A conical tank leaks water with a radius of 5m and a height of 15m. The tank is full of water before the hole on the bottom was opened and it took a day to be emptied. Assume discharge coefficient to be $1$ and g is $9.81\frac{m}{s^2}$
So I what I did is. First ratio and proportion $\frac{5}{12}$=$\frac{r}{h}$ so r=$\frac{5}{12}h$ then substituting it to the Volume equation of a cone and deriving both sides I get $\frac{dv}{dt}$=$\frac{3h^20.1818dh}{dt}$ where I think that $\frac{dv}{dt}$ is $981\frac{cm}{s^2}$ and h, since mentioned full is the same as the height $1500cm$ finally I get $0.0008\frac{cm}{s}$ and now I'm stuck on how do I relate on the diameter of the hole?
According to a number of sources the liquid outlet velocity ($v$) when draining a tank or a container can be calculated$^{\text{*}}$ by $v=C\sqrt{2gh}$, where $C$ is the velocity coefficient, $g$ is the acceleration of gravity, and $h$ is the height of the water level in the vessel. The formula is independent of the shape of the container. In our case, $C\approx 1$ and the height of the water level ($h(t)$) in the container and (as a result) the outlet velocity depends on $t$, the time. So, we are going to work with the following formula:
$$v(t)=\sqrt{2gh(t)}.$$
Let $A$ denote the surface area of the opening at the bottom of our vessel. Considering a very short time interval, say $\Delta t$, the volume of water flowing out$^{\text{*}}$ during $\Delta t$ is
$$\Delta V=A\Delta t \sqrt{2gh(t)}.$$
During the same period of time the amount of water is decreasing in the container.
The container is of conical shape (with the given dimensions) and the radius of the surface of the water is changing according to $r(t)=\frac13h(t)$ then
$$\Delta V=-\Delta h \pi\frac19 h^2(t).$$
(The sign is negative because $\Delta h$ is negative.) That is, we have the following equation:
$$A\Delta t \sqrt{2gh(t)}=-\Delta h \pi\frac19 h^2(t).$$
Dividing both sides by $\Delta t$, and letting $\Delta t\rightarrow 0$ we get the following:
$$A\sqrt{2gh(t)}=-\frac{dh}{dt}\frac19\pi h^2(t).$$
After rearranging the expression above, we have
$$\frac{dh}{dt}= -\frac A{\pi}\sqrt{162g}\times h^{-\frac32}(t).$$
The following function is obviously a solution:
$$h(t)=-\frac52\frac A{\pi}\sqrt{162g}\times t^{\frac25}+\text{c} \tag 1$$
Now, $c=15$ because we know that $h(0)=15$. One day is $86,400$ sec, and $h(86,400)=0$ so we have an equation for $A$:
$$15=\frac52\times\frac A{3.14}\sqrt{162\times 9.81}\times 86,400^{\frac25}.$$
Finally
$$A\approx 50 \ \text{cm}^2,$$
corresponding to a hole of a radius of $4\ \text{cm}$.
The conic tank contained about $392500 \ \text{dm}^3$ of water. Had the water been discharged uniformly through the opening then $392500/86400 \approx 4.5 \text{dm}^3$ water would have been poured out in every second of the day. I did this latter calculation to see if the result obtained above was realistic. I needed this because I did not have any physical intuition supporting my math. I am still not convinced that the result ($A\approx 50 \ \text{cm}^2$) is OK. Sorry.
$^*$Note: We could use another formula
$$\frac{dV}{dt}=A C' \sqrt{2gh(t)}$$
where $C'\approx 1$ is the discharge coefficient.