A right circular cone is cut into two parts of equal height by a horizontal plane.

Let the original cone have height = H and radius = R.

The cone is cut by a horizontal plane into two parts of equal height, so the smaller cone has height:

$h = \frac{H}{2}$

Since the cones are similar, the radius of the smaller cone will also be half:

$r = \frac{R}{2}$

Volume of the original cone

$V = \frac{1}{3}\pi R^2 H$

Volume of the smaller cone

$V_1 = \frac{1}{3}\pi \left(\frac{R}{2}\right)^2 \left(\frac{H}{2}\right)$

$V_1 = \frac{1}{3}\pi \frac{R^2}{4} \cdot \frac{H}{2}$

$V_1 = \frac{1}{3}\pi \frac{R^2H}{8}$

So,

$V_1 = \frac{1}{8}V$

Volume of the frustum

$V_2 = V - V_1$

$V_2 = V - \frac{1}{8}V = \frac{7}{8}V$

Ratio (smaller cone : frustum)

$\frac{1}{8}V : \frac{7}{8}V = 1 : 7$

Correct Answer: 1 : 7