1. Speed of efflux from the hole

If the liquid level above the hole is $h$, then by Torricelli’s theorem the speed of the emerging liquid is

$v = \sqrt{2gh}.$

2. Mass of liquid emerging per second

The hole has area $a$, so volume flow rate is

$Q = av.$

Hence mass flow rate is

$\dot m = \rho av,$

where $\rho$ is the density of the liquid.

3. Horizontal reaction on the vessel due to jet

The jet carries horizontal momentum. So the reaction force on the vessel is

$F = \dot m \, v = (\rho av)(v) = \rho a v^2.$

Using $v^2 = 2gh$,

$F = \rho a(2gh) = 2\rho agh.$

4. Normal reaction on the vessel

The vessel is light, so only the liquid contributes significantly to the weight.

Weight of liquid in the vessel:

$W = \rho (Ah) g = \rho Agh.$

So normal reaction from the ground is

$N = \rho Agh.$

5. Condition for no sliding

Maximum static friction is

$f_{\max} = \mu N = \mu \rho Agh.$

To prevent sliding,

$f_{\max} \ge F.$

So,

$\mu \rho Agh \ge 2\rho agh.$

Cancel $\rho gh$:

$\mu A \ge 2a$

$\mu_{\min} = \frac{2a}{A}.$

6. Match with options

$\boxed{\mu_{\min} = \frac{2a}{A}}$

So the correct option is C.