A hollow steel ball of diameter 3.0 m just barely floats in water. What is the wall thickness of the ball? $\left( \rho \mathrm { Fe } = 7.87 \mathrm {~g} / \mathrm { cm } ^ { 3 } , \rho _ { \text {water } } = 1.00 \mathrm {~g} / \mathrm { cm } ^ { 3 } \right)$
A) 6.6 cm
B) 131 cm
C) 37 cm
D) 79 cm

Question

A hollow steel ball of diameter  3.0 m just barely floats in water. What is the wall thickness of the ball? $\left( \rho \mathrm { Fe } = 7.87 \mathrm {~g} / \mathrm { cm } ^ { 3 } , \rho _ { \text {water } } = 1.00 \mathrm {~g} / \mathrm { cm } ^ { 3 } \right)$
A)  6.6 cm
B)  131 cm
C)  37 cm
D)  79 cm

Quizplus · Accepted Answer

For the steel ball to just barely float in water, the weight of the water displaced by the ball must equal the weight of the ball itself. The volume of water displaced is equal to the volume of the ball, and the weight of the water displaced can be found using the density of water. 1. Calculate the volume of the ball:$$ V_{\text{ball}} = \frac{4}{3} \pi r^3 $$Given the diameter is 3.0 m, the radius $ r = 1.5 $ m. Therefore,$$ V_{\text{ball}} = \frac{4}{3} \pi (1.5)^3 = 14.137 \text{ m}^3 $$2. Calculate the weight of the water displaced (which equals the weight of the ball for it to float):$$ \text{Weight} = \rho_{\text{water}} \cdot V_{\text{ball}} \cdot g $$Given $\rho_{\text{water}} = 1.00 \text{ g/cm}^3 = 1000 \text{ kg/m}^3$, and ignoring $g$ since we'll compare weights directly,$$ \text{Weight}_{\text{water}} = 1000 \cdot 14.137 = 14137 \text{ kg} $$3. The steel ball consists of a shell of steel; let's denote the thickness of this shell as $t$. The volume of the steel $V_{\text{steel}}$ can be calculated by subtracting the volume of the inner sphere (with radius $1.5 - t$) from the volume of the outer sphere:$$ V_{\text{steel}} = \frac{4}{3} \pi (1.5)^3 - \frac{4}{3} \pi (1.5 - t)^3 $$4. The weight of the steel ball must equal the weight of the water displaced for it to just float:$$ \rho_{\text{Fe}} \cdot V_{\text{steel}} = \text{Weight}_{\text{water}} $$Given $\rho_{\text{Fe}} = 7870 \text{ kg/m}^3$,$$ 7870 \cdot V_{\text{steel}} = 14137 $$5. Solving the equation for $t$ involves equating the volume of the steel to the volume of water displaced, divided by the density of steel, and then solving for $t$. This calculation involves algebraic manipulation and solving a cubic equation, which is not shown here in detail due to the complexity. However, when you perform these calculations correctly, you find that the thickness that satisfies this condition is closest to the value given in option A, 6.6 cm. This is a simplified explanation. The exact calculation involves more detailed algebra, especially in solving for $t$, but the key concept is balancing the weight of the steel ball (which depends on the volume of steel and its density) with the weight of the water displaced (which depends on the volume of the ball and the density of water).

A Hollow Steel Ball of Diameter 3 A) 66 Cm B) 131 Cm C) 37 Cm

A Hollow Steel Ball of Diameter 3 A) 66 Cm
B) 131 Cm
C) 37 Cm