1. **State the problem:** Water is flowing into a conical vessel with a depth of 15 cm and a top radius of 3.75 cm. The water level is rising at a rate of 12 cm/min. We need to find the rate at which water is flowing into the vessel (volume rate) when the water depth is 4 cm. 2. **Relevant formula:** The volume $V$ of a cone is given by $$V = \frac{1}{3} \pi r^2 h,$$ where $r$ is the radius of the water surface and $h$ is the depth of the water. 3. **Relate radius and height:** Since the vessel is a cone, the radius and height are proportional. The full cone has radius 3.75 cm and height 15 cm, so $$\frac{r}{h} = \frac{3.75}{15} = \frac{1}{4}.$$ Thus, $$r = \frac{h}{4}.$$ 4. **Express volume in terms of height only:** Substitute $r = \frac{h}{4}$ into the volume formula: $$V = \frac{1}{3} \pi \left(\frac{h}{4}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{16} h = \frac{\pi}{48} h^3.$$ 5. **Differentiate volume with respect to time $t$:** $$\frac{dV}{dt} = \frac{\pi}{48} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{16} h^2 \frac{dh}{dt}.$$ 6. **Plug in known values:** When $h = 4$ cm and $\frac{dh}{dt} = 12$ cm/min, $$\frac{dV}{dt} = \frac{\pi}{16} \times 4^2 \times 12 = \frac{\pi}{16} \times 16 \times 12 = \pi \times 12 = 12\pi \approx 37.7 \text{ cm}^3/\text{min}.$$ 7. **Check options:** None of the options match 37.7 cm³/min exactly, but the closest is option d. 6.28 cm³/min which is $2\pi$. This suggests a possible misinterpretation or typo in the problem or options. **Final answer:** The rate at which water is flowing into the vessel when the depth is 4 cm is $$\boxed{12\pi \approx 37.7 \text{ cm}^3/\text{min}}.$$