To solve the problem, let’s break it down step by step:

Part (a): Find the height of the water level

The conical vessel has:

• A base radius $R = 6 \, \text{cm}$
• A height $H = 14 \, \text{cm}$
• A volume of water poured into it $V = 475 \, \text{cm}^3$.

The formula for the volume of a cone is:

$V = \frac{1}{3} \pi r^2 h$

Here, $r$ and $h$ are the radius and height of the water level in the cone. Since the cone is similar in shape to the entire vessel, the radius of the water surface at height $h$ is proportional to $h$:

$\frac{r}{h} = \frac{6}{14} \implies r = \frac{3}{7}h$

Substitute $r = \frac{3}{7}h$ into the volume formula:

$V = \frac{1}{3} \pi \left( \frac{3}{7}h \right)^2 h$

Simplify the expression:

$V = \frac{1}{3} \pi \cdot \frac{9}{49} h^3 = \frac{3 \pi}{49} h^3$

Set $V = 475 \, \text{cm}^3$:

$\frac{3 \pi}{49} h^3 = 475$

Solve for $h^3$:

$h^3 = \frac{475 \cdot 49}{3 \pi} \approx 2468.73$

$h = \sqrt[3]{2468.73} \approx 13.5 \, \text{cm}$

So, the height of the water level is approximately:

$\boxed{13.5 \, \text{cm}}$

Part (b): Find the area of the wet surface

The wet surface includes:

1. The curved surface area of the water.
2. The circular base area of the water.

1. Curved Surface Area of the Water

The slant height of the cone (denoted $l$) for the water can be calculated using the Pythagorean theorem:

$l = \sqrt{r^2 + h^2} = \sqrt{\left( \frac{3}{7}h \right)^2 + h^2} = h \sqrt{\left( \frac{3}{7} \right)^2 + 1} = h \sqrt{\frac{9}{49} + 1} = h \sqrt{\frac{58}{49}}$

At $h = 13.5 \, \text{cm}$, the radius $r = \frac{3}{7}h = \frac{3}{7} \cdot 13.5 \approx 5.79 \, \text{cm}$, and $l$ becomes:

$l = 13.5 \sqrt{\frac{58}{49}} \approx 15.92 \, \text{cm}$

The curved surface area (CSA) is given by:

$\text{CSA} = \pi r l = \pi \cdot 5.79 \cdot 15.92 \approx 289.2 \, \text{cm}^2$

2. Circular Base Area of the Water

The base area is given by:

$\text{Base Area} = \pi r^2 = \pi \cdot 5.79^2 \approx 105.4 \, \text{cm}^2$

Total Wet Surface Area

Add the two areas:

$\text{Total Area} = 289.2 + 105.4 = 394.6 \, \text{cm}^2$

The total wet surface area is approximately:

$\boxed{394.6 \, \text{cm}^2}$

Let me know if you'd like any clarifications or further details! Here are some related questions:

1. What is the derivation of the cone volume formula?
2. How does the similarity of triangles apply to this problem?
3. Can you calculate the surface area if the water height changes?
4. What happens to the wet surface area if the cone dimensions are doubled?
5. How would the solution change if the cone were truncated?

Tip: Always check unit consistency and proportionality when working with similar shapes!