Thicc toilet paper

The volume enclosed between the outer boundary and the inner boundary of the roll is given by $V = \pi ( R^2 - r^2 ) L$ , where $R$ is the unknown outer radius, and $r = 2 \text{ cm}$ is the inner radius, and $L$ is the length of the roll. Now the volume of the $200$ sheets is $200 \times 12 \times 0.04 \times L \text{ cm}^3$ . Equating the two expressions, results in

$\pi ( R^2 - 4 ) = 96$

From which, $R = \sqrt{ 4 + \dfrac{96}{\pi}} = 5.879 \text{ cm}$ , hence $D = 2 R = 11.758 \text{ cm}$ .

The first layer of toilet paper we wrap around the roll contains $4\pi = 12.57$ cm worth of toilet paper. The second layer is now wrapped on top of the first layer, which is why the diameter is slightly bigger now. It is 0.08 cm bigger to be exact, since the thickness of the toilet paper contributes once to the radius, hence twice to the newer diameter. Now we can wrap $4.08\pi = 12.82$ cm worth of toilet paper around on top of the first layer. This pattern continues like that.

Since the roll contains 200 sheets we need to find the amount of layers where the circumferences of each circle add up to a value above $12*200=2400$ cm.

With the knowledge of the pattern above, I wrote a programm that would tell me exactly that.

The answer is, that after 98 layers one can wrap around 2426 cm worth of toilet paper around the roll in total. Now we just have to calculate the diameter: $2*98*0.04 + 4=\boxed{11.84}$ cm, since the thickness of the toilet paper contributes twice to the diameter.

Let be $r=2$ the radius of the inner cylinder and $s=0.04$ the thickness of the sheet. The paper rolled upon it could be modelled as an archimedean spiral of general equation

$\displaystyle r(\theta)=a+b\theta$

The total lenght of the spiral will be $L=12 \cdot 200=2400$ . For $\theta=0$ we'll have that

$\displaystyle r(0)=a+b \cdot 0=r=2 \hspace{5pt} \Longrightarrow \hspace{5pt} a=2$

For two generic point we'll have that

$\displaystyle r(\theta+2\pi)-r(\theta)=s \hspace{5pt} \Longrightarrow \hspace{5pt} 2+b(\theta+2\pi)-2-b\theta =s \hspace{5pt} \Longrightarrow \hspace{5pt} b=\frac{s}{2\pi}$

Hence

$\displaystyle r(\theta)=2+\frac{s}{2\pi}\theta$ .

The lenght of the spiral is $dl=r(\theta)d\theta$ , so

$\displaystyle \int_{0}^{{\theta}_0} r(\theta)d\theta=L \hspace{5pt} \Longrightarrow \hspace{5pt} \int_{0}^{{\theta}_0} \bigg(2+\frac{s}{2\pi}\theta\bigg)d\theta=2400 \hspace{5pt} \Longrightarrow \hspace{5pt} {\theta}_0=609.24 \hspace{3pt}\text{rad}$

Eventually the diameter $D$ we're looking for is

$\displaystyle D=2r({\theta}_0)=\bigg(2+\frac{0.04}{2\pi}\cdot 609.24\bigg)=\boxed{11.75}$