JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Application of Derivatives

Water is being filled at the rate of

1 cm^{3} / sec

in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm . When the height of the water level is 10 cm , the rate (

in cm^{2} / sec

) at which the wet conical surface area of the vessel increases is

Answer: (C)

Solution

From the cone dimensions,

\frac{r}{h} = \frac{7}{35} = \frac{1}{5}

, so

h = 5 r

.

Given

\frac{d V}{d t} = 1

. Since

V = \frac{π r ^{2} h}{3} = \frac{π r ^{2} ( 5 r )}{3} = \frac{5 π r ^{3}}{3}

\frac{d}{d t} (\frac{5 π r ^{3}}{3}) = 1 \Rightarrow 5 π r^{2} \frac{d r}{d t} = 1 \Rightarrow r^{2} \frac{d r}{d t} = \frac{1}{5 π}

The wet conical surface area is

S = π r ℓ = π r h^{2} + r^{2} = π r 25 r^{2} + r^{2} = 26 π r^{2}

\frac{d S}{d t} = 226 π r \frac{d r}{d t}

When

h = 10

r = 2

. Thus:

\frac{d S}{d t} = 226 π (2) \cdot \frac{1}{5 π ( 4 )} = \frac{4 26 π}{20 π} = \frac{26}{5}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option