Question

A steel ball of vol. 1 cm3 is recat in a hollow sphere to float in water.what should be the minimum volume of this sphere.

Answer 1

Answer on Question#39642 – Physics – Other

A steel ball of vol. 1 cm³ is recat in a hollow sphere to float in water. What should be the minimum volume of this sphere.

Solution:

$V_{\text{ball}} = 1 \, \text{cm}^3$ – volume of the steel ball;

$\rho_{\text{steel}} = 7.8 \frac{\text{g}}{\text{cm}^3}$ – density of the steel;

$\rho_{\text{water}} = 1 \frac{\text{g}}{\text{cm}^3}$ – density of the water;

Minimum volume should be enough to satisfy the condition of buoyancy (Archimedes force compensates for the weight of the ball):

$$m_{\text{ball}} \, \text{g} = \rho_{\text{water}} \, \text{g} \, V_{\text{min}}$$ $$m_{\text{ball}} = V_{\text{ball}} \cdot \rho_{\text{steel}}$$

(2)in(1):

$$V_{\text{ball}} \cdot \rho_{\text{steel}} = \rho_{\text{water}} \, V_{\text{min}}$$ $$V_{\text{min}} = \frac{V_{\text{ball}} \cdot \rho_{\text{steel}}}{\rho_{\text{water}}} = \frac{1 \, \text{cm}^3 \cdot 7.8 \, \frac{\text{g}}{\text{cm}^3}}{1 \, \frac{\text{g}}{\text{cm}^3}} = 7.8 \, \text{cm}^3$$

Answer: minimum volume of this sphere is $7.8 \, \text{cm}^3$