A hydrometer of mass y kg and volume 2y x 10$^{-5}$m$^3$ floats in a fluid with 20% of its volume above the fluid, what is the density of the fluid?

Analysis

To solve this problem, we apply the Law of Flotation, which states that for an object floating in equilibrium, the weight of the object is equal to the weight of the fluid displaced (Archimedes' Principle).

Concept

• Weight of the hydrometer ($W$): $W = mg = y \times g$
• Volume of fluid displaced ($V_{disp}$): Since 20% of the volume is above the fluid, 80% of the volume is submerged. • $V_{submerged} = 0.80 \times V_{total}$ • $V_{total} = 2y \times 10^{-5} \, \text{m}^3$ • $V_{submerged} = 0.80 \times (2y \times 10^{-5}) = 1.6y \times 10^{-5} \, \text{m}^3$
• Weight of fluid displaced ($W_{fluid}$): $W_{fluid} = \text{mass of fluid} \times g = (\rho \times V_{submerged}) \times g$ • Where $\rho$ is the density of the fluid.

Solution

1. Set the weight of the object equal to the weight of the displaced fluid: $$y \times g = \rho \times V_{submerged} \times g$$

2. Cancel $g$ from both sides: $$y = \rho \times V_{submerged}$$

3. Substitute the known values: $$y = \rho \times (1.6y \times 10^{-5})$$

4. Solve for $\rho$: $$\rho = \frac{y}{1.6y \times 10^{-5}}$$The$y$cancels out:$$\rho = \frac{1}{1.6 \times 10^{-5}}$$ $$\rho = \frac{1}{1.6} \times 10^5$$ $$\rho = 0.625 \times 10^5$$

5. Convert to standard scientific notation: $$\rho = 6.25 \times 10^4 \, \text{kgm}^{-3}$$

Conclusion

Comparing our result to the given options, we find that the calculated density matches option D.

Exam Tip

When dealing with "floating" problems, always remember that the density of the object relative to the fluid is equal to the fraction of the object that is submerged. In this case, $\frac{\rho_{object}}{\rho_{fluid}} = 0.8$. You can use this ratio shortcut to save time in the exam hall!