A cubical metal block with an edge of 12 cm floats in mercury such...

Let the edge of the cubical block be $L = 12 \text{ cm}$ and its base area be $A = L^2$.

Initially, the block floats in mercury with $\dfrac{1}{5}$ of its height submerged. Equating the weight of the block to the buoyant force provided by mercury:

$L A \rho_b g = \left(\dfrac{L}{5}\right) A \rho_m g$

$\rho_b = \dfrac{\rho_m}{5} = \dfrac{13.6}{5} = 2.72 \text{ g/cm}^3$

When water is poured until the upper surface of the block is just immersed, let the height of the water column be $h$. The block is now submerged in water by a height $h$ and in mercury by a height $(L - h)$.

Equating the total buoyant force from both liquids to the weight of the block:

$h A \rho_w g + (L - h) A \rho_m g = L A \rho_b g$

Dividing by $A g$ throughout, we get:

$h \rho_w + (L - h) \rho_m = L \rho_b$

Substituting the known values ($\rho_w = 1 \text{ g/cm}^3$, $\rho_m = 13.6 \text{ g/cm}^3$, $L = 12 \text{ cm}$, $\rho_b = 2.72 \text{ g/cm}^3$):

$h(1) + (12 - h)(13.6) = 12 \times 2.72$

$h + 163.2 - 13.6h = 32.64$

$12.6h = 163.2 - 32.64$

$12.6h = 130.56$

$h = \dfrac{130.56}{12.6} \approx 10.36 \text{ cm}$

Rounding to one decimal place, the height of the water column is approximately $10.4 \text{ cm}$.

Answer: $10.4 \text{ cm}$