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Условие Решение

IZhO 2026, Теория — Задача 1. Problem 1

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XXII International Zhautykov Olympiad/Theoretical Competition · 11 януари 2026 г. · 10 т.

THEORETICAL COMPETITION

January 11, 2026

Please read this first:

The time available for the theoretical competition is 4 hours. There are three questions.
You can use your own calculator for numerical calculations. If you don’t have one, please ask for it from Olympiad organizers.
You are provided with Writing sheet and additional paper. You can use the additional paper for drafts of your solutions but these papers will not be checked. Your final solutions which will be evaluated should be on the Writing sheets. Please use as little text as possible. You should mostly use equations, numbers, figures and plots.
Use only the front side of Writing sheets. Write only inside the boxed area.
Each task should be solved starting from a new page of the Writing sheets
At the top of each sheet of paper your country (Country) and your student code (Student Code) are printed, you only need to fill in the progressive number of each sheet (Page Number), and the total number of Writing sheets used (Total Number of Pages). If you use some blank Writing sheets for notes that you do not wish to be evaluated, put a large X across the entire sheet and do not include it in your numbering.

Условие

This problem consists of three independent parts.

1.1 A thin cylindrical core of radius $R$ has a strip of foil tightly wound around it in a thin layer. The foil has thickness $\delta$ , mass $m$ , and length $L = 10.0\ \mathrm{m}$ . One end of the foil is fixed to the core, while the other end is attached to the edge of a thin horizontal platform $\Pi$ . Initially, the roll (the core with the wound foil) is held so that its axis $O$ is horizontal and at the same level as the platform. The roll is then released, and under the action of gravity it begins to unwind. Assuming that $\delta \ll R \ll L$ , find the length $x_0$ of the unwound part of the foil for which the force $T_0$ acting on the platform is equal to the weight of the roll, $mg$ . Also determine the acceleration $a$ and the velocity $v$ of the downward motion of the roll's center of mass at that instant. [3 т.]

A roll with centre O falling under gravity g, unwinding from a foil strip attached to a horizontal platform Π; the tension T₀, the unwound length x and the downward velocity v of the centre are labelled.

1.2 The readings of a multimeter ( $\Omega$ ) operating in ohmmeter mode and connected to a circuit change with time according to the law $R_\Omega(t) = A - Be^{-t/\tau_0}$ , where $A = 100\ \mathrm{k\Omega}$ , $B = 40\ \mathrm{k\Omega}$ , and $\tau_0 = 100\ \mathrm{s}$ . Using these data, determine the values of the resistances $r$ and $R$ , as well as the capacitance $C$ of the capacitor, which is initially uncharged. Also find the amount of heat $Q$ released in the resistor $R$ during the measurement time $t \gg \tau_0$ .

Circuit diagram: an ohmmeter (Ω) connected across a parallel combination of resistor r and a series R–C branch.

Note: In ohmmeter mode, the multimeter measures the voltage drop $U_x$ across an unknown resistor $R_x$ when a constant current $I_0 = 0.1\ \mathrm{mA}$ flows through it. The instrument displays the value $R_x = U_x / I_0$ . [4 т.]

1.3 An observer (the eye) is located at a distance $L = 55\ \mathrm{cm}$ from a convex–concave lens and observes images of a distant street lamp formed due to reflections from both surfaces of the lens (see photo). The lamp and the eye are positioned close to the principal optical axis of the lens on the side of the concave surface. The radius of curvature of the concave surface is $r = 7\ \mathrm{cm}$ , and the radius of curvature of the convex surface is $R = 28\ \mathrm{cm}$ . Upon observation, two images of the lamp are visible: one real and the other virtual. The ratio of the apparent angular sizes of these images is $\gamma = \varphi_1/\varphi_2 = 0.1$ , where $\varphi_1$ is the angular size of the real image and $\varphi_2$ is the angular size of the virtual image. Find the optical power $D$ of the lens. [3 т.]

Photograph showing a large bright circular image of a street lamp and a small secondary reflected image inside a dark lens.

Решение

Attention. Points in grading are not divided!

