### Solution:

Let us draw the figure of the sector from the circle and the view of the cone.AOB is the sector of the circle with centre O and radius of 12 cm. The angle of the sector is 120°.

If we twist the sector, the arc length of the sector will be equal to the circumference of the base of the cone, and the radius of the sector will be the height of the slope of the cone.

We know that for 120° or radians is .

Then we find the arc length of the sector by comparison.

Where θ is the sector angle in radians.

The sector angle is 120°.

We know that 180° is radians.

$\n \n \n \n \n \n 180\n \n 120\n \n =\n \pi \n \theta \n \n \n \n \n \n \n \theta =\n \n 2\pi \n 3\n \n \n \n \n \n \n $

Therefore, we will substitute $\n \n \theta =\n \n 2\pi \n 3\n \n \n $ and r = 12 in $\n \n \n l\n s\n \n =r\theta \n $ to find the length of the arc of the sector.

$\n \n \n \n \u21d2\n l\n s\n \n =(12)\n \n \n \n 2\pi \n 3\n \n \n \n \n \n \n \n \u21d2\n l\n s\n \n =8\pi \n \n \n \n \n $

This length will be equal to the circumference of the base circle. Let the radius of the base circle be r.

$\n \n \n \n \u21d22\pi r=8\pi \n \n \n \n \n \u21d2r=4\n \n \n \n \n $

Therefore, the radius of the base is 4 cm.

As we can see in the picture, a right triangle is created with height, base radius and slope height as three sides.

We apply the Pythagorean theorem to this triangle to find the height of the cone.

where l is the slope height, h is the height and r is the radius of the cone.

$\n \n \n \n \u21d2\n (12)\n 2\n \n =\n h\n 2\n \n +\n (4)\n 2\n \n \n \n \n \n \n \u21d2144=\n h\n 2\n \n +16\n \n \n \n \n \u21d2\n h\n 2\n \n =132\n \n \n \n \n \u21d2h=\n \n 132\n \n \n \n \n \n \n $

We know that the volume of a cone is given by the relation $\n \n \n 1\n 3\n \n \pi \n r\n 2\n \n h\n $ .

$\n \n \n \n \u21d2V=\n 1\n 3\n \n \pi \n (4)\n 2\n \n (\n \n 132\n \n )\n \n \n \n \n \u21d2V=\n \n 16\pi \n \n 132\n \n \n 3\n \n \n \n \n \n \n $

Hence, the volume of the cone is

$c{m}^{3}$.