A right angled triangle with sides …

AB = 3 cm, AC = 4 cm
In $\triangle$BAC, by pythagoras theorem
BC² = AB² + AC² 
$\Rightarrow$BC² = 3² + 4² 
$\Rightarrow$BC² = 25 
$\Rightarrow$BC = $\sqrt {25}$ = 5 cm
In $\triangle$AOB and $\triangle$CAB
$\angle$ABO = $\angle$ABC [common]
$\angle$AOB = $\angle$BAC [each 90^o] 
Then, $\triangle$AOB - $\triangle$CAB [by AA similarity]
$\therefore$ $\frac { A O } { C A } = \frac { O B } { A B } = \frac { A B } { C B }$ [c.p.s.t]
$\Rightarrow$ $\frac { A O } { 4 } = \frac { O B } { 3 } = \frac { 3 } { 5 }$
Then, AO = $\frac{{4 \times 3}}{5}$ and OB = $\frac{{3 \times 3}}{5}$
$\Rightarrow$ AO = $\frac{12}{5}$ cm and OB = $\frac{9}{5}$ cm
$\therefore$OC = 5 - $\frac{9}{5}$ = $\frac{16}{5}$cm
$\therefore$ Volume of double cone thus generated = volume of first cone + volume of second cone
$= \frac { 1 } { 3 } \pi ( A O ) ^ { 2 } \times B O + \frac { 1 } { 3 } \pi ( A O ) ^ { 2 } \times O C$
$= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 12 } { 5 } \right) ^ { 2 } \times \frac { 9 } { 5 } + \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 12 } { 5 } \right) ^ { 2 } \times \frac { 16 } { 5 }$
$= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \frac { 12 } { 5 } \times \frac { 12 } { 5 } \left[ \frac { 9 } { 5 } + \frac { 16 } { 5 } \right]$
$= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \frac { 12 } { 5 } \times \frac { 12 } { 5 } \times 5$
=$\frac{1056}{35}$ = $30 \frac { 6 } { 35 } \mathrm { cm } ^ { 3 }$.