IZhO 2023, Теория — Задача 1

• 1.1 312 J

• 1.2 $r_C = \frac{R_1R_2}{r_L} = 2\ \mathrm{M\Omega},\quad C = \frac{L}{R_1R_2} = 0.5\ \mu\mathrm{F}$

• 1.3 $1.2 \times 10^{7}$

  Official result: $R_{\min} = \frac{M_S}{M_J}\left(\frac{T_J}{T_E}\right)^{1/3}\frac{c}{2v_E} = 1.20\cdot 10^{7}$.

Problem 1 (10.0 points)

Problem 1.1 (3.0 points)

It follows from the first law of thermodynamics that

where $\delta Q$ is the amount of heat supplied, $dU$ is the change in internal energy, $dA$ is the work done by the gas.

For one mole of an ideal gas, these quantities can be written in terms of a change in volume $dV$ and temperature $dT$ at a known pressure $p$ in the following form

By definition of heat capacity, we have

then from relations (1)-(4) one obtains

at the molar heat capacity of a monatomic gas at a constant volume equal to

From the graph given in the problem statement, it can be seen that at a temperature

the heat capacity is $C = C_V$ and, accordingly, $\frac{dV}{dT} = 0$. When passing through this temperature, the derivative sign changes from plus to minus. This means that at this temperature the gas volume reaches a local maximum: $T_{\max} = T_1^* = 350\ \mathrm{K}$.

At a temperature

the derivative $\frac{dV}{dT}$ also equals zero, and when passing through this point, the sign of the derivative changes from minus to plus. This means that $T_2^*$ is the point of the local minimum of the volume: $T_{\min} = T_2^* = 500\ \mathrm{K}$.

In the section from $T_1^* = 350\ \mathrm{K}$ to $T_2^* = 500\ \mathrm{K}$, the gas receives heat $Q$, numerically equal to the area under the dependence $C(T)$, i.e. the area of the figure ABCDE. The change in internal energy $\Delta U = C_V(T_2^* - T_1^*)$ is numerically equal to the area of the rectangle ABDE. According to the first law of thermodynamics, therefore, the work on the gas from $T_1^*$ to $T_2^*$ is numerically equal to the difference in the areas of the rectangle ABDE and the figure ABCDE, i.e. area of the shaded figure BDC:

Note: Exact dependence $V(T)$:

$\displaystyle\frac{V}{V_1} = \left(\frac{T}{T_1}\right)^{7/2}\exp\left(-\frac{T - T_1}{\Delta T_1}\right)$, at $T_1 = 300\ \mathrm{K} \le T \le T_0 = 400\ \mathrm{K}$ and $\Delta T_1 = 100\ \mathrm{K}$;

$\displaystyle\frac{V}{V_0} = \left(\frac{T}{T_0}\right)^{-5/2}\exp\left(\frac{T - T_0}{\Delta T_2}\right)$, at $T_0 = 400\ \mathrm{K} \le T \le T_2 = 600\ \mathrm{K}$ and $\Delta T_2 = 200\ \mathrm{K}$.

Dependences $V(T)$ and $P(V)$ in the process of gas heating are shown in the figures below.

Problem 1.2 (3.0 points)

The equivalent circuit of the bridge is shown in the figure below, which takes into account that the non-ideal inductance circuit is equivalent to an ideal coil $L$ and resistor $r_L$ connected in series, whereas the equivalent circuit of a leaky capacitor is a resistor $r_C$ connected in parallel to an ideal capacitor $C$.

Solution 1. The bridge balance condition in complex numbers is written as

where the impedances are respectively

and

After some transformation we get from expressions (1)-(3):

While varying the frequency, this equality is not violated if both sides of the equation are equal to zero, therefore

Solution 2. Let the voltage across the capacitor be

then current through it is found as

and the current through its leakage resistance is

The total current through the upper arm containing the capacitor is

and since the bridge is balanced, the same current flows through the resistance $R_1$, therefore

On the other hand, this voltage is equal to the voltage drop across the arm with the inductance

for which the voltage drop is given by

in which the current is determined by the balance equation

since

Collecting equations (1)-(9) together, we obtain

It can be seen from this equality that the frequency-independent balance condition is satisfied if both sides of the equation are equal to zero, that is, one obtains the final answer

Problem 1.3 (4.0 points)

Let a planet of mass $m$ move around the Sun in a circular orbit of radius $R$ with a speed $v$, then the equation of motion of the planet in the projection onto the radial direction is written as

which results in

with $G$ being the gravitational constant.

Writing formula (2) for Jupiter with the index J and Earth with the index E, we get after dividing

and, on the other hand, we have according to Kepler's third law for the ratio of rotation periods

The motion of Jupiter cannot be detected with a spectrometer, but it can be done for the Sun, since it also moves around the center of mass of the Sun-Jupiter system. The speed of the Sun is easy to find from the expression

Since the Sun moves around the common center of mass of the system, and the observer is located in the same plane, according to the Doppler effect formula, the following condition is satisfied for detection

Putting together equations (3)-(6), we get the final answer

Such resolution is achievable for many modern spectrometers manufactured in different countries of the world.