QJEEM22Ma390a: JEE Main 2022 Mathematics - Class 12 JEEM22Ma390a Solution: A water tank has the shape of a right circular cone with axis…

$v = {1 \over 3}\pi {r^2}h$ ..... (i)

And $\tan \theta = {3 \over 4} = {r \over h}$ ...... (ii)

i.e. if $h = 4,\,r = 3$

$v = {1 \over 3}\pi {r^2}\left( {{{4r} \over 3}} \right)$

${{dv} \over {dt}} = {{4\pi } \over 9}3{r^2}{{dr} \over {dt}} \Rightarrow 6 = {{4\pi } \over 3}(9){{dr} \over {dt}}$

$\Rightarrow {{dr} \over {dt}} = {1 \over {2\pi }}$

Curved area $= \pi r\sqrt {{r^2} + {h^2}}$

$= \pi r\sqrt {{r^2} + {{16{r^2}} \over 9}}$

$= {5 \over 3}\pi {r^2}$

${{dA} \over {dt}} = {{10} \over 3}\pi r{{dr} \over {dt}}$

$= {{10} \over 3}\pi \,.\,3\,.\,{1 \over {2\pi }}$

$= 5$