Q. The height of a cone …

Let the radius of original cone be r₂
Radius of cut off cone be r₁
According to the question
Height of the original cone = 10 cm (given)
The cone is cut off from the midpoint of the height,
therefore, height the cone cut off = 5 cm
{tex}\Delta A O C \sim \Delta AO' D{/tex}
OA = Radius of original cone = r₂
O'A' = Radius of cutoff cone = r₁
Ratio of radius of two cones = Ratio of the height of cones
{tex}\therefore \quad \frac { A O } { A' O ^ { \prime } } = \frac { r _ { 2 } } { r _ { 1 } } = \frac { 10 } { 5 }{/tex}
{tex}\Rightarrow \quad r _ { 2 } = 2 r _ { 1 }{/tex}
Radius of original cone = 2 (radius of the cut off cone)
Volume of the cut off cone {tex}= \frac { 1 } { 3 } \pi r _ { 1 } ^ { 2 } \times 5{/tex}
{tex}= \frac { 5 } { 3 } \pi r _ { 1 } ^ { 2 }{/tex} cubic units
Volume of original cone {tex}= \frac { 1 } { 3 } \pi \left( 2 r _ { 1 } \right) ^ { 2 } \times 10{/tex}
{tex}= \frac { 40 } { 3 } \pi r _ { 1 } ^ { 2 }{/tex} cubic units
Volume of frustum = Volume of an original cone - Volume of cut off cone
{tex}= \frac { 40 } { 3 } \pi r _ { 1 } ^ { 2 } - \frac { 5 } { 3 } \pi r _ { 1 } ^ { 2 }{/tex}
{tex}= \frac { 35 } { 3 } \pi r _ { 1 } ^ { 2 }{/tex} cubic units
Required ratio = Volume of frustum: Volume of cut off cone
{tex}= \frac { 35 \pi r _ { 1 } ^ { 2 } } { 5 \pi r _ { 1 } ^ { 2 } } = \frac { 7 } { 1 }{/tex}
Therefore, the required ratio = 7: 1.