question_answer

A right angled sector of radius \[r\,\,cm\]is rolled up into a cone in such a way that two binding radii are joined together. Then, the curved surface area of the cone is                                                                            [SSC (10+2) 2011]

A) \[\pi {{r}^{2}}c{{m}^{2}}\]              

B) \[4\pi {{r}^{2}}c{{m}^{2}}\]

C) \[\frac{\pi {{r}^{2}}}{4}c{{m}^{2}}\]

D) \[2\pi {{r}^{2}}c{{m}^{2}}\]

Correct Answer: C

Solution :

Length of arc, \[x=\frac{\theta }{360{}^\circ }\times 2\pi r\]

\[x=\frac{\theta }{360{}^\circ }\times 2\pi r\]

\[\Rightarrow \]   \[x=\frac{90{}^\circ }{360{}^\circ }\times 2\pi r=\frac{\pi r}{2}\]

Now, length of arc will be circumference of base and let radius of base be \[{{r}_{1}}.\]

\[\therefore \]      \[2\pi {{r}_{1}}=x,\]\[2\pi {{r}_{1}}=\frac{\pi r}{2}\]

\[{{r}_{1}}=\frac{r}{4}\]and slant height \[=r\]

\[\therefore \]Curved surface area\[=\pi {{r}_{1}}l=\pi \times \frac{r}{4}\times r=\frac{\pi {{r}^{2}}}{4}c{{m}^{2}}\]