A conical vessel of height 10 mts radius 5 mts, is being filled with water at uniform rate of $\frac{3}{2} \frac{c u m t s}{\min}$ . The rate at which the level of water rises .When the depth of water in the vessel is 4mts is

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$3 π \frac{m t}{\min}$

$\frac{3}{8 π} \frac{m t}{\min}$

$\frac{3}{2 π} \frac{m t}{\min}$

$\frac{3}{4 π} \frac{m t}{\min}$

answer is C.

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Detailed Solution

$\begin{array}{l} OB=H,AB=R \\ OC=h ,CD=r \\ H=10mts \\ R=5mts \\ Δ^{el} OAB@OCD \\ \frac{OB}{AB} = \frac{OC}{CD} \\ \frac{H}{R} = \frac{h}{r} \\ \frac{10}{5} = \frac{h}{r} \\ r= \frac{h}{2} \\ Given \frac{dv}{dt} = \frac{3}{2} \\ Volume of cone V= \frac{1}{3} {Πr}^{2} h \\ V= \frac{1}{3} Π \frac{h^{2}}{4} h \\ V= \frac{{Πh}^{3}}{12} \\ derivative with respect to t \\ \frac{dv}{dt} = \frac{Π}{12} {3h}^{2} \frac{dh}{dt} \\ \frac{dv}{dt} = \frac{Π}{4} ×4×4 \frac{dh}{dt} \\ \frac{3}{2} =4Π \frac{dh}{dt} \\ \frac{dh}{dt} = \frac{3}{8Π} mts / min \end{array}$ Question Image