Problem 1 (10.0 points)

Problem 1.1 (3.0 points)

Let the $x$ -axis be directed downwards, and the initial position of the roll's center of mass correspond to the origin. If the velocity of the roll's center of mass at a given moment in time is $v$ , then its kinetic energy is

E_k = \frac{m}{L}(L-x)v^2. \qquad (1)

On the other hand, the potential energy of the roll relative to the initial position is equal to

E_p = -\frac{m}{L}g\frac{x^2}{2} - \frac{m}{L}g(L-x)x. \qquad (2)

At the initial moment of time the total energy is zero, therefore, according to the conservation law we get

E_k + E_p = 0, \qquad (3)

from which the dependence of the speed of the roll center of mass on the coordinate is given by

v^2(x) = g\left[x + \frac{x^2}{2(L-x)}\right]. \qquad (4)

The acceleration of the roll is determined by the expression

a(x) = \frac{dv}{dt} = v\frac{dv}{dx}, \qquad (5)

which after substitution (4) gives the following dependence

a(x) = \frac{g}{4}\left[1 + \frac{L^2}{(L-x)^2}\right]. \qquad (6)

The total momentum of the roll is directed along the $x$ -axis and is equal to

p = \frac{m}{L}(L-x)v, \qquad (7)

and therefore the change in momentum over time takes the form

\frac{dp}{dt} = -\frac{m}{L}v^2 + \frac{m}{L}(L-x)a. \qquad (8)

According to the conditions of the problem, the force acting on the platform from the roll is equal to the force of gravity, which means that the total external force on the roll itself is zero, and we obtain

\frac{dp}{dt} = 0. \qquad (9)

Solving together (4), (6), (8) and (9), we derive the quadratic equation

x + \frac{x^2}{2(L-x)} = \frac{1}{2}\left(L + \frac{x^2}{2(L-x)}\right), \qquad (10)

whose positive root equals to

x_0 = \frac{3-\sqrt{3}}{3}L = 4.23\ \mathrm{m}. \qquad (11)

Substituting into (4) and (6) provides the final answers

v(x_0) = \frac{\sqrt{gL}}{\sqrt[4]{3}} = 7.52\ \mathrm{m/s}, \qquad (12)

a(x_0) = g = 9.80\ \mathrm{m/s^2}. \qquad (13)

Content	Points
Formula (1): $E_k = \frac{m}{L}(L-x)v^2$	0.2
Formula (2): $E_p = -\frac{m}{L}g\frac{x^2}{2} - \frac{m}{L}g(L-x)x.$	0.2
Formula (3): $E_k + E_p = 0$	0.1
Formula (4): $v^2(x) = g\left[x + \frac{x^2}{2(L-x)}\right]$	0.2
Formula (5): $a(x) = \frac{dv}{dt} = v\frac{dv}{dx}$	0.2
Formula (6): $a(x) = \frac{g}{4}\left[1 + \frac{L^2}{(L-x)^2}\right]$	0.2
Formula (7): $p = \frac{m}{L}(L-x)v$	0.2
Formula (8): $\frac{dp}{dt} = -\frac{m}{L}v^2 + \frac{m}{L}(L-x)a$	0.2
Formula (9): $\frac{dp}{dt} = 0$	0.2
Formula (10): $x + \frac{x^2}{2(L-x)} = \frac{1}{2}\left(L + \frac{x^2}{2(L-x)}\right)$	0.1
Formula (11): $x_0 = \frac{3-\sqrt{3}}{3}L$	0.2
Numerical value in formula (11): $x_0 = 4.23\ \mathrm{m}$	0.2
Formula (12): $v(x_0) = \frac{\sqrt{gL}}{\sqrt[4]{3}}$	0.2
Numerical value in formula (12): $v(x_0) = 7.52\ \mathrm{m/s}$	0.2
Formula (13): $a(x_0) = g$	0.2
Numerical value in formula (13): $a(x_0) = 9.80\ \mathrm{m/s^2}$	0.2
Total	3.0

Problem 1.2 (4.0 points)

At the initial moment of time $t = 0$ the capacitor is not charged and the voltage drops across the connected resistances are the same, that is, they are connected in parallel, therefore

