Animated Solution for Physics - Properties of Solids and Liquids: An air bubble of radius $1\text{ cm}$ in water has an upward acceleration $9.8{\text{ cm s}}^{-2}$. The density of water is $1{\text{ g/cm}}^3$ and water offers negligible drag force on the bubble. The mass of the bubble is (Take, $g=980{\text{ cm/s}}^2$)

## The Physics of a Rising Air Bubble

Have you ever watched an air bubble gracefully rise to the surface of a glass of water? It seems so simple, yet beneath that smooth motion lies a beautiful interplay of forces. In this problem, we are going to dive into the mechanics of a rising air bubble and calculate its mass using Newton's laws and Archimedes' principle.

Imagine our air bubble, with a radius of $1\text{ cm}$, submerged in water. It's not just floating; it's accelerating upwards at $9.8{\text{ cm/s}}^2$. To understand why, we need to look at the forces acting on it.

First, there is the buoyant force ($B$). According to Archimedes' principle, the water pushes the bubble upwards with a force equal to the weight of the water displaced by the bubble.

Second, there is the bubble's own weight ($mg$), pulling it downwards.

Since the bubble is accelerating upwards, the upward buoyant force must be winning the tug-of-war against the downward weight.

Let's translate this physical picture into mathematics using Newton's Second Law of Motion, which states that the net force on an object equals its mass times its acceleration ($F_{\text{net}}=ma$).

For our bubble, the net force is the buoyant force minus the weight:

We know that the buoyant force $B$ is given by the density of the fluid (${\rho }_w$), the volume of the displaced fluid ($V$), and the acceleration due to gravity ($g$):

Substituting this into our Newton's law equation, we get:

Our goal is to find the mass $m$. Let's group the terms with $m$ on one side:

Now, we can isolate $m$:

Now comes the fun part—plugging in the numbers! We must be careful to keep all our units consistent. The problem provides values in the CGS (centimeter-gram-second) system, so we'll stick with that.

- Density of water, ${\rho }_w=1{\text{ g/cm}}^3$ - Acceleration due to gravity, $g=980{\text{ cm/s}}^2$ - Upward acceleration, $a=9.8{\text{ cm/s}}^2$ - Radius of the bubble, $r=1\text{ cm}$

First, let's find the volume of the spherical bubble:

Now, substitute everything into our mass equation:

Let's simplify the numerator and the denominator:

Dividing these values gives us our final answer:

Rounding to two decimal places, the mass of the air bubble is $4.15\text{ g}$.

The problem stated that the water offers negligible drag force. But what if it didn't? As the bubble moves faster, the water would resist its motion more strongly. This drag force would act downwards, alongside the weight. Our initial equation would become $B-mg-F_{\text{drag}}=ma$. This would mean a smaller net upward force, and consequently, a different calculated mass for the same acceleration. Physics is all about understanding these subtle interactions!