuums in these circumstances, but the fundamentals remain the same. The spontaneous separation of air and water, or oil and water, are examples of buoyancy-driven flows. The center of buoyancy of an object is the center of gravity of the displaced volume of fluid.