A cube with a side length of 5 cm floats on the surface of water. If 4/5 of the cube volume floats above the surface of the water, what is the density of the cube? (Note: Density of water = 1,000 kg/m³.)

Question

Quizplus · Accepted Answer

11ecba2d_131b_662e_a3bf_edc6133c9477_MD0008_00 Archimedes' principle states that the magnitude of the buoyant force (F_B) that acts on a floating object is equal to the weight of the fluid displaced by the object. Consequently, the buoyant force can be expressed in terms of the acceleration of gravity (g) and the displaced fluid's density (ρ_fluid) and volume (V_fluid): 11ecba2d_131b_662f_a3bf_63d555e1b53b_MD0008_00 Furthermore, the buoyant force has an equal magnitude and opposite direction to the weight (W_O) of the floating object such that: 11ecba2d_131b_8d40_a3bf_b7fc9eb72209_MD0008_00 11ecba2d_131b_8d41_a3bf_e954c96108f1_MD0008_00 Given that the mass of the floating object (m_object) can be expressed as a product of the object density (ρ_object) and object volume (V_object), it follows that: 11ecba2d_131b_8d42_a3bf_c16222f599f5_MD0008_00 11ecba2d_131b_8d43_a3bf_9d30ef08ec63_MD0008_00 Rearranging the equation above yields an expression that relates the ratio of the object and fluid densities to the ratio of the displaced fluid volume and the object volume: 11ecba2d_131b_8d44_a3bf_9367148e82fd_MD0008_00 Because V_fluid depends on the buoyancy of the object, V_fluid = x · V_object, where x is the fraction of the object volume that is submerged in the fluid. Therefore, the V_fluid/V_object ratio equals x. 11ecba2d_131b_8d45_a3bf_f5f47536d596_MD0008_00 In this question, 4/5 of the cube volume floats above the surface of the water. This means that 1/5 of the cube volume is submerged below the surface of the water, and the cube displaces a volume of water that is 1/5 its own volume: 11ecba2d_131b_b456_a3bf_e1d6d037539a_MD0008_00 Given that the density of water is 1,000 kg/m³, the density of the object can be calculated as: 11ecba2d_131b_b457_a3bf_5d31b8b1da6b_MD0008_00 11ecba2d_131b_b458_a3bf_f17c028901bc_MD0008_00 Therefore, the density of a floating cube that displaces 1/5 its volume in water is 200 kg/m³ (Choice A). Educational objective: Archimedes' principle states that the buoyant force acting on a floating object is equal to the weight of the fluid displaced by the object. The fraction of the object submerged is related to the density of the object and the density of the fluid on which the object floats.

A Cube with a Side Length of 5 Cm Floats