A metallic wire right circular cone …

According to the question,A metallic right circular cone 20 cm high and whose vertical angel is 60° is cut into two parts at the middle of its height by a plane parallel to its base if the frustum so obtained be drawn into a wire of uniform diameter {tex}\frac { 1 } { 16 }{/tex} cm.
Total height of cone = 20 cm and Vertex angle = 30°
Let the radius of cone be r{tex}_2{/tex}.
{tex}\therefore \frac { r _ { 2 } } { 20 } = \tan 30 ^ { \circ } \Rightarrow \frac { 1 } { \sqrt { 3 } }{/tex} 
{tex}r _ { 2 } = \frac { 20 } { \sqrt { 3 } } \mathrm { cm }{/tex} 
The height of the cone cut off = 10 cm Let its radius be r₁
{tex}\Rightarrow \frac { r _ { 1 } } { 10 } = \tan 30 ^ { \circ } \Rightarrow r _ { 1 } = \frac { 10 } { \sqrt { 30 } } \mathrm { cm }{/tex} 
Let the length of wire be l
Its radius {tex}= \frac { 1 } { 32 } \mathrm { cm }{/tex} 
{tex}\therefore{/tex} Volume of frustum = Volume of wire
{tex}\Rightarrow \frac { 1 } { 3 } \pi \times h \left[ \left( r _ { 1 } \right) ^ { 2 } + \left( r _ { 2 } \right) ^ { 2 } + \left( r _ { 1 } r _ { 2 } \right) \right] = \pi r ^ { 2 } l{/tex} 
{tex}\Rightarrow \frac { 1 } { 3 } \times 10 \times \pi \left[ \left( \frac { 10 } { 3 } \right) ^ { 2 } + \left( \frac { 20 } { 3 } \right) ^ { 2 } + \frac { 10 } { 3 } \times \frac { 20 } { 3 } \right]{/tex} 
{tex}= \pi \left( \frac { 1 } { 32 } \right) ^ { 2 } \times l{/tex} 
{tex}\Rightarrow \frac { 1 } { 3 } \times 10 \left[ \frac { 100 } { 9 } + \frac { 400 } { 9 } + \frac { 200 } { 9 } \right] = \frac { 1 } { 32 \times 32 } \times l{/tex} 
{tex}\Rightarrow \frac { 1 } { 3 } \times 10 \times \frac { 700 } { 9 } = \frac { 1 } { 32 } \times \frac { 1 } { 32 } \times l{/tex} 
{tex}\Rightarrow l = \frac { 32 \times 32 \times 700 \times 10 } { 3 \times 9 }{/tex}
= 796444.4 cm.
Hence, the length of wire = 7964.44 m