\frac{1}{R_0} = \frac{1}{R} + \frac{1}{r}, \qquad (1)

and the ohmmeter readings are

R_0 = R(0) = A - B. \qquad (2)

At time $t = \infty$ the capacitor is fully charged and no current flows through it, therefore

R_\infty = r, \qquad (3)

and the ohmmeter readings are

R_\infty = R(\infty) = A, \qquad (4)

Solving (1)-(4) together, we obtain

r = 100\ \mathrm{k\Omega}, \qquad (5)

R = 150\ \mathrm{k\Omega}. \qquad (6)

At any given moment in time, the voltage drop $U_r$ across the resistance $r$ is equal to the voltage drop $U_R$ across the resistance $R$ and the voltage drop $U_C$ across the capacitor $C$ , that is,

U_r = U_R + U_C. \qquad (7)

On the other side according to Ohm's law we write

U_R = I_c R, \qquad (8)

U_r = I_r r, \qquad (9)

where $I_r$ denotes the current flowing through the resistor $r$ , and $I_c$ designates the current flowing through the capacitor $C$ and the resistance $R$ .
The voltage across the capacitor is equal to

U_C = \frac{q}{C}. \qquad (10)

Note that the capacitor is charged due to the flow of current $I_0$ generated by the ohmmeter, that is,

I_0 = I_r + I_C, \qquad (11)

and the current flowing through the capacitor is equal to the derivative of its charge over time

I_C = \frac{dq}{dt}. \qquad (12)

From (7)-(12) we derive the differential equation for $q$

\frac{dq}{dt} + \frac{q}{\tau_0} = I_0\frac{r}{R+r}, \qquad (13)

whose solution yields the current through the capacitor

I_c = I_0\frac{r}{R+r}e^{-t/\tau_0}. \qquad (14)

Here $\tau_0 = C(R+r)$ .
Thus, the capacitance is found as

C = \frac{\tau_0}{R+r} = 400\ \mathrm{\mu F}. \qquad (15)

The amount of heat released by the resistor is determined by the Joule-Lenz law and is equal to

Q = \int_0^\infty I_c^2 R\,dt = \frac{I_0^2 r^2 R \tau_0}{2(R+r)^2} = 12\ \mathrm{mJ}. \qquad (16)

Note that, in principle, there is no need to consider equations (1)-(4) separately; to find unknown resistances, it is sufficient to find the dependence $R(t)$ , which is obtained as

R(t) = \frac{U_r}{I_0} = r - \frac{r^2}{R+r}e^{-t/\tau_0}. \qquad (17)

Content	Points
Formula (1): $\frac{1}{R_0} = \frac{1}{R} + \frac{1}{r}$	0.2
Formula (2): $R_0 = R(0)\ = A - B$	0.2
Formula (3): $R_\infty = r$	0.2
Formula (4): $R_\infty = R(\infty) = A$	0.2
Formula (5): $r = 100\ \mathrm{k\Omega}$	0.2
Formula (6): $R = 150\ \mathrm{k\Omega}$	0.2
Formula (7): $U_r = U_R + U_C$	0.2
Formula (8): $U_R = I_c R$	0.2
Formula (9): $U_r = I_r r$	0.2
Formula (10): $U_C = \frac{q}{C}$	0.2
Formula (11): $I_0 = I_r + I_C,$	0.2
Formula (12): $I_C = \frac{dq}{dt}$	0.2
Formula (13): $\frac{dq}{dt} + \frac{q}{\tau_0} = I_0\frac{r}{R+r}$	0.2
Formula (14): $I_c = I_0\frac{r}{R+r}e^{-t/\tau_0}$	0.4
Formula (15): $C = \frac{\tau_0}{R+r}$	0.2
Numerical value in equation (15): $C = 400\ \mathrm{\mu F}$	0.2
Formula (16): $Q = \frac{I_0^2 r^2 R\tau_0}{2(R+r)^2}$	0.4
Numerical value in equation (16): $Q = 12\ \mathrm{mJ}$	0.2
Total	4.0

Problem 1.3 (3.0 points)

A real image with a smaller angular size $\varphi_1$ is formed as a result of reflection at the air-glass boundary on the concave surface of the lens as from a concave mirror with a radius of curvature $r$ at a distance equal to the focal length $F_1$ of the mirror

