a metallic right circular cone 20cm height and whose Vertical angle is 60 degree is cut into two parts at the middle of its height by a plane parallel to its base if frustum so obedient be drawn into a wire of diameter 1 by 16 cm find the length of the wire
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In △AEG
tan30=
AG
EG
3
1
=
10
EG
EG=
3
10
=
3
10
3
cm
In △ABD
tan30=
AD
BD
3
1
=
20
BD
EG=
3
20
=
3
20
3
cm
Radius (r
1
) of upper end of frustum=
3cm
10
3
Radius (r
2
) of lower end of container=
3cm
20
3
Height (h) of container=10cm
Volume of frustum=
3
1
πh(r
1
2
+r
2
2
+r
1
r
2
)
=
3
1
×
7
22
×10
⎣
⎢
⎡
(
3
10
3
)
2
+(
3
20
3
)
2
+
3
10
3
×
3
20
3
⎦
⎥
⎤
=
21
220
[
3
100
+
3
400
+
3
200
]
=
21
220
[
3
700
]
=
3
220
[
3
100
]
=
9
22000
cm
3
Radius (r) of wire=
16
1
×
2
1
=
32
1
cm
Let the length of wire=l
Volume of wire= Area of cross-section\times Length
=πr
2
l
=π(
32
1
)
2
l
Now,
Volume of frustum=Volume of wire
9
22000
=π(
32
1
)
2
l
⇒
9
22000
=
7
22
(
32
1
)
2
l
⇒
9
22000
×
22
7
×(32)
2
=l
⇒l=
9
7000
×1024
⇒l=796444.44cm
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