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UPSC CDS Math Volume and Surface Area Questions
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© examsnet.com
Question : 26
Total: 104
A right circular cone is sliced into a smaller cone and a frustum of a cone by a plane perpendicular to its axis. The volume of the smaller cone and the frustum of the cone are in the ratio 64 : 61. Then their curved surface areas are in the ratio
[2018 CDS-I]
4 :1
16 : 9
64 : 61
81 : 64
Validate
Solution:
Volume of cone ADE
Volume of frustum DEBC
=
64
61
∴
Volume of cone ADE
Volume of cone ABC
=
64
61
+
64
=
64
125
1
3
π
r
1
2
h
1
1
3
π
r
2
2
h
2
=
64
125
⇒
r
1
2
h
1
r
2
2
h
2
=
64
125
⇒
r
1
3
r
2
3
=
64
125
(Since
,
r
1
r
2
=
h
1
h
2
=
l
1
l
2
)
∴
r
1
r
2
=
4
5
Now,
Curved surface area of cone ADE
Curved surface area of cone ABC
=
π
r
1
l
1
π
r
2
l
2
=
r
1
2
r
2
2
=
16
25
∴
Curved surface area of cone ADE
Curved surface area of frustum DEBC
=
16
25
−
16
=
16
9
© examsnet.com
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