Question

Ankit has a right circular cylinder which he inserted completely into a right circular cone of height 20 cm. The vertical angle of the cone is 60º and the diameter of the cylinder is 10√3 cm. The volume of the cone is

(a)\(\frac{4000}{7} \pi \,\,{cm}^3\) 

(b) \(\frac{8000}{3} \pi \,\,{cm}^3\)

(c) \(\frac{8000}{9} \pi \,\,{cm}^3\)

(d) \(\frac{3000}{7} \pi \,\,{cm}^3\)

Answer 1

Answer : (c) = \(\frac{8000}{9} \pi \,\,{cm}^3\)

The cylinder inside the cone is shown in Fig. (a). 

Here, AQ = 20 cm, PG = QE = \(\frac{1}{2}\)DE = \(5\sqrt{3}\) cm  and ∠CAB = 60º.

The cross-section as a plane figure is shown in Fig. (b).

\(\frac{PG}{AP} =\) tan 30⁰ ⇒ \(\frac{5\sqrt{3}}{AP} = \frac{1}{\sqrt{3}}\) 

⇒ AP = 15 cm.

Let QB = r be the radius of the cone. 

Then, in similar triangles APG and AQB, 

\(\frac{AP}{PG} = \frac{AQ}{QB} \)  ⇒ QB = \(\frac{AQ.PG}{AP}\)  

= \(\frac{20 \times 5\sqrt{3}}{15}\)  

= \(\frac{20}{\sqrt{3}}\) cm.

∴ Radius of cone = \(\frac{20}{\sqrt{3}}\)  cm , Height of cone = 20 cm.

∴ The volume of cone = \(\frac{1}{3} \pi r^2h\) 

= \(\frac{1}{3}\times \pi \times \big(\frac{20}{\sqrt{3}}\big)^2\times 20\) 

= \(\frac{8000}{9} \pi \,\,{cm}^3\)