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JEE Main Math Class 12 Application of Derivatives Part 1 Questions
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© examsnet.com
Question : 2
Total: 100
A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan
–
1
. Water is poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in m/min.), at which the level of water is rising at the instant when the depth of water in the tank is 10m; is:
[April 09, 2019 (II)]
1/15π
1/10π
2/π
1/5π
Validate
Solution:
Given that water is poured into the tank at a constant rate of
5
m
3
/
minute.
∴
d
v
d
t
=
5
m
3
∕
m
i
n
Volume of the tank is,
V
=
1
3
π
r
2
h
.,......(i)
where r is radius and h is height at any time.
By the diagram,
tan
θ
=
r
h
=
1
2
⇒
h
=
2
r
⇒
d
h
d
t
=
2
d
r
d
t
.......(ii)
Differentiate eq. (i) w.r.t. ‘t’, we get
d
V
d
t
=
1
3
(
π
2
r
d
r
d
t
h
+
π
r
2
d
h
d
t
)
Putting
h
=
10
,
r
=
5
and
d
V
d
t
=
5
in the above equation.
5
=
75
π
3
d
h
d
t
⇒
d
h
d
t
=
1
5
π
m
∕
m
i
n
.
© examsnet.com
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