\frac{1}{F_1} = \frac{2}{r}, \qquad (1)

which means the image is located at a distance from the observer

a_1 = L - F_1. \qquad (2)

Let $\varphi_0$ be the apparent angular size of the street lamp from the lens location. Then the linear size of the image of the lamp at the focus is

l_1 = \varphi_0 F_1, \qquad (3)

and therefore, the angular size of the visible image is found as

\varphi_1 = \frac{l_1}{a_1}. \qquad (4)

The second image, with a larger angular size of $\varphi_2$ , is a virtual image of the lamp, formed as a result of reflection at the glass-air boundary from the convex surface of the lens. Considering that the rays that formed this image passed twice through a diverging lens with optical power $D$ , and the focal length of the convex surface is $R/2$ , the focal length of the system is

-\frac{1}{F_2} = 2D + \frac{2}{R}, \qquad (5)

and the image itself is located at a distance from the observer

a_2 = L + F_2. \qquad (6)

The linear size of the image of the lantern at focus is equal to

l_2 = \varphi_0 F_2, \qquad (7)

and therefore the angular size of the visible image is finally obtained as

\varphi_2 = \frac{l_2}{a_2}. \qquad (8)

Using the condition $\gamma = \varphi_1/\varphi_2$ , the answer is derived as

D = \frac{\gamma+1}{2L} - \frac{1}{R} - \frac{\gamma}{r} = -4\ \mathrm{dpt}. \qquad (9)

Content	Points
Formula (1): $\frac{1}{F_1} = \frac{2}{r}$	0.2
Formula (2): $a_1 = L - F_1$	0.2
Formula (3): $l_1 = \varphi_0 F_1$	0.4
Formula (4): $\varphi_1 = \frac{l_1}{a_1}$	0.4
Formula (5): $-\frac{1}{F_2} = 2D + \frac{2}{R}$	0.2
Formula (6): $a_2 = L + F_2$	0.2
Formula (7): $l_2 = \varphi_0 F_2$	0.4
Formula (8): $\varphi_2 = \frac{l_2}{a_2}$	0.4
Formula (9): $D = \frac{\gamma+1}{2L} - \frac{1}{R} - \frac{\gamma}{r}$	0.2
Numerical value in formula (9): $D = -4\ \mathrm{dpt}$	0.4
Total	3.0

Mathematical hint for the theoretical competition

The following formulas may be useful:

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$ , where $n \neq -1$ is a fixed number, $C$ refers to an arbitrary constant;

$\int \frac{dx}{x} = \ln|x| + C$ , where $C$ stands for an arbitrary constant;

$\int \frac{dx}{(a^2+x^2)^{3/2}} = \frac{x}{a^2\sqrt{a^2+x^2}} + C$ , where $a$ – constant, $C$ – arbitrary constant;

$(1+x)^{\gamma} \approx 1 + \gamma x + \frac{\gamma(\gamma-1)}{2}x^2$ , for $|x| \ll 1$ and any value of $\gamma$ ;

$\tan x \approx \sin x \approx x$ , for $|x| \ll 1$ ;

$\ln(1+x) \approx x$ , for $|x| \ll 1$ .

List of physical constants

Acceleration of gravity $g = 9.80\ \mathrm{m/s^2}$ ;

Universal gas constant $R = 8.31\ \mathrm{J/(mole\cdot K)}$ ;

Speed of light in vacuum $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$ ;

Reduced Planck constant $\hbar = 1.05 \cdot 10^{-34}\ \mathrm{J\cdot s}$ ;

Magnetic constant $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{H/m}$ ;

Electric constant $\varepsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{F/m}$ ;

Stefan-Boltzmann constant $\sigma = 5.67 \cdot 10^{-8}\ W/(m^2 \cdot K^4)$ ;

Elementary charge $e = 1.60 \cdot 10^{-19}\ \mathrm{C}$ ;

Electron mass $m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}$ .

Оригинал в Архива: IZhO-2026-Theory_eng.pdf · официални решения: IZhO-2026-Theory_eng_sol.pdf