The volume $V$ of a cone is given by the formula:

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

where $r$ is the radius, and $h$ is the height.

Step 1: Relate $r$ and $h$

We are given that the radius $r$ is twice the height $h$, so:

$r = 2h$

Step 2: Substitute $r = 2h$ into the volume formula

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

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

Simplify $(2h)^2$:

$V = \frac{1}{3} \pi (4h^2) h = \frac{1}{3} \pi (4h^3)$

$V = \frac{4}{3} \pi h^3$

Step 3: Solve for $h$

We are given that $V = \frac{\pi}{6}$, so:

$\frac{4}{3} \pi h^3 = \frac{\pi}{6}$

Cancel $\pi$ from both sides:

$\frac{4}{3} h^3 = \frac{1}{6}$

Multiply through by 3:

$4h^3 = \frac{1}{2}$

Divide by 4:

$h^3 = \frac{1}{8}$

Take the cube root of both sides:

$h = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$

Step 4: Find the radius $r$

Since $r = 2h$, substitute $h = \frac{1}{2}$:

$r = 2 \times \frac{1}{2} = 1$

Final Answer:

The radius of the cone is:

$\boxed{1 \, \text{cm}}$

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Do you want more details or have any questions?
Here are 5 related questions to explore:

1. How do you derive the volume formula for a cone?
2. What is the relationship between radius and height in this cone?
3. How would the solution change if the cone’s volume was doubled?
4. How do you find the cube root of a fraction?
5. Can you generalize this problem to cones where $r = k \cdot h$?

Tip: Always simplify expressions before solving equations to avoid unnecessary complications.