A small hollow vessel which has a small circular hole of radius r in its base, is immersed in a tank of oil of density $\rho$ and surface tension T. The oil will penetrate into the vessel at a depth of

Let the depth at which the oil penetrates the vessel be $h$.

The hydrostatic pressure exerted by the oil at depth $h$ is given by $P = h\rho g$.

For the oil to enter the vessel through the small hole of radius $r$, this hydrostatic pressure must overcome the excess pressure due to surface tension at the meniscus formed at the hole.

The maximum excess pressure for a meniscus of radius $r$ is $\Delta P = \dfrac{2T}{r}$.

Equating the hydrostatic pressure to the excess pressure:

$h\rho g = \dfrac{2T}{r}$

$h = \dfrac{2T}{r\rho g}$

Answer: $\dfrac{2T}{r\rho g}$