Question-About Frustum

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 diameter1/16 cm, find the length of the wire. Solution=Let VAB be the solid metallic right circular cone of height 20 cm. Suppose, this cone is cut by a plane parallel to its base at a point O' such that V O^ prime =0^ prime 0,i i.e. O' is the mid-point of VO. Let r1and r2 be the radii of circular ends of the frustum ABB'A'. In right triangles VOA and VO'A', we have

tan 30 deg = (OA)/(VO) and tan 30 deg = O^ prime A^ prime VO^ prime 1(sqrt(3)) = r 1 20 : and 1/(sqrt(3)) = r_{2}/10 r_{1} = 20/(sqrt(3)) * cmi and r_{2} = 10/(sqrt(3)) cm Volume of the frustum = 1/3 * pi(r_{1} ^ 2 + r_{2} ^ 2 + r_{1}*r_{2}) * h Volume of the frustum = pi/3 * (400/3 + 100/3 + 200/3) * 10c * m ^ 3 = 7000/9 * pi*c * m ^ 3 Let the length of the wire of 1/16cm diameter be l. Then, Volume of the metal used in wire = pi * (1/32) ^ 2 * l Volume of the metal used in wire = (pi*l)/1024 Since the frustum is recast into a wire of length 1 cm and diameter 1/16 * cm Volume of the metal used in wire = Volume of the frustum pi/1024 = (7000pi)/9 介 l = (7000pi)/9 * 1024/pi * cm = 7000/9 * 1024cm = 796444.4 cm = 7964.44 m

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