$1.5.20.$ The reel of tape is played back during time $t$ at the film drawing speed $v$. The initial radius of the reel (with film) is $R$, and the final radius (without film) is $r$. What is the thickness of the film?
Film wound on a reel takes up an area
$$ S = \pi (R^2 - r^2) $$
If the film is completely unwound at a speed of $v$ in time $t$, then the length of the film
$$ l = vt $$
In cross-section, the film is a rectangle whose area is $S$ and length is $l$. Then the width of this rectangle (it is also the thickness of the film)
$$ \fbox{$d = \frac{S}{l} = \frac{\pi (R^2 - r^2)}{vt}$} $$
$$d = \frac{\pi (R^2 - r^2)}{vt}$$