A cone of radius 10 cm …

Let height of cone = 2x cm
Radius of cone = 10 cm
In {tex}\triangle{/tex}OAB and {tex}\triangle{/tex}OCD 
{tex}\angle{/tex}AOB = {tex}\angle{/tex}COD [ common ]
{tex}\angle{/tex}OAB = {tex}\angle{/tex}OCD [ each 90^o ]
Then, {tex}\triangle{/tex}OAB {tex}\sim{/tex} {tex}\triangle{/tex}OCD [ by AA simiarity ] 
{tex}\therefore{/tex} {tex}\frac { O A } { O C } = \frac { A B } { C D }{/tex} [ corresponding sides of similar triangle are proportional]
{tex}\Rightarrow \quad \frac { X } { 2 X } = \frac { r } { 10 }{/tex}
{tex}\Rightarrow{/tex} r = 5 cm
{tex}\therefore{/tex} {tex}\frac { \text { Volume of small cone } } { \text { Volume of frustum } } = \frac { \frac { 1 } { 3 } \pi ( 5 ) ^ { 2 } \times x } { \frac { 1 } { 3 } \pi ( x ) \left[ 10 ^ { 2 } + 10 \times 5 + 5 ^ { 2 } \right] }{/tex}
{tex}= \frac { 25 x } { x \times 175 }{/tex}
={tex}\frac{1}{7}{/tex}