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 \[{{V}_{1}}:{{V}_{2}}:{{V}_{3}}=1:7:19\]You need to login to perform this action.
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