A test tube of radius 1.0cm is loaded to 8.8g. If is placed upright in water, find the depth to which it would sink

The depth to which the test tube sinks in water depends on the buoyant force acting on it, which in turn depends on the weight of water displaced by the test tube. The weight of water displaced is equal to the weight of the test tube, so we can use the density of water and the weight of the test tube to find the volume of water displaced. The formula for the buoyant force is given by: Buoyant force = (density of fluid) x (volume of fluid displaced) x (acceleration due to gravity) The weight of the test tube is given by: Weight of test tube = (mass of test tube) x (acceleration due to gravity) Since the test tube is placed upright, the height to which it sinks is the same as the depth of water displaced. Therefore, the depth to which the test tube sinks in water is given by: Depth = (Weight of test tube) / (Buoyant force) Substituting the values given in the question and simplifying, we get: Depth = (Weight of test tube) / ((density of water) x (volume of water displaced) x (acceleration due to gravity)) Depth = (0.0088 kg x 9.8 m/s^2) / ((1000 kg/m^3) x (pi x (0.01 m)^2 x d) x (9.8 m/s^2)) where 'd' is the depth to which the test tube sinks in meters. Solving for 'd', we get: d = 0.0282 m or 2.82 cm Therefore, the depth to which the test tube sinks in water is approximately 2.8 cm. So, the correct option is: 2.8cm.