Three immiscible transparent liquids with refractive indices $3 / 2,4 / 3$ and $6 / 5$ are arranged one above the other in a container. The depths of the liquids are $3 \mathrm{~cm}, 4 \mathrm{~cm}$ and 6 cm respectively. The apparent depth of the vessel is

Apparent depth of liquid layers

For a liquid layer of refractive index $n$ and thickness $d$, the apparent depth is given by $d_{\text{app}}$ = $\frac{d}{n}$.

Given three distinct layers: the apparent depths are
$\frac{3\ \text{cm}}{3/2} = 2\ \text{cm}$;
$\frac{4\ \text{cm}}{4/3} = 3\ \text{cm}$;
$\frac{6\ \text{cm}}{6/5} = 5\ \text{cm}$.

The total apparent depth of the vessel is the sum of the contributions from each layer: $2 + 3 + 5 = 10$ cm.

$\boxed{10\ \text{cm}}$