JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Application of Derivatives

A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semivertical angle is

tan^{- 1} \frac{3}{4}

. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is ____.

Answer: 5

Solution

Given:

tan θ = \frac{3}{4} = \frac{r}{h}

and

\frac{d V}{d t} = 6

m³/hr.

Volume of cone:

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π h^{3} tan^{2} θ = \frac{3 π}{16} h^{3}

.

Differentiating:

\frac{d V}{d t} = \frac{9 π}{16} h^{2} \frac{d h}{d t} = 6

.

At

h = 4

\frac{d h}{d t} = \frac{2}{3 π}

m/hr.

Curved surface area:

S = π r ℓ = π r \cdot \frac{5 h}{4} = \frac{15 π}{16} h^{2}

(since

ℓ = \frac{5 h}{4}

from the geometry).

Differentiating:

\frac{d S}{d t} = \frac{15 π}{8} h \frac{d h}{d t}

.

At

h = 4

\frac{d S}{d t} = \frac{15 π}{8} \cdot 4 \cdot \frac{2}{3 π} = 5

m²/hr.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Application of Derivatives

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