If water is poured into an inverted hollow cone whose semivertical angle is given by argument of $z_{1}$ . The depth of water increases at a rate equal to the least value of $| z_{2} + z_{3} | c m / s$ . The rate at which the volume of water increases when the depth is $9 c m$ is $m π c m^{3} / s$ . Where $z_{1} = 11 + (\sqrt{2} i)^{2} + \sqrt{27} i, z_{2} = 8 + 15 i, | z_{3} | = 7$ , $(i = \sqrt{- 1})$ . The value of $m$ is

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Solution

$α = {tan}^{- 1} (\frac{\sqrt{27}}{11 + (\sqrt{2} i)^{2}})$
$= {tan}^{- 1} (\frac{3 \sqrt{3}}{11 - 2})$
$= {tan}^{- 1} \frac{1}{\sqrt{3}} = 30^{\circ}$
$∴ tan 30^{\circ} = \frac{O^{'} D}{h}$
$\Rightarrow O^{'} D = \frac{h}{\sqrt{3}} \Rightarrow r = \frac{h}{\sqrt{3}}$
$| z_{2} + z_{3} | \geq | | z_{2} | - | z_{3} | | \geq | | 8 + 15 i | - 7 |$
$| z_{2} + z_{3} | \geq | 17 - 7 | \geq 10$
$\Rightarrow$ Least value of $| z_{2} + z_{3} |$ is $10.$
$\Rightarrow \frac{d h}{d t} = 10 c m / s$
The volume of water in the cone is,
$V = \frac{1}{3} π (O^{'} D)^{2} (A O^{'}) = \frac{1}{3} π (\frac{h}{\sqrt{3}})^{2} h$
$V = \frac{π h^{3}}{9} \Rightarrow \frac{d V}{d t} = \frac{π}{3} h^{2} \frac{d h}{d t}$
$\frac{d V}{d t} = \frac{π}{3} 9^{2} \times 10 = 270 π c m^{3} / s ∴ m = 270$

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