Deadline for submission: Oct. 5, 2025, 23:59, CET.

Problem Statement of Series 1, Year 39

About the Competition Rules How to Write Solutions Results

1... water goblin

3 points

A water goblin lives at the bottom of the Dead Sea. He, as is common for such a folklore creature, traps the souls of people who have drowned in the sea, keeping them in small mugs.

However, the jars are quite small and therefore cannot contain the souls, he has to reduce the volume five-fold by an isothermic process.

One day, the goblin got bored of the Dead Sea and decided to move to the bottom of the Orlík reservoir in southern Bohemia. What is the minimum depth he has to live in at his new place such that the souls are not able to escape by lifting the physical lids?

Before moving, the water goblin used to live at the depth of $33\,\mathrm{m}$ such that ascending by any amount would allow the jars to open. Consider the souls to be an ideal gas with the same temperature as that of the surrounding water. At the moment of drowning, just before their compression, the souls are under atmospheric pressure.

The water temperature of the Dead Sea equals $23\,\mathrm{^\circ\mskip-2mu\mathup{C}}$, water temperature of the Orlík reservoir equals $6\,\mathrm{^\circ\mskip-2mu\mathup{C}}$, density of the Dead sea equals $1.24\,\mathrm{g\cdot cm^{-3}}$, while the density of the Orlík reservoir equals $0.99\,\mathrm{g\cdot cm^{-3}}$. Consider water properties to be constant in both bodies of water. The jars were properly sealed for the duration of moving from the Dead Sea to the Orlík reservoir.

We will publish the solution to this problem soon.
Terka would like to go to a vacation by the sea from which she will be able to return alive.

2... turbocyclist

3 points

At what maximal velocity can a cyclist go through a left turn with a radius of $r = 7.5\,\mathrm{m}$, for him not to touch the ground with his pedal? The cyclist and his bicycle weigh $m = 80\,\mathrm{kg}$, their collective center of gravity is in the plane of symmetry of the bicycle. The pedal clique is at its lowest point the whole time. While the pedal itself is not turned either to the front or backwards, the position of the pedal in relation to the clique stays the same as depicted in the picture. Pedal is $a = 2.0\,\mathrm{cm}$ high and $b = 10\,\mathrm{cm}$ wide, the dimensions of the clique are $d = 15\,\mathrm{cm}$ and $l = 6.5\,\mathrm{cm}$, as can be seen from the picture. When the bicycle is in vertical position, the horizontal axis of the rotation of the clique around the frame of the bicycle $o$ is $h = 30\,\mathrm{cm}$ above the ground. The vertical dashed line marks the vertical symmetry of the bicycle. The center of gravity also lies on this line, length $l$, therefore it is measured from the center of the bottom bracket. Friction is sufficient for the bicycle not to slip.

We will publish the solution to this problem soon.
Terka was thinking whether she is going to almost crash on her bicycle (again.

3... resisting spiderwebs

6 points

Consider a spider web formed by a circle with radius $r$ with two perpendicular crossbars, which go through the middle of the web. We put $N-1$ cocentric circles inside the original circle in a certain way, so that the crossbars always intersect at the distance $r/N$. How much resistance is between the ends of one crossbar? Material of the web has linear resistivity $\lambda$ and knots are conductively connected.

Bonus: Determine resistance for three crossbars, which are rotated by $120^\circ$.

We will publish the solution to this problem soon.
Jarda likes animal-themed problems.

4... cake problems

8 points

A cake on Jarda's plate sometimes topples over to its other side. What is the work needed for such misfortune? Consider polar symetric cake of weight $M$ and of radius $R$, its center of mass is located in height $h$; solve universally for its cutting to $n$ same pieces.

Bonus: Bake your own cake and send us a photo and a recipe. The best will be shared in the problem solution!

We will publish the solution to this problem soon.
You are more likely to find Jarda in a patisserie than in school.

