JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Application of Derivatives

A water tank has the shape of an inverted right circular cone, whose semi vertical angle is

tan^{- 1} (\frac{1}{2})

. Water is poured in 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 10 m is:

Answer: (B)

Solution

The semi-vertical angle gives

tan θ = \frac{1}{2} = \frac{r}{h}

, so

r = \frac{h}{2}

and the volume is

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π \frac{h ^{2}}{4} h = \frac{π h ^{3}}{12}

. Differentiating with respect to time,

\frac{d V}{d t} = \frac{π h ^{2}}{4} \frac{d h}{d t}

. With

\frac{d V}{d t} = 5

and

h = 10

5 = \frac{π ( 100 )}{4} \frac{d h}{d t}

, so

\frac{d h}{d t} = \frac{20}{100 π} = \frac{1}{5 π}

m/min.

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option