Application of Derivatives 2 Question 3

####3. A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan112. Water is poured into it at a constant rate of 5cum/min. 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 10 m is

(2019 Main, 9 April II)

(a) 2π

(b) 15π

(c) 115π

(d) 110π

Show Answer

Answer:

(b)

Solution:

Key Idea Use formula: Volume of cone =13πr2h, where r= radius and h= height of the cone.

Given, semi-vertical angle of right circular cone

=tan112

Let α=tan112

tanα=12

rh=12 [from fig. tanα=rh ]

r=12h

Volume of cone is (V)=13πr2h

V=13π12h(h)=112πh3 [from Eq. (i)]

On differentiating both sides w.r.t. ’ t ‘, we get

dVdt=112π(3h2)dhdtdhdt=4πh2dVdtdhdt=4πh2×5[ given dVdt=5 m3/min]

Now, at h=10 m, the rate at which height of water level is rising =dhdt|h=10

=4π(10)2×5=15πm/min



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