5... eliptical lens

9 points

Consider a stretched rotary ellipsoid from a seethrough material with a refraction index $n=1.5$. Consider a plane that is parallel to the rotary axis and intersects the ellipsoid near the adjacent vertex. We cut through this plane and make the analogous cut at the other side. We stick both of the cut pieces together at the flat sides such that we get a narrow lens. How does a point at the distance of $d=4\,\mathrm{m}$ that lies at the optical axis display from the center of the lens? Projection of the original body is an ellipse on the given scheme with semi-major axis $a=3\,\mathrm{m}$ and semi-minor axis $b=2\,\mathrm{m}$.

We will publish the solution to this problem soon.
Jarda was inspired by nurofen medicine.

P... riding a bike

10 points

This problem has an open solution, so be sure to cite all sources used.

Imagine that you are riding a bike during a windy day in wavy terrain. Discuss the characteristics of the movement in different directions and different velocities qualitatively. Also, discuss how the description of the movement would change if we neglected the following parameters:

  • air resistance,
  • rolling resistance of the wheels,
  • slope of the terrain,
  • other resisting forces.
Using the equations for corresponding physical quantities, determine under which conditions you can neglect the individual forces in the description of the movement.

We will publish the solution to this problem soon.
Dodo's bike ride away from the dormitory went surprisingly smoothly.

E... lost object

11 points

Sometimes, maybe even some of you are wondering how far a small item can get right after it slips from your hands and falls to the ground. Choose a small item (screw, nail, and so on) and measure the distance between the place where you dropped the item and the place where you have found it, in relation to the height (from the ground) from which you dropped it. For each initial height, repeat the measurement enough times to decrease the statistical error of your measurement. Think about suitable surfaces and items.

Instructions for Experimental Tasks
We will publish the solution to this problem soon.

S... X-ray methods

10 points

  1. A layer of iridium $\ce{Ir}$ was sputtered onto a ruthenium $\ce{Ru}$ monocrystal, the process took $240\,\mathrm{min}$. The attenuation of the sublayer signal was measured after the deposition using XPS (Al K$\alpha$, $hf=1486.71\,\mathrm{eV}$), see the final spectrum in the figure below.
    Intensity of the signal from the binding energy of the ruthenium 3d orbital. The dashed line represents the signal before deposition; the solid line is the ruthenium signal after iridium deposition. The unit $\mathrm{cps}$ denotes the number of detected electrons per second.

    Determine the thickness of the deposited layer of $\ce{Ir}$. Assume that the emission was perpendicular to the surface and the effective photoelectron mean free path in $\ce{Ir}$ is $\lambda_{\ce{Ir}}=1.6\,\mathrm{nm}$. – 4 points

  2. Consider an iridium sample covered by $2\,\mathrm{nm}$ thick layer of oxide $\ce{IrO_2}$ on its surface. Our goal is to study only the active $\ce{IrO_2}$. This could be achieved by measuring on a synchrotron with the primary photon energy set to $\approx 150\,\mathrm{eV}$, such that we would reach the minimum of photoelectron energy as shown on Figure.
    Dependence of the electrons' mean free path $\lambda$ on the electron energy with an approximate equation for the curve. Both axes are logarithmic.

    How could we solve this without a synchrotron? Try to find the specific conditions under which this the result can be reached with X-ray tube capadle of energy $1486.71\,\mathrm{eV}$, resulting in the photoelectron mean free path of $\lambda_{\ce{IrO_2}} = 3.2\,\mathrm{nm}$. Does this approach have any disadvantages? – 2 points

    Hint: Use the formula \begin{equation*} I(d) = I_0 \exp\!\left(-\frac{d}{\lambda \cos\theta}\right) \,. \end{equation*}

  3. Figure  shows a diffractogram of polycrystalline cerium dioxide measured at photon energy of $77\,\mathrm{keV}$. To which interplanar spacing do the first 3 maxima correspond to? How do these planes look like in the crystal lattice? And at what angle $2\theta$ would we measure them if we used X-ray with energy of $90\,\mathrm{keV}$? – 4 points

    Diffractogram for a powder sample of crystalline cerium dioxide, $\ce{CeO2}$, measured at an X-ray photon energy of $77\,\mathrm{keV}$.

We will publish the solution to this problem soon.