Animated Solution for Physics - Properties of Solids and Liquids: Comprehension Passage

A Classic Physics Toy

Imagine you are holding a soft plastic bottle completely filled with water. Inside, there's an inverted glass test tube with a little bit of air trapped inside. This is a classic physics toy called a Cartesian diver!

Initially, the test tube is just floating there, perfectly balanced. The trapped air has a volume of $3.3\text{ cc}$ at normal atmospheric pressure. But what happens when we squeeze the bottle? The pressure inside the water increases, which compresses the trapped air. As the air bubble shrinks, it displaces less water, so the upward buoyant force decreases.

The test tube will begin to sink exactly at the moment when the buoyant force becomes equal to its weight. Let's dive into the math behind this critical condition.

To find out when the test tube sinks, we need to equate the downward gravitational force with the upward buoyant force.

The downward force is simply the weight of the test tube, $W=mg$. The upward buoyant force, $F_B$, depends on the total volume of water displaced. This displaced volume is the sum of the volume of the glass material itself and the volume of the trapped gas.

We know the mass of the glass is $5\text{ g}$ and its density is $2.5\text{ g/cc}$. By dividing mass by density, we can find the volume of the glass:

The total buoyant force is the density of water times this total volume times the acceleration due to gravity:

Now, we set the buoyant force equal to the weight to find the critical volume of the trapped gas at which sinking begins:

Notice how the acceleration due to gravity, $g$, beautifully cancels out from both sides. Substituting the known values—the density of water is $1\text{ g/cc}$ and the mass of the test tube is $5\text{ g}$—we get:

Solving this simple equation, we find that the new volume of the trapped gas must be exactly $3\text{ cc}$ for the tube to start sinking:

We are almost there for the first part! The initial volume of the air was given as $3.3\text{ cc}$. We just found that it shrinks to $3\text{ cc}$.

So, the change in volume, $\Delta v$, is simply the initial volume minus the final volume:

The problem defines $\Delta v$ as $X\text{ cc}$, so our value for $X$ is exactly $0.3$.

Now, let's figure out the change in pressure, $\Delta p$. The problem explicitly states that the bottle is squeezed at a constant temperature.

This means the trapped air undergoes an isothermal process. For an ideal gas at a constant temperature, we can confidently apply Boyle's Law, which states that the product of pressure and volume remains constant.

Let's set up our equation. Initially, the pressure $P_1$ is the atmospheric pressure, $10^5\text{ Pa}$, and the volume $V_1$ is $3.3\text{ cc}$. Finally, the volume $V_2$ is $3\text{ cc}$, and we need to find the new pressure $P_2$.

Plugging these into Boyle's Law, we get:

Solving for $P_2$ is straightforward. We divide $3.3$ by $3$, which gives us exactly $1.1$.

So, the new pressure inside the trapped air, $P_2$, is $1.1\times 10^5\text{ Pa}$. This makes physical sense—the volume decreased, so the pressure had to increase!

Finally, we need the change in pressure, $\Delta p$. We subtract the initial pressure $P_1$ from the final pressure $P_2$:

To match the format given in the question, we rewrite this by shifting the decimal point:

Comparing this with the given expression $Y\times 10^3\text{ Pa}$, we get our final answer: $Y=10$.

And that completes our beautiful Cartesian diver problem!