question_answer

If the right circular cone is separated into three solids of volume \[{{\upsilon }_{1}},{{\upsilon }_{2}},{{\upsilon }_{3}}\] by two planes parallel to the base and trisect the altitude, then \[{{\upsilon }_{1}},{{\upsilon }_{2}},{{\upsilon }_{3}}\]is

A)  1: 2 : 3            

B)         1 : 4 : 6

C)  1 : 6 : 9            

D)         1:7:19  

Correct Answer: D

Solution :

 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 \[\Delta \,CPQ=\Delta \,CAB,\]  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 \[\Delta \,CRS=\Delta \,CAB\] \[\therefore \] \[\frac{{{O}_{2}}S}{{{O}_{1}}B}=\frac{C{{O}_{2}}}{{{O}_{1}}C}\] or            \[{{r}_{2}}=\frac{2{{r}_{1}}}{3}\]                                                ?..(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,                 \[{{V}_{2}}=\frac{\pi .\frac{h}{3}}{3}\left[ {{\left( \frac{2r}{3} \right)}^{2}}+{{\left( \frac{r}{3} \right)}^{2}}+\frac{r}{3}.\frac{2r}{3} \right]\]                 \[=\frac{\pi {{r}^{2}}h}{81}\times 7\]                                      ?..(iv) Volume of frustum ABSR,                 \[{{V}_{3}}=\frac{\pi .\frac{h}{3}}{3}\left[ {{r}^{2}}+{{\left( \frac{2r}{3} \right)}^{2}}+r.\frac{2r}{3} \right]\]                 \[=\frac{\pi {{r}^{2}}h}{81}\times 91\]                                    ?..(v) From equations (Hi), (iv) and (u), we get                 \[{{V}_{1}}:{{V}_{2}}:{{V}_{3}}=1:7:19\]