# Question af0b9

Dec 15, 2014

Solving this problem requires applying Archimedes Principle which states that "a solid that is partially or fully submerged in a fluid experiences an upward buoyant force whose magnitude is equal to the weight of the fluid displaced by the solid".

1. The apparent weight of the solid inside water is equal to its true weight (weight outside) minus the buoyant force. Thus the buoyant force is equal to the the difference between the true and apparent weights.
W_{app} = W - F_B; => F_B = W - W_{app}.
2. Next apply the Archimedes Principle to relate the buoyant force to the weight of the water displaced. This will let us find the volume of the water displaced.
F_B = \rho_w V g; => V = F_B/(\rho_w g),
$V = \frac{W - {W}_{a p p}}{\setminus {\rho}_{w} g} , = \frac{36 g m - 34 g m}{1.0 g m . c {m}^{- 3}} = 2 c {m}^{3}$
3. The volume of the water displaced is equal to the volume of the submerged solid. So the total volume of the submerged solid is $2.0 c {m}^{3}$. From this one can find the average density of the solid.
$\setminus \overline{\setminus \rho} = \frac{M}{V} = \frac{36 g m}{2.0 c {m}^{3}} = 18 \frac{g m}{c {m}^{3}}$
4. Density of copper is $\setminus {\rho}_{c} = 8.9 \frac{g m}{c {m}^{3}}$,
Density of gold is $\setminus {\rho}_{g} = 19.3 \frac{g m}{c {m}^{3}}$.

This average density is less than the density of gold and more than the density of copper, clearly indicating that the solid is neither pure gold nor copper but a mixture. From the information of average density and the densities of copper and gold we have to find the ratio of gold to copper.

$M$ - Total mass; ${m}_{c}$ - mass of copper; ${m}_{g}$ - mass of gold;
$V$ - Total volume; ${v}_{c}$ - volume of copper; ${v}_{g}$ - volume of gold;
$\setminus {\rho}_{c} = {m}_{c} / {v}_{c}$; $\setminus {\rho}_{g} = {m}_{g} / {v}_{g}$.

Volume fractions of gold and copper are : X_{g} = v_g/V; X_c = v_c/V

 X_c+X_{g}=1; => X_{g} = 1- X_c
\bar{\rho} = M/V = (m_g+m_c)/V = (\rho_{g} v_g + \rho_c v_c)/V = \rho_{g} v_{g}/V + \rho_c v_c/V
\bar{\rho} = \rho_g} X_{g} + \rho_c X_c = \rho_{g}(1-X_c)+\rho_c X_c

Solving for ${X}_{c} = \frac{\setminus {\rho}_{g} - \setminus \overline{\setminus \rho}}{\setminus {\rho}_{g} - \setminus {\rho}_{c}} = \frac{19.3 - 18.0}{19.3 - 8.9} = 0.125$

So the solid is made of 12.5% copper (by volume) and 87.5% gold.

The mass of copper is then,
m_c = \rho_c v_c = \rho_c X_c V = 8.9 (gm)/(cm^3)\times(0.125)\times(2.0 gm) = 2.225 gms.#