Question

A metallic right circular cone 20 cm high and whose vertical angle 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 diameter 1/16 cm, find the length of the wire.

Answer 1

$\huge\bf{Question:-}$

A metallic right circular cone 20 cm high and whose vertical angle 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 diameter 1/16 cm, find the length of the wire.

$\huge\bf{Solution:-}$

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Answer 2

${ \large{ \sf{ \underbrace{\underline{\bigstar \:Answer}}}}}$

➦ Let ABC is the metal lic cone and DE C B is the required frust um.

Let the radii of frus tum are${ \large{ \boxed{ \red{ \underline{ \bf \:r1 \:and \:r2}}}}}$

$\mapsto\bf{i.e.\: DP \:= \:r1 \:and \:BO\:=\: r2}$

Now from ΔADP and ΔABO,

r2 = h1 tan30

r2 = 101/√3

r2 = 10/√3

r1 = (h1 + h2 )tan30

r1 = 20 1/√3

r1 = 20/√3

$\textbf{Now volume of the fr ustu m DECB }$

$\sf{= (π h2 /3)(r12 + r1 r2 + r22 )}$

$\sf{ = (π 10/3){(20/√3)2 + 10/√3 20/√3 + (10/√3)2 }$

$\sf{= (π 10/3){400/3 + 200/3 + 100/3 }$

$\sf{= π 10/3 700/3}$

$/sf{ = π 7000/9}$

$\sf\color{red}Now\: let \:l\: is\: the\: length\: of \:the\: wire.$

Given diameter of the wire d = 1/16

So radius of the wire R = d/2 = 1/16 * 1/2 = 1/32

$\sf\color{red}Now \:volume \:of \:the\: frustum\: = /:volume \:of\: the\: wire\: drawn/: from\: it$

➦ $\sf{(π*7000)/9 = Π R2 l}$

➦ $\sf{l = (Π*7000)/(ΠR2 9)}$

➦ $\sf {l = (7000/{(1/32)2 9}$

➦ $\sf {l = (7000 32 32)/9}$

➦ $\sf {l = 7168000/9}$

➦ $\sf{l= 796444.444 cm}$

➦ $\sf {l = 796444.444/100 m}$

➦ $\sf {l = 7964.444 m}$

$\bold\pink{\fbox{\sf{(sin ce\: 100\: cm\: =\: 1 m)}}}$

${ \large{ \boxed{ \red{ \underline{ \bf \:So\: length\: of \:wire\: is\: 7964.444 m}}}}}$

Thank you :)