The following question was presented as a screening question in an interview for an engineering start-up:
A cylinder is of an outer diameter of 100 mm.It is necessary to roll around it a sheet of paper with a thickness of 30 micron.What would be the diameter of the cylinder once a sheet of said thickness with a length of 500 meters is rolled around it ?
This may seem as a simple question at first, however, I noticed that as we add thickness and further layers to the cylinder it's diameter and perimeter are increasing, and so is it's capacity to to hold length which also increases with each curl....therefore the simple formula of total length of the sheet divided by the outer perimeter times twice the thickness added to the initial diameter wont do....is this supposed to be solved with a differential equation of some sort, a geometric series or otherwise ?
In reality, the cross-section of the paper around the cylinder is a sort of spiral (though not an Archimedean one - the first, say, $90^\circ$of the first of paper layer lies directly on the cylinder's surface and has constant radius). To model this perfectly, even ignoring physical aspects like compressibility of the paper, would be very hard, and probably isn't what they're looking for in an interview.
So we need a way to approximate the shape. Here are a couple of approaches.
Firstly, you can imagine the process of winding the paper on. Every time the paper wraps around the cylinder, the radius (and circumference) increase slightly. To approximate the shape, we'll assume the radius goes up by steps. If $r(n)$ is the radius of the outermost surface after $n$ layers, we have $$r(n)=50+\frac{3n}{100}$$ (I'll keep measurements in mm for simplicity.) So we're looking at an arithmetic progression. The circumference of the $n^{th}$ layer is $2\pi r(n-1)$, so we need to sum the series and solve $$L = \sum_{n=0}^{N-1} 2\pi r(n)$$
for the final number of layers, $N$, where $L$ is the length of the paper. In this case, $L=500000$ and solving the above we get $$N\approx 1176.6$$
Plugging this value into the formula for the radius, we find $r(N) \approx 85.3$ giving a final diameter of $170.6$mm.
But I don't think this is what they're after. Instead of thinking about winding the paper on, consider the whole, wrapped cylinder. Its cross-section is an annulus of paper, with inner radius $50$mm and outer radius $R$. The area of the cross-section is $$A=\pi \left(R^2-50^2\right)$$
Another way to work out this area is the length of the paper multiplied by its thickness; ie $$500000\cdot \frac{3}{100} = \pi \left(R^2-50^2\right)$$
Solving this we get $R\approx 85.3$ and again find the final diameter to be $170.6$mm.
These two answers are not exactly the same (remember both are approximations) - they differ by about $12\mu m$, a bit less than half the thickness of the paper.
I suspect the point of the question is probably to know the first method but realise you can do the second, but I'd be interested in what others made of it.