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UPSC 2016 CDS II Math Paper
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© examsnet.com
Question : 50
Total: 100
A building is in the form of a cylinder surmounted by a hemispherical dome on the diameter of the cylinder. The height of the building is three times the radius of the base of the cylinder. The building contains
67
1
21
m
3
of air. What is the height of the building ?
[2016 CDS-II]
6 m
4 m
3 m
2 m
Validate
Solution:
Let r be the radius of the base of the cylinder.
∴ Height of the building = 3r
Height of the hemispherical dome = Radius of the hemispherical dome = r
∴ Height of the cylindrical portion of the building
=
3
r
−
r
=
2
r
Volume of air in the building
=
67
1
21
=
1408
21
m
3
∴
2
3
π
r
3
+
π
r
2
×
2
r
=
1408
21
⇒
8
3
π
r
3
=
1408
21
⇒
r
3
=
1408
×
7
×
3
21
×
22
×
3
=
8
⇒
r
=
2
m
∴ Height of the building
=
3
×
2
=
6
m
© examsnet.com
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