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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 is drawn into a wire of diameter 1/16 cm, find the length of the wire [use π=22/7]
Solution
In ΔAEG,
EG/AG = tan 30º
`EG = 10/sqrt3 cm = (10sqrt3)/3`
In ΔABD,
BD/AD = tan 30º
`BD = 20/sqrt3 = (20sqrt3)/3cm`
Radius (r1) of upper end of frustum = `(10sqrt3)/3 cm`
Radius (r2) of lower end of container = `(20sqrt3)/3cm `
Height (h) of container = 10 cm
Volume of frustum = `1/3pih(r_1^2+r_2^2+r_1r_2)`
`=1/3xxpixx10[((10sqrt3)/3)^2 + ((20sqrt3)/3)^2 + ((10sqrt3)(10sqrt3))/(3xx3)]`
`= 10/3pi[100/3+400/3+200/3]`
`=10/3xx22/7xx700/3=22000/9 cm^3`
Radius (r) of wire = = `1/16xx1/2 = 1/32cm`
Let the length of wire be l.
Volume of wire = Area of cross-section × Length
= (πr2) (l)
=pi(1/32)^2 xx l
Volume of frustum = Volume of wire
`22000/9 = 22/7xx(1/32)^2xxl`
`7000/9 xx1024 = l`
l = 796444.44 cm
= 7964.44 meters
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