A hollow steel ball of diameter 3.0 m just barely floats in water.What is the wall thickness of the ball? (ρ_Fe = 7.87 g/cm³,ρ_water= 1.00 g/cm³) A)6.6 cm B)37 cm C)131 cm D)79 cm

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Quizplus · Accepted Answer

For the steel ball to just barely float, the weight of the water displaced by the ball must equal the weight of the ball itself. The volume of the ball (V_ball) can be calculated using the formula for the volume of a sphere: V = 4/3 * π * r^3, where r is the radius of the ball. Since the diameter is 3.0 m, the radius is 1.5 m. The volume of the ball is thus 4/3 * π * (1.5)^3 m^3.The volume of the steel (V_steel) that makes up the ball is the total volume of the ball minus the volume of the hollow part inside. Since the ball just barely floats, the mass of the steel must equal the mass of the water displaced by the ball. The mass of the water displaced can be found by multiplying the volume of the ball by the density of water (ρ_water = 1.00 g/cm^3 = 1000 kg/m^3).Let's denote the thickness of the steel wall as t. The outer radius of the ball is 1.5 m, so the inner radius is (1.5 - t) m. The volume of the steel can be calculated as the volume of the outer sphere minus the volume of the inner sphere (which is hollow).The density of steel (ρ_steel) is given as 7.87 g/cm^3 = 7870 kg/m^3. Using the principle that the mass of the displaced water equals the mass of the steel, we set up the equation:ρ_steel * V_steel = ρ_water * V_ballSolving this equation for t (the thickness of the steel wall) involves algebraic manipulation and substitution of the known values. After solving, we find that the thickness of the steel wall that satisfies this condition is approximately 6.6 cm.

A Hollow Steel Ball of Diameter 3