COMEDK 2025 Morning Solved Paper

Concept:Use similar triangles to relate the radius of the oil surface to its depth, then apply differentiation to the cone volume formula to find the rate of change of height.Explanation:Let $r$ be the radius of the oil surface and $h$ its vertical distance from the tip of the cone.From similar triangles, $\frac{r}{h} = \frac{10}{20}$, so $r = \frac{h}{2}$.The volume of oil inside the cone is $V = \frac{1}{3}\pi r^2 h = \frac{\pi}{12}h^3$.Differentiate with respect to time $t$: $\frac{dV}{dt} = \frac{\pi}{4}h^2 \cdot \frac{dh}{dt}$.Given the oil drips out at $5$ cm$^3$/s, so $\frac{dV}{dt} = -5$ (negative because volume decreases).The oil level is $5$ cm from the top, meaning its distance from the tip is $h = 20 - 5 = 15$ cm.Substitute: $-5 = \frac{\pi}{4}(15)^2 \cdot \frac{dh}{dt}$.Solve for $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{-5}{\frac{\pi}{4} \cdot 225} = -\frac{20}{225\pi} = -\frac{4}{45\pi}$ cm/s.Answer:$-\frac{4}{45\pi}$ cm/s, which corresponds to option D.