To solve the problem of how deep a test tube will sink when placed upright in water, we need to apply the principles of buoyancy and Archimedes' principle. Let's break down the solution step-by-step. ### Step 1: Understand the Problem We have a test tube with a radius of 1.0 cm and a mass of 8.8 g. We need to find out how deep it will sink when placed upright in water. ### Step 2: Calculate the Volume of the Test Tube The volume \( V \) of a cylinder (which is the shape of the test tube) can be calculated using the formula: \[ V = \pi r^2 h \] where: - \( r \) is the radius, - \( h \) is the height (or depth submerged in this case). ### Step 3: Calculate the Weight of the Test Tube The weight \( W \) of the test tube can be calculated using the formula: \[ W = mg \] where: - \( m \) is the mass (8.8 g, which we convert to kg for standard SI units: \( 8.8 \, \text{g} = 0.0088 \, \text{kg} \)), - \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). Calculating the weight: \[ W = 0.0088 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 0.0863 \, \text{N} \] ### Step 4: Apply Archimedes' Principle According to Archimedes' principle, the buoyant force \( F_b \) acting on the submerged part of the test tube is equal to the weight of the water displaced by that volume. The buoyant force can be expressed as: \[ F_b = \rho V_{displaced} g \] where: - \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)), - \( V_{displaced} \) is the volume of water displaced, - \( g \) is the acceleration due to gravity. ### Step 5: Set Up the Equation For the test tube to be in equilibrium (not sinking or floating), the weight of the test tube must equal the buoyant force: \[ W = F_b \] Substituting the expressions we have: \[ 0.0863 \, \text{N} = 1000 \, \text{kg/m}^3 \times \pi (0.01 \, \text{m})^2 h g \] Here, we convert the radius from cm to meters (1.0 cm = 0.01 m). ### Step 6: Solve for \( h \) Rearranging the equation to solve for \( h \): \[ h = \frac{0.0863 \, \text{N}}{1000 \, \text{kg/m}^3 \times \pi (0.01 \, \text{m})^2 \times 9.81 \, \text{m/s}^2} \] Calculating the denominator: \[ 1000 \times \pi \times (0.01)^2 \times 9.81 \approx 3.08 \, \text{N/m} \] Now substituting back: \[ h = \frac{0.0863}{3.08} \approx 0.0280 \, \text{m} = 2.80 \, \text{cm} \] ### Conclusion The depth to which the test tube would sink is approximately **2.8 cm**. Therefore, the correct option is **A**. ### Explanation of Other Options - **B. 5.2 cm**: This value is too high because it would imply a greater volume of water displaced than the weight of the test tube can support. - **C. 25.5 cm**: This is significantly higher than the calculated depth and does not satisfy the equilibrium condition. - **D. 28.0 cm**: This is also too high and would suggest that the test tube is displacing an unrealistic volume of water compared to its weight. ### Revision Summary - The test tube's weight must equal the buoyant force for it to float or sink to a certain depth. - Use Archimedes' principle to relate the weight of the object to the weight of the displaced fluid. - The volume of a cylinder is calculated using \( V = \pi r^2 h \). - The final depth calculation shows that the test tube sinks to approximately 2.8 cm in water.