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EAMCET Engineering 2010 Solved Paper
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© examsnet.com
Question : 39
Total: 160
The height of the cone of maximum volume inscribed in a sphere of radius
R
is
R
3
2
R
3
4
R
3
4
R
√
3
Validate
Solution:
Let the height of the cone
=
h
and the radius of the cone
=
r
Given, radius of the sphere
=
R
Now, In
△
O
P
B
⇒
R
2
=
r
2
+
(
h
−
R
)
2
⇒
r
2
=
R
2
−
(
h
−
R
)
2
=
(
R
+
h
−
R
)
(
R
−
h
+
R
)
⇒
r
2
=
h
(
2
R
−
h
)
The volume of the cone is
V
=
1
3
π
r
2
h
⇒
V
=
1
3
π
h
(
2
R
−
h
)
h
⇒
V
=
π
3
(
2
R
h
2
−
h
3
)
Differentiating with
r
to
h
d
V
d
h
=
π
3
(
4
R
h
−
3
h
2
)
For maximum or minimum value of volume
d
V
d
h
=
0
⇒
π
3
(
4
R
h
−
3
h
2
)
=
0
⇒
h
(
4
R
−
3
h
)
=
0
⇒
h
=
0
,
h
=
4
R
3
(Not possible)
Now,
d
2
V
d
h
2
=
π
3
(
4
R
−
6
h
)
(
d
2
V
d
h
2
)
(
at
h
=
4
R
3
)
=
π
3
(
4
R
−
6
⋅
4
R
3
)
=
π
3
(
4
R
−
8
R
)
=
−
4
π
3
R
⇒
Negative
ie., Maximum
Hence, the height of the cone of maximum volume is
(
4
R
3
)
.
© examsnet.com
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