Spherical rain drop gaining mass 
A spherical rain drop, falling in a constant gravitational field, grows by the absorption of moisture from the surrounding at a rate proportional to its surface area. If it starts with zero radius, find its acceleration.

My attempt
My assumptions are $m$ is mass of the rain drop, $r$ is the radius of the rain drop, $\rho_w$ is density of the rain drop, $\vec{g}$ is gravity, $\vec{F}$ is the downward force, $t$ is time and $v$ is the instantaneous speed.
$\frac{dm}{dt} = 4\pi r^2 \cdot K = \frac{d}{dt} m = \frac{d}{dt}(\frac{4}{3}\pi r^3 \rho_w) = \frac{4}{3}\pi \rho_w \frac{dr^3}{dt}  = 4\pi \rho_w r^2  \frac{dr}{dt}.$\
This  implies $dt = dr \rho_w/K $.
Now, $\vec{F} = \frac{d}{dt}(mv)  =  \frac{4}3 \pi \rho_w\frac{d}{dt}(r^3v )$.  By substituting $dt$ by $dr \rho_w/K $ on this equation, one can see that  $\vec{F} = \frac{4}3 \pi K\frac{d}{dt}(r^3v) $.
Here comes the confusing bit. I don't  know what to do with $\vec{F}$. First I thought taking $\vec{F} = m\vec{g} = \frac{4}3 \pi \rho_w r^3 \cdot \vec{g} $. When I do that I end up with $\dfrac{d}{dt}r^3 v = \frac{\rho_w r^3 v }{K} \implies v = \frac{\rho_w \vec{g} r }{K}. $ and ultimately $dv/dt = \vec{g}/4$. But then I realized, if there is something that can reduce the acceleration (from gravitational acceleration), then the magnitude  of $ \vec{F}$ should be less than the magnitude  of $ m\vec{g} = \frac{4}3 \pi \rho_w r^3 \cdot \vec{g}$ because of the drag force reducing the magnitude of $\vec{g}$. Can anyone say anything about this or even propose another solution? I apologize for the errors or informalities in the question in advance.
 A: It looks to me like you got the right answer but were missing the physical intuition and mathematical confidence necessary to believe your answer:
Let’s assume that the problem meant us to start with an infinitesimal speck of a raindrop, rather than truly zero radius (which could never grow and would make for a pretty boring situation!)
Let us also observe that the radius of the raindrop must increase at a constant rate.    How will the raindrop move?   The answer might not be immediately obvious.
There is a clear reason for the acceleration to be less than g.  All of the absorbed water which is hitching onto the raindrop as the raindrop falls is doing so with zero initial momentum!  When the raindrop is small, it is also falling slowly.  So while the presently absorbed water is massive relative to the drop, it does not need to be accelerated much.  When the raindrop is large, it is also falling quickly.  So while the presently absorbed water is less massive relative to the drop, it must be accelerated significantly.
It turns out that when you do the math, as you already did, these effects balance each other perfectly and the result is constant acceleration at one quarter g!!   So your answer was correct, although I think I see a couple typos in the intermediate steps.  What a beautifully unexpected result!
I will be adding this to my “very cool AP physics problems” file.
P.S. Another potential surprise here is the fact that the movement of the drop is independent of K!      This is caused by another case of perfectly balanced opposing effects.  At the same drop size, the drop will be moving faster (at a later time) for smaller K, which balances the reduced rate of absorbtion perfectly.
A: You mention you were not sure to do with $\vec{F}$. The guiding equation in this problem is Newton's second law of motion, $\sum \vec{F} =m\vec{a}$. If you want to included drag, then you have two forces on the object, giving $\sum \vec{F} = \vec{F_g}+\vec{F_d}$. This may then be equated to $m\vec{a}$. The drag equation gives $\vec{F_d}=-\frac{1}{2}C_d \rho A v^2 \hat{v}$. The factor $-\hat{v}$ means that this force is directed opposite the motion of the object. $A$ is the maximum cross sectional area of the object (in this case $\pi r^2$, which grows with time). $C_d$ is the drag coefficient which is a dimensionless parameter determined empirically, but you might be able to use 0.47 if you desire a numeral result (valid for a sphere in a fluid at Reynold's Numbers around $10^4$). $\rho$ is the density of the fluid medium (not the object). You should now be equipped to carry out your calculation.
