2025 EJC P1 Q11

A cylindrical container of height \(h\) metres is originally completely filled with water. The water is flowing out through an opening at the base of the container into a pipe. The rate of flow of water through the opening at time \(t\) seconds is proportional to the square root of the height of the water in the container, \(y\) metres. Since the container is cylindrical, the amount of water in the container is also proportional to the height of water, so the differential equation relating \(y\) and \(t\) can be written as

\[\frac{\mathrm{d}y}{\mathrm{d}t} = -k\sqrt{y}\]

where \(k\) is a positive constant.

1. Solve this differential equation to find \(\sqrt{y}\) in terms of \(t\), \(h\) and \(k\).[4]
2. Hence find the time taken for the container to empty, in terms of \(h\) and \(k\).[2]

It is found that the water was draining too fast. To regulate the flow of water, a mechanical device is placed at the opening. This device slows the flow of water into the pipe by covering all or part of the opening, where the area covered varies with time. It is proposed to model this situation using the revised differential equation

\[\frac{\mathrm{d}y}{\mathrm{d}t} = -k\left( \frac{1 + \cos t}{2} \right)\sqrt{y},\]

where \(k\) is the same constant as before.

3. With reference to the possible values of \(\cos t\), explain how the revised differential equation corresponds to a model with slower flow of water.[1]
4. Given again that the container is originally completely filled with water, solve the revised differential equation to find \(\sqrt{y}\) in terms of \(t\), \(h\) and \(k\).[3]
5. In the case where \(h = 10\) and \(k = 0.1\), find the ratio of the time taken in the revised model to the time taken in the original model for the container to empty. Give your answer to \(3\) significant figures.[2]