question_answer

The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lower part) of the cone is 44 cu cm. The radius of the upper circular surface of the frustum \[(taking\,\pi =\frac{22}{7})\] is

A) \[\sqrt[3]{12}\,cm\]

B) \[\sqrt[3]{13}\,cm\]

C) \[\sqrt[3]{6}\,cm\]

D) \[\sqrt[3]{6}\,cm\]

Correct Answer: B

Solution :

[b] Here, \[\Delta \,AO'B'\] and \[\Delta \,AOB\]are similar, let 00' = h and OB' = r \[\therefore \]      \[\frac{AO}{AO'}=\frac{OB}{O'B'}\] \[\Rightarrow \]   \[\frac{9}{9-h}=\frac{3}{r}\] \[\Rightarrow \]   \[3r=9-h\] \[\Rightarrow \]   \[h=9-3r\] Now, volume of frustum = 44 \[\Rightarrow \]   \[\frac{1}{3}\pi h\,({{R}^{2}}+{{r}^{2}}+Rr)=44\] \[\Rightarrow \]   \[\frac{1}{3}\times \frac{22}{7}\times (9-3r)(9+{{r}^{2}}+3r)=44\] \[\Rightarrow \]   \[(3-r)(9+{{r}^{2}}+3r)=2\times 7\] \[\Rightarrow \]   \[{{3}^{3}}-{{r}^{3}}=14\] \[\Rightarrow \]   \[{{r}^{3}}=27-14\] \[\Rightarrow \]   \[{{r}^{3}}=13\] \[\Rightarrow \]   \[r=\sqrt[3]{13}\,cm\]