a solid metallic right circular cone 20 cm high with vertical angle 60 is cut into two parts at the - Maths - Surface Areas and Volumes

Question

a solid metallic right circular cone 20 cm high with vertical
angle 60 is cut into two parts at the middle point of its height by
a plane parallel to the base its frustum, so obtained be drawn into
a wire if diameter 1/16 cm, find the length of the wire

Gursheen kaur. · Accepted Answer

In Î”AEG, $\tan 30^{0} = \frac{E G}{A G} \Rightarrow \frac{1}{\sqrt{3}} = \frac{E G}{10} \Rightarrow E G = \frac{10}{\sqrt{3}} = \frac{10 \sqrt{3}}{3} c m$ In Î”ABD, $\tan 30^{0} = \frac{B D}{A D} \Rightarrow \frac{1}{\sqrt{3}} = \frac{B D}{20} \Rightarrow B D = \frac{20}{\sqrt{3}} = \frac{20 \sqrt{3}}{3} c m$ Radius (r1) of upper end of frustum = $\frac{10 \sqrt{3}}{3} c m$ Radius (r2) of lower end of container = $\frac{20 \sqrt{3}}{3} c m$ Height (h) of container = 10 cm Volume of frustum = $\frac{1}{3} π h ({r_{1}}^{2} + {r_{2}}^{2} + r_{1} r_{2})$ $= \frac{1}{3} \times \frac{22}{7} \times 10 [{(\frac{10 \sqrt{3}}{3})}^{2} + {(\frac{20 \sqrt{3}}{3})}^{2} + \frac{10 \sqrt{3}}{3} \times \frac{20 \sqrt{3}}{3}] = \frac{220}{21} [\frac{100}{3} + \frac{400}{3} + \frac{200}{3}] = \frac{220}{21} [\frac{700}{3}] = \frac{220}{3} [\frac{100}{3}] = \frac{22000}{9} {c m}^{3}$ Radius (r) of wire = $\frac{1}{16} \times \frac{1}{2} = \frac{1}{32} c m$ Let the length of wire =l Volume of wire = Area of cross-section Ã— Length = (Ï€r2) (l) $= π {(\frac{1}{32})}^{2} l$ Now, Volume of frustum = Volume of wire $\frac{22000}{9} = π {(\frac{1}{32})}^{2} l \Rightarrow \frac{22000}{9} = \frac{22}{7} {(\frac{1}{32})}^{2} l \Rightarrow l = \frac{22000}{9} \times \frac{7}{22} \times {(32)}^{2} \Rightarrow l = \frac{7000}{9} \times 1024 \Rightarrow l = 796444 44 c m$

a solid metallic right circular cone 20 cm high with vertical angle 60 is cut into two parts at the middle point of its height by a plane parallel to the base. its frustum, so obtained be drawn into a wire if diameter 1/16 cm, find the length of the wire .