Water is dropping out at the steady rate of $2 c c / s e c$ through a tiny hole at the vertex of a conical funnel, whose axis is vertical. When the slant height of the water is 4 cm then the rate of decrease of the slant height of water is [Given that the vertex angle of the funnel is $120^{0}$ ].

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$\frac{1}{3 π} c m / s e c$

$\frac{1}{π} c m / s e c$

$\frac{2}{3 π} c m / s e c$

$\frac{4}{3 π} c m / s e c$

answer is D.

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Detailed Solution

$\frac{d v}{d t} = 2 c c / s, l = 4 c m$

$r = l \sin 60^{0} = \frac{\sqrt{3} l}{2}$

$h = l \cos 60^{0} = \frac{l}{2}$

$v = \frac{l}{3} {πr}^{2} h$

$v = \frac{1}{3} π . \frac{3 l^{2}}{4} . \frac{l}{2}$

$v = \frac{π}{8} l^{3} \Rightarrow \frac{d v}{d t} = \frac{π}{8} 3 l^{2} . \frac{d l}{d t}$

$\frac{d v}{d t} = \frac{π}{8} (3 \times 4^{2}) \frac{d l}{d t}$

$2 = \frac{π}{8} \times 3 \times 16 . \frac{d l}{d t}$

$\frac{d l}{d t} = \frac{l}{3 π} c m / s$

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