Question

If the right circular cone is separated into three solids of volume $V_1,V_2,V_3$ by two planes parallel to the base and intersect to altitude, then $V_1:V_2:V_3$ is

• A. $1:2:3$
• B. $1:4:6$
• C. $1:6:9$
• D. $1:7:19$

Answer 1

Answer: (D)

$1:7:19$

Let $C{O}_1=h$. Then $C{O}_3=O_3{O}_2=O_2{O}_1=\frac{h}{3}$ ...(given)

Let $O_{1}B=r_1$, $O_{2}S=r_2$ and $O_{3}Q=r_3$

Since $\Delta CPQ\sim \Delta CAB$,

$\therefore \frac{O_{3}Q}{O_{1}B}=\frac{O_{3}C}{O_{1}C}$

or $r_3=\frac{r_1}{3}$ ......(i)

Also, since $\Delta CRS\sim \Delta CAB$

$\therefore \frac{O_{2}S}{O_{1}B}=\frac{C{O}_2}{O_{1}C}$

or $r_2=\frac{2r_1}{3}$ .....(ii)

Now, volume of cone CPQ
$V_1=\frac{1}{3}\pi .r_3^2.\frac{h}{3}=\frac{1}{81}\pi r_1^{2}h$ ...(iii)
Volume of frustum PQRS,

$V_2=\frac{\pi .\frac{h}{3}}{3}\left[\right(\frac{2r}{3}{)}^2+(\frac{r}{3}{)}^2+\frac{r}{3}.\frac{2r}{3}]$
$=\frac{\pi r^{2}h}{81}\times 7$

Volume of frustum ABSR,

$V_3=\frac{\pi .\frac{h}{3}}{3}[r^2+(\frac{2r}{3}{)}^2+r.\frac{2r}{3}]$
$=\frac{\pi r^{2}h}{81}\times 91$ ..........(v)

From equations (iii), (iv) and (v), we get

$V_1:V_2:V_3=1:7:19$