question_answer

A sector of a circle of radius 12 cm has the angle \[{{120}^{\text{o}}}\]. It is rolled up so that two bounding radii are joined together to form a cone. Find the volume of the cone.    

A)  \[189.61\text{ }c{{m}^{3}}\]   

B)         \[169.51\text{ }c{{m}^{3}}\]    

C)         \[179.61\text{ }c{{m}^{3}}\]   

D)         \[125.51\text{ }c{{m}^{3}}\]         

Correct Answer: A

Solution :

Length of arc of circle  \[=\frac{\theta }{{{360}^{o}}}\times 2\pi r\] \[=\frac{{{120}^{o}}}{{{360}^{o}}}\times 2\times \pi \times 12=8\pi \,cm\]             \[\therefore \] Circumference of the base of the cone \[=8\pi \] \[\Rightarrow \]  \[2\pi {{r}_{1}}=8\pi \,\,\,\,\Rightarrow \,\,\,\,{{r}_{1}}=4\,cm\] Now, Slant height of cone \[(l)=12\,cm\] Height of the cone \[=\sqrt{{{l}^{2}}-r_{1}^{2}}=8\sqrt{2}\] Volume of the cone             \[=\frac{1}{3}\pi r_{1}^{2}h=\frac{1}{3}\times \frac{22}{7}\times 4\times 4\times 8\sqrt{2}\] \[=\frac{2816\times 1.414}{21}=189.61c{{m}^{3}}\]