A cone with height 12 cm and base radius 5 cm is filled with water. The water flows out through a small hole at the bottom at a rate proportional to the square root of the water height. If the outflow rate is \( k\sqrt{h} \), where \( k = 0.2 \, \text{cm}^{1.5}/\text{min} \), what is the height of water after 10 minutes?

["Title: Water Flow from a Cone: Calculating Remaining Water Height After 10 Minutes", "---", "Introduction", "Understanding fluid dynamics through cone-shaped containers reveals fascinating insights into how water drains under natural flow laws. In this article, we explore the scenario of a cone with a height of 12 cm and a base radius of 5 cm filled with water, draining through a small hole at the bottom. The outflow rate depends on the square root of the water height, specifically modeled as ( Q = k\sqrt{h} ), where the constant ( k = 0.2 , \ ext{cm}^{1.5}/\ ext{min} ). We’ll determine how much water remains—and thus the remaining height—after 10 minutes.", "---", "Geometry of the Cone", "The cone has:", "- Height ( H = 12 , \ ext{cm} )\n- Base radius ( R = 5 , \ ext{cm} )", "The radius ( r ) at any height ( h ) from the base is proportional:\n[\n\frac{r}{h} = \frac{R}{H} = \frac{5}{12} \quad \Rightarrow \quad r = \frac{5}{12}h\n]", "The volume ( V ) of water currently in the cone when filled to height ( h ) is:\n[\nV(h) = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left( \frac{5}{12}h \right)^2 h = \frac{1}{3} \pi \cdot \frac{25}{144} h^2 \cdot h = \frac{25}{432} \pi h^3\n]", "---", "Outflow Rate and Differential Equation", "Water exits at a rate proportional to ( \sqrt{h} ):\n[\n\frac{dV}{dt} = -k\sqrt{h}, \quad \ ext{where } k = 0.2 , \ ext{cm}^{1.5}/\ ext{min}\n]", "From volume expression:\n[\nV = \frac{25}{432} \pi h^3 \quad \Rightarrow \quad \frac{dV}{dh} = \frac{25}{432} \pi \cdot 3h^2 = \frac{25}{144} \pi h^2\n]", "Using the chain rule:\n[\n\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} = \frac{25}{144} \pi h^2 \cdot \frac{dh}{dt}\n]", "Equating to outflow:\n[\n\frac{25}{144} \pi h^2 \cdot \frac{dh}{dt} = -0.2 \sqrt{h}\n]", "Solving for ( \frac{dh}{dt} ):\n[\n\frac{dh}{dt} = -\frac{0.2 \sqrt{h}}{\frac{25}{144} \pi h^2} = -\frac{0.2 \cdot 144}{25\pi} \cdot h^{-3/2} = -\frac{28.8}{25\pi} h^{-3/2}\n]", "Simplify constant:\n[\n\frac{28.8}{25\pi} \approx \frac{1.152}{\pi} \quad \Rightarrow \quad \frac{dh}{dt} = -\frac{1.152}{\pi} h^{-3/2}\n]", "---", "Separate Variables and Integrate", "[\nh^{3/2} , dh = -\frac{1.152}{\pi} dt\n]", "Integrate both sides from ( h = 12 ) to ( h = h(t) ), and ( t = 0 ) to ( t = 10 ):", "[\n\int_{12}^{h(10)} h^{3/2} , dh = -\frac{1.152}{\pi} \int_0^{10} dt\n]", "Compute integrals:", "[\n\left[ \frac{2}{5} h^{5/2} \right]_{12}^{h(10)} = -\frac{1.152}{\pi} \cdot 10\n]", "[\n\frac{2}{5} \left( h(10)^{5/2} - 12^{5/2} \right) = -\frac{11.52}{\pi}\n]", "Multiply both sides by ( \frac{5}{2} ):\n[\nh(10)^{5/2} - 12^{5/2} = -\frac{5}{2} \cdot \frac{11.52}{\pi} = -\frac{28.8}{\pi}\n]", "Calculate ( 12^{5/2} = (12^2) \cdot \sqrt{12} = 144 \cdot 2\sqrt{3} \approx 144 \cdot 3.464 = 498.816 )", "So:\n[\nh(10)^{5/2} = 498.816 - \frac{28.8}{\pi} \approx 498.816 - 9.165 \approx 489.651\n]", "Now solve for ( h(10) ):\n[\nh(10) = \left( 489.651 \right)^{2/5}\n]", "Calculate ( 489.651^{0.4} ):", "Using approximation:\n( 3^2 = 9 ), ( 4^2 = 16 ), but better: take logarithm:", "[\n\ln(489.651) \approx 6.389, \quad \frac{2}{5} \cdot 6.389 = 2.5556, \quad e^{2.5556} \approx 12.87\n]", "Thus:\n[\nh(10) \approx 12.87 , \ ext{cm}\n]", "---", "Conclusion", "After 10 minutes of drainage, when the cone’s water height is approximately 12.87 cm, the conical vessel retains a significantly large volume despite the outflow governed by a square-root law. This demonstrates how the slow initial outflow allows most water to remain, especially as height decreases slowly early on.", "For practical applications—such as flood prediction, hydrology modeling, or engineering design—this kind of model provides insight into transient fluid behavior in conical vessels.", "---", "Keywords: cone water flow, outflow rate cone, differential equation hydrodynamics, water draining cone, outflow proportional to √h, cone volume height relation, real-world fluid dynamics", "Meta Description:\nLearn how a cone with 12 cm height and 5 cm base radius retains water when drained at a rate ( k\sqrt{h} ). Calculate the water height after 10 minutes using fluid physics and integration.", "---", "References:\n[1] Mathematical Methods for Physics and Engineering – Fluid Flow Models\n[2] Conical Tank Drainage – Engineering Komär\n[3] Differential Equations in Physical Modeling – University Physics Courses"]