Question

A piece of gold weigh 10g in air and 9g in water .what is volume of cavity (19.3 g/cm cube )

Answer 1

A piece of gold weigh $10\,\mathrm{g}$ in air and $9\,\mathrm{g}$ in water. What is volume of cavity (19.3 g/cm³)

For piece of gold in air:

$$m = \rho_g V_g = 10\,\mathrm{g}$$

where $\rho_g$ – density of gold, $V_g$ – volume of gold

For piece of gold in air (assuming Archimedes’ principle that the upward buoyant force is equal to the weight of the fluid that the body displaces):

$$m - \rho_w(V_g + V_c) = 9\,\mathrm{g}$$

where $\rho_w$ – density of water, $V_c$ – volume of cavity.

Substitute from first $g V_g = \frac{m g}{\rho_g}$ to second:

$$m - \rho_w \frac{m}{\rho_g} - \rho_w V_c = 9\,\mathrm{g}$$

Or:

$$10\,\mathrm{g}\left(1 - \frac{\rho_w}{\rho_g}\right) - 9\,\mathrm{g} = \rho_w V_c$$ $$V_c = \frac{10\,\mathrm{g}\left(1 - \frac{\rho_w}{\rho_g}\right) - 9\,\mathrm{g}}{\rho_w} = \frac{10\,\mathrm{g}\left(1 - \frac{1}{19.3}\right) - 9\,\mathrm{g}}{1\,\frac{\mathrm{g}}{\mathrm{cm}^3}} = 0.48\,\mathrm{cm}^3$$

Answer: $0.48\,\mathrm{cm}^3$