di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Optics

Figure 1.1

<img_4357>

A plane monochromatic light wave, wavelength $\lambda$ and frequency $f$, is incident normally on two identical narrow slits, separated by a distance $d$, as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance $x$ in a direction $\theta$ at time $t$, by

$$
y=a \cos [2 \pi(f t-x / \lambda)]
$$

where the amplitude $a$ is the same for both waves. (Assume $x$ is much larger than $d$ ).
Context question:
(i) Show that the two waves observed at an angle $\theta$ to a normal to the slits, have a resultant amplitude A which can be obtained by adding two vectors, each having magnitude $a$, and each with an associated direction determined by the phase of the light wave.

Verify geometrically, from the vector diagram, that

$$
A=2 a \cos \theta
$$

where

$$
\beta=\frac{\pi}{\lambda} d \sin \theta
$$
Context answer:
\boxed{证明题}


Context question:
(ii) The double slit is replaced by a diffraction grating with $N$ equally spaced slits, adjacent slits being separated by a distance $d$. Use the vector method of adding amplitudes to show that the vector amplitudes, each of magnitude $a$, form a part of a regular polygon with vertices on a circle of radius $R$ given by

$$
R=\frac{a}{2 \sin \beta}
$$

Deduce that the resultant amplitude is

$$
\frac{a \sin N \beta}{\sin \beta}
$$



and obtain the resultant phase