A metallic right circular cone 20cm high and whose vertical angle is 90° is cut into two parts at the middle point of its axis by a plane parallel to base. If frustum so obtained bee
drawn into a wire of diameter
(1/16) cm find length of the wire?
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 of a point O' such that VO' = O'O i.e O' is the mid-point of VO.
Let r1 and r2 be the radii of circular ends of the frustum ABB'A'
In triangle VOA we have
tan 45° = `(OA)/(VO)`
`rArr r_1 = 20 cm`
Now, in Δ VO'A' we have
`rArr 1= r_2/10`
`rArr r_2 = 10 cm`
∴ volume of the frustum
Volume of the frustum
Let the length of the wire of 1/16 cm
diameter be l cm. then,
radius of wire = 1/32 cm
Volume of the metal used in wire `= (pil)/1024 cm^2`
Since the frustum is recast into a wire of lenght l cm and diameter 1/16 cm.
∴ Volume of the metal used in wire = Volume of the frustum
`rArr (pil)/1024 = 7000(pi)/3`
`rArr l= 700/3xx1024`
= 2389333.33 cm
= 23893.33 m
Hence, the lenght of wire is 23893.33 m.