The problem states that y is directly proportional to sqrt(x). This can be written as:

The problem states that $y$ is directly proportional to $\sqrt{x}$. This can be written as:

$$y = k\sqrt{x}$$

where $k$ is the constant of proportionality.

We are given that $y=12$ when $x=4$. Substitute these values into the equation:

$$12 = k\sqrt{4}$$ $$12 = k(2)$$

Divide by 2 to find $k$:

$$k = \frac{12}{2}$$ $$k = 6$$

Substitute the value of $k$ back into the variation equation from Step 1:

$$y = 6\sqrt{x}$$

Substitute $x=25$ into the law found in Step 3:

$$y = 6\sqrt{25}$$ $$y = 6 \times 5$$ $$y = 30$$

The law connecting $y$ and $x$ is $\boxed{y = 6\sqrt{x}}$.
When $x=25$, $y = \boxed{30}$.