<p>A spherical container of radius $5$ m is formed by rotating the following circle $C$ about the $y$-axis.</p>
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<p>The container has negligible thickness, and the circle $C$ passes through the origin $O$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">State a cartesian equation of $C$.</div><p style="text-align: right;">[1]</p></li></ol>
<p>Initially the spherical container is completely filled with water. Two engineers are calculating the time needed for the container to be completely drained from a small circular hole at the bottom. The volume of water in the container at time $t$ seconds is denoted by $V$ m$^{3}$.</p>
<p>[The volume of a sphere of radius $r$ is $\frac{4}{3}\pi {{r}^{3}}$.]</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">The first engineer proposes that the rate of change of $V$ with respect to $t$ is a constant $k$.</div><p style="text-align: right;">[1]</p><ol class="i-list"><li><div class="list-item-paragraph">Write down a differential equation relating $V$, $t$ and $k$.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Determine, with justification, the sign of $k$.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Find $V$ in terms of $t$ and $k$, leaving your answer in exact form.</div><p style="text-align: right;">[2]</p></li></ol></li><li><div class="list-item-paragraph">The second engineer argues that the rate at which water flows out from the hole will be at its greatest in the beginning, and decreases as the depth of water in the container decreases. He suggests using Torricelli’s law, which says that</div><div class="list-item-paragraph" style="text-align: center">$\frac{\mathrm{d}V}{\mathrm{d}t}=-\alpha \sqrt{20h}$,</div><div class="list-item-paragraph">where $\alpha $ m$^{2}$ is the area of the circular hole at the bottom of the container, and $h$m is the depth of water in the container at time $t$ seconds. The radius of the hole at the bottom of the container is found to be constant at $1$ cm.</div><ol class="i-list"><li><div class="list-item-paragraph">Show that $\left( 10{{h}^{\frac{1}{2}}}-{{h}^{\frac{3}{2}}} \right)\frac{\mathrm{d}h}{\mathrm{d}t}=-\frac{\sqrt{5}}{5000}$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Hence find the numerical value of $t$ when the container is completely drained.</div><p style="text-align: right;">[4]</p></li></ol></li></ol>

A spherical container of radius $5$ m is formed by rotating the following circle $C$ about the $y$-axis.

The container has negligible thickness, and the circle $C$ passes through the origin $O$.

Initially the spherical container is completely filled with water. Two engineers are calculating the time needed for the container to be completely drained from a small circular hole at the bottom. The volume of water in the container at time $t$ seconds is denoted by $V$ m$^{3}$.

[The volume of a sphere of radius $r$ is $\frac{4}{3}\pi {{r}^{3}}$.]

The first engineer proposes that the rate of change of $V$ with respect to $t$ is a constant $k$.

The second engineer argues that the rate at which water flows out from the hole will be at its greatest in the beginning, and decreases as the depth of water in the container decreases. He suggests using Torricelli’s law, which says that

$\frac{\mathrm{d}V}{\mathrm{d}t}=-\alpha \sqrt{20h}$,

where $\alpha $ m$^{2}$ is the area of the circular hole at the bottom of the container, and $h$m is the depth of water in the container at time $t$ seconds. The radius of the hole at the bottom of the container is found to be constant at $1$ cm.

2022 EJC P1 Q10

2022 EJC P1 Q10

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