1. **State the problem:** Water is pumped into an inverted conical tank at a rate of 300000 cm³/min. The tank is 6 m high with a top diameter of 4 m. We need to find the rate at which the water level $h$ is rising when $h=2$ m. 2. **Convert units:** Since the volume rate is in cm³/min, convert tank dimensions to cm: height $H=600$ cm, diameter $D=400$ cm, radius $R=200$ cm. 3. **Relate radius and height:** The radius $r$ of the water surface relates to height $h$ by similar triangles: $$\frac{r}{h} = \frac{R}{H} = \frac{200}{600} = \frac{1}{3} \implies r = \frac{h}{3}$$ 4. **Volume formula for a cone:** $$V = \frac{1}{3} \pi r^2 h$$ Substitute $r = \frac{h}{3}$: $$V = \frac{1}{3} \pi \left(\frac{h}{3}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{9} h = \frac{\pi h^3}{27}$$ 5. **Differentiate volume with respect to time $t$:** $$\frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi h^2}{9} \frac{dh}{dt}$$ 6. **Given:** $$\frac{dV}{dt} = 300000 \text{ cm}^3/\text{min}, \quad h=200 \text{ cm}$$ 7. **Solve for $\frac{dh}{dt}$:** $$300000 = \frac{\pi (200)^2}{9} \frac{dh}{dt}$$ $$300000 = \frac{\pi \times 40000}{9} \frac{dh}{dt}$$ 8. **Isolate $\frac{dh}{dt}$:** $$\frac{dh}{dt} = \frac{300000 \times 9}{\pi \times 40000} = \frac{2700000}{40000 \pi} = \frac{67.5}{\pi} \approx 21.5 \text{ cm/min}$$ 9. **Convert to meters per minute:** $$21.5 \text{ cm/min} = 0.215 \text{ m/min}$$ **Final answer:** The water level is rising at approximately **0.2 m/min** when the water height is 2 m (to 1 decimal place).