1. **Stating the problem:** We are given the mass of an empty measuring cylinder, the combined mass of the cylinder and water, the volume of water, the weight of a stone in air and in water, and the acceleration due to gravity. We need to find relevant quantities such as the mass of water, density of water, volume of the stone, and density of the stone. 2. **Given data:** - Mass of empty cylinder $m_c = 10.9$ g - Mass of cylinder + water $m_{cw} = 51.1$ g - Volume of water $V_w = 43$ cm$^3$ - Weight of stone in air $W_a = 0.9$ N - Weight of stone in water $W_w = 0.5$ N - Acceleration due to gravity $g = 9.8$ m/s$^2$ 3. **Find mass of water:** $$m_w = m_{cw} - m_c = 51.1 - 10.9 = 40.2 \text{ g}$$ 4. **Calculate density of water:** Density $\rho = \frac{\text{mass}}{\text{volume}}$ $$\rho_w = \frac{m_w}{V_w} = \frac{40.2}{43} \approx 0.935 \text{ g/cm}^3$$ 5. **Find volume of stone using buoyancy principle:** Buoyant force $= W_a - W_w = 0.9 - 0.5 = 0.4$ N Buoyant force equals weight of displaced water: $$F_b = \rho_w \times V_s \times g$$ where $V_s$ is volume of stone in m$^3$, $\rho_w$ in kg/m$^3$. Convert density of water to kg/m$^3$: $$0.935 \text{ g/cm}^3 = 935 \text{ kg/m}^3$$ Rearranging for $V_s$: $$V_s = \frac{F_b}{\rho_w g} = \frac{0.4}{935 \times 9.8} \approx 4.37 \times 10^{-5} \text{ m}^3$$ Convert to cm$^3$: $$V_s = 4.37 \times 10^{-5} \times 10^6 = 43.7 \text{ cm}^3$$ 6. **Calculate mass of stone:** $$m_s = \frac{W_a}{g} = \frac{0.9}{9.8} \approx 0.092 \text{ kg} = 92 \text{ g}$$ 7. **Calculate density of stone:** $$\rho_s = \frac{m_s}{V_s} = \frac{92}{43.7} \approx 2.1 \text{ g/cm}^3$$ **Final answers:** - Mass of water = 40.2 g - Density of water = 0.935 g/cm$^3$ - Volume of stone = 43.7 cm$^3$ - Density of stone = 2.1 g/cm$^3$