Question

a solid steel ball of volume 1 cm3 is recast into a hollow sphere so as to float on water. What should be the minimum volume of this hollow sphere when density of steel is 7.71 g/cm3 and that of water is 1g/cm3 ?

Answer 1

QUESTION:

A solid steel ball of volume 1 cm³ is recast into a hollow sphere so as to float on water. What should be the minimum volume of this hollow sphere when density of steel is 7.71 g/cm³ and that of water is 1 g/cm³?

SOLUTION

Let's draw a free-body diagram for the ball in the hollow sphere:

We assume that the sphere is massless, and as sphere is floating in the water:

$$F _ {B} - m _ {\text {b a l l}} g = 0 \text{ (y - axis projection)}$$ $$F _ {B} = \rho_{\text{water}} g \cdot V_{\text{sphere}} \text{ (the buoyancy force)}$$ $$m _ {\text {ball}} = \rho_{\text{steel}} V _ {\text {ball}}$$

Hence,

$$F _ {B} - m _ {\text {ball}} g = 0$$ $$\rho_{\text{water}} g \cdot V_{\text{sphere}} - \rho_{\text{steel}} V_{\text{ball}} g = 0$$ $$\rho_{\text{water}} \cdot V_{\text{sphere}} - \rho_{\text{steel}} V_{\text{ball}} = 0$$ $$V _ {\text {sphere}} = \frac {\rho_{\text {steel}}}{\rho_{\text {water}}} V _ {\text {ball}}$$ $$V _ {\text {sphere}} = \frac {7.71}{1} \cdot 1 = 7.71 \text{ cm}^3$$

ANSWER

$$V _ {\text {sphere}} = 7.71 \text{ cm}^3$$