This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
To find the relative density of the liquid, we will use the principle of flotation, which states that for a floating object, the buoyant force equals the weight of the object. Step 1: Calculate the volume of the cube. The side of the cube is L = 10 cm. The volume of the cube (V_cube) is: V_cube = L^3 = (10 cm)^3 = 1000 cm^3 Convert the mass of the cube to grams for consistency with cm^3: m_cube = 0.5 kg = 500 g Step 2: Determine the volume of the cube submerged in the liquid. The problem states that (1)/(5) of the cube's height is above the liquid surface. This means that the fraction of the cube's height submerged in the liquid is 1 - (1)/(5) = (4)/(5). The height submerged (h_submerged) is: h_submerged = (4)/(5) × L = (4)/(5) × 10 cm = 8 cm The volume of the cube submerged (V_submerged) is the base area multiplied by the submerged height: V_submerged = L^2 × h_submerged = (10 cm)^2 × 8 cm = 100 cm^2 × 8 cm = 800 cm^3 Step 3: Apply the principle of flotation to find the density of the liquid. For a floating object, the weight of the object is equal to the weight of the displaced liquid. Weight of cube = Weight of displaced liquid m_cube × g = _liquid × V_submerged × g We can cancel g from both sides: m_cube = _liquid × V_submerged Now, solve for the density of the liquid (_liquid): _liquid = m_cubeV_submerged Substitute the values: _liquid = 500 g800 cm^3 _liquid = (5)/(8) g/cm^3 _liquid = 0.625 g/cm^3 Step 4: Calculate the relative density of the liquid. Relative density (RD) is the ratio of the density of the substance to the density of water. The density of water (_water) is 1 g/cm^3. RD_liquid = _liquid_water RD_liquid = 0.625 g/cm^31 g/cm^3 RD_liquid = 0.625 The final answer is 0.625.