Correspondence between time and the level of water while filling a hemisphere with water at a constant rate I am trying some question in quantitative aptitude section and I couldn't reason for those question.
A hemispherical bowl is being filled with water at a constant volumetric rate. The level of water in the bowl increases

*

*A. in direct proportion to time,


*B. in inverse proportion to time,


*C. faster than direct proportion to time,


*D. slower than direct proportion to time.
I thought as time increases level of water rising decreases as at a height above it has a large surface area. So, I chose B.
But this is not the case.
Answer is ->

D

So, please help. Also, it would be really helpful if someone can give a purely mathematical treatment of the reasoning used.
 A: Here is an idea for a possible answer. Suppose you hemispheric bowl is
$$
x^2 + y^2 + z^2 \leq 1, \quad z \leq 0.
$$
The bowl is filled at constant volumetric rate $K > 0 $ so the volume of water in the bowl at time $t$ is $V(t) = Kt$. Let $z(t)$ be the level of the water at time $t$, with $z(t) = -1$ the initial level when the bowl is empty. The volume of the bowl when it is filled at $z(t)$ is
$$
V(t) = Kt = \int_{z=-1}^{z(t)} \underbrace{\pi (1-z^2)}_{\text{area of the disk $x^2 + y^2 
\leq 1 - z^2$ in the plane ($z$ fixed)}} dz
$$
So differentiating in $t$ we get
$$
K = \pi(1-z^2(t))z'(t) \qquad \text{i.e.} \qquad z'(t) = \frac{K}{\pi(1-z^2(t))}.
$$
Now you have to integrate this ODE with initial condition $z(0) = -1$ and compare $z(t)$ to $K$. Unfortunately I am not too good with ODEs so I am letting you look for how to solve this!
EDIT: As TonyK says in the comments, you can already see from the differential equation that $z(t)$ is decreasing in time, which gives the answer to the initial question.
Now concerning intuition, the bowl gets larger and larger in radius as the water fills it so adding a small volume $dV$ to the bowl when it is already partially filled will make the level rise by much less than if the the bowl is almost empty. Therefore the speed at which the level rises decreases in time. But filling an erlenmeyer you would get the opposite.
A: (Images and formulas are from math24)
The question is "how much volume of water do you need to raise the bowl water level of $1$" (say $1$ centimeter), from the bottom to the top of the bowl (that is, half the sphere).
The volume of a sphere is $\dfrac{4}{3}\pi R^3$.
The volume of a spherical segment is ${\large\frac{{\pi h\left( {3r_1^2 + 3r_2^2 + {h^2}} \right)}}{6}\normalsize}$, and $R = {\large\frac{{{r_i^2} + {h_i^2}}}{{2h_i}}\normalsize}$, ($r_1$ being the radius at the height of the bottom segment, $r_2$ at the top segment, we have $r_i=\sqrt{2Rh_i-h_i^2}$.

Now, thanks to these formulas, let see for a bowl of height $R=10$ (the radius of the sphere is $10$ centimeters), how much volume of water you need to reach level $1$, then level $2$ from level $1$... up to level $10$ from level $9$.
0 to  1 =  30.36 cm³
1 to  2 =  86.91 cm³
2 to  3 = 137.18 cm³
3 to  4 = 181.16 cm³
4 to  5 = 218.86 cm³
5 to  6 = 250.28 cm³
6 to  7 = 275.41 cm³
7 to  8 = 294.26 cm³
8 to  9 = 306.82 cm³
9 to 10 = 313.11 cm³

So as you can see, the higher the level, the more water you need to reach $1$ more centimeter level. Thus the answer D, where it takes more and more time to reach one more centimeter level ; the constant volume of water fills the bowl, level wise, slower and slower.
(Note: the sum of these $10$ volumes gives $2094.40$, which is half of $4188.80$, the volume of a sphere of radius $10$)
