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An inverted cone has a depth 10 cmts and a base of radius 5 cmts. Water is poured into it at the rate of 1.5 cubic cmts per second. The rate at which water is rising when the depth is 4 cmts is

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a

$\frac{8}{3 π}$ cmts/sec

b

$\frac{3}{8 π}$ cmts/sec

c

$\frac{5}{x}$ cmts/sec

answer is A.

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Detailed Solution

$h = 10 r = 5 \frac{dv}{dt} = \frac{3}{2} \frac{h}{r} = 2 \Rightarrow r = \frac{h}{2} h = 4 V = \frac{1}{3} {πr}^{2} h = \frac{1}{3} π \times \frac{h^{2}}{4} h \Rightarrow V = \frac{1}{2} {πh}^{3} \frac{dv}{dt} = \frac{1}{4} {πh}^{2} \frac{dh}{dt} \Rightarrow \frac{3}{2} = \frac{1}{4} \times π \times 16 \times \frac{dh}{dt} \frac{dh}{dt} = \frac{3}{8 π}$

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