Question

A 400 g solid cube having an edge of length \(10\) cm floats in water. How much volume of the cube is outside the water? (Given: density of water = \(1000 { kg/m}^3\))

• A. \( 600 { cm}^3 \)
• B. \( 4000 { cm}^3 \)
• C. \( 1400 { cm}^3 \)
• D. \( 400 { cm}^3 \)

Hint:

The floating condition follows Archimedes’ principle: the buoyant force equals the weight of the displaced liquid.

Answer 1

The Correct Option is D

The total volume of the cube is: \[ V_{{total}} = (10 { cm})^3 = 1000 { cm}^3 \] The mass of the cube is: \[ m = 400 { g} = 0.4 { kg} \] The density of the cube is: \[ \rho_{{cube}} = \frac{m}{V_{{total}}} = \frac{0.4}{1000 \times 10^{-6}} = 400 { kg/m}^3 \] Since the cube floats, the submerged volume is given by: \[ V_{{submerged}} = V_{{total}} \times \frac{\rho_{{cube}}}{\rho_{{water}}} \] \[ V_{{submerged}} = 1000 \times \frac{400}{1000} = 600 { cm}^3 \] Thus, the volume outside the water is: \[ V_{{outside}} = V_{{total}} - V_{{submerged}} \] \[ V_{{outside}} = 1000 - 600 = 400 { cm}^3 \] Thus, the correct answer is (4) 400 cm³.