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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$3 π \frac{m t}{\min}$

$\frac{3}{8 π} \frac{m t}{\min}$

$\frac{3}{2 π} \frac{m t}{\min}$

$\frac{3}{4 π} \frac{m t}{\min}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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