FLUIDS CYU 9

Solution

Let $V$ be the external volume of the bottle and $M$ be its mass.
Let the density of water be $\rho$.
Let the initial volume of water in the vessel be $V_0$.

Step 1: Analyzing the Initial State

Initially, the bottle floats. We are given that the bottle begins to sink only when a mass $m = 300 \text{ g}$ of water is added to it. This implies that the total mass $(M + m)$ equals the maximum buoyant force the bottle can generate.

$$ (M + m)g = \rho V g \implies M = \rho V – m $$

In the initial floating state (without the added mass $m$), the volume of water displaced $V_{disp1}$ supports the weight of the empty bottle $M$.

$$ V_{disp1} = \frac{M}{\rho} = \frac{\rho V – m}{\rho} = V – \frac{m}{\rho} $$

The initial level of water $H_1$ in the vessel (area $S$) is given by volume conservation:

$$ S H_1 = V_0 + V_{disp1} = V_0 + V – \frac{m}{\rho} \quad \text{…(i)} $$

Step 2: Analyzing the Final State

In the final state, the bottle has sunk. The mass $m$ of water was added from an external source into the system (inside the bottle). The bottle is “thin glass,” so its material volume is negligible ($V_{glass} \approx 0$).

The total volume of water in the tank is now $V_0 + \frac{m}{\rho}$ (initial water + added water).

$$ S H_2 = (V_0 + \frac{m}{\rho}) + V_{glass} \approx V_0 + \frac{m}{\rho} \quad \text{…(ii)} $$

Step 3: The Change in Water Level

The change in level is the difference between final and initial volumes divided by area $S$.

$$ S(H_2 – H_1) = (V_0 + \frac{m}{\rho}) – (V_0 + V – \frac{m}{\rho}) $$

$$ S(H_2 – H_1) = \frac{2m}{\rho} – V $$

Let $\Delta h$ be the magnitude of the level change ($0.6 \text{ cm}$). Since the problem does not specify if the level rises or falls, we evaluate both cases.

Numerical Values:
$m = 300 \text{ g}, \rho = 1 \text{ g/cm}^3 \implies \frac{2m}{\rho} = 600 \text{ cm}^3$
$S = 250 \text{ cm}^2, \Delta h = 0.6 \text{ cm} \implies S\Delta h = 150 \text{ cm}^3$