Question

A tank contains water on top of mercury. A cube of Fe, 60 mm along each edge, is a sitting upright in equilibrium in the liquids. Find how much of it is in each liquid, The rho Fe= 7.7x10^3 kg/m^3 & rho Hg= 13.6x10^3 kg/m^3.

Answer 1

Answer on Question #72418, Physics / Other

A tank contains water on top of mercury. A cube of Fe, 60 mm along each edge, is a sitting upright in equilibrium in the liquids. Find how much of it is in each liquid.

The rho Fe = 7.7x10³ kg/m³ & rho Hg = 13.6x10³ kg/m³.

Solution:

Suppose $x_1$ is depth in water and $x_2$ in mercury

$$x_1 + x_2 = h = 60 \text{ mm}$$

The equilibrium equation is

Weight = total buoyant force for each medium,

$$W = B = B_w + B_{Hg}$$

where weight is

$$W = mg = \rho_{Fe}Vg = \rho_{Fe}Ahg,$$

where area A is $h^2$.

The buoyant forces are:

in water

$$B_w = \rho_w(Ax_1)g$$

in mercury

$$B_{Hg} = \rho_{Hg}(Ax_2)g$$

Substituting

$$\rho_{Fe}Ahg = \rho_w(Ax_1)g + \rho_{Hg}(Ax_2)g$$

after cancelling terms, we will get

$$\rho_{Fe}h = \rho_w x_1 + \rho_{Hg} x_2$$

solve for $x_1$ and $x_2$ using first equation

$$\rho_{Fe}h = \rho_w(h - x_2) + \rho_{Hg} x_2$$

So, depth in mercury is

$$x_2 = \frac{\rho_{Fe}h - \rho_w h}{\rho_{Hg} - \rho_w} = \frac{\rho_{Fe} - \rho_w}{\rho_{Hg} - \rho_w} h = \frac{7700 - 1000}{13600 - 1000} \times 60 \text{ mm} = 31.9 \text{ mm}$$

Thus, depth in water

$$x_1 = 60 - 31.9 = 28.1 \text{ mm}$$

Answer: in water $x_1 = 28.1 \text{ mm}$; in mercury $x_2 = 31.9 \text{ mm}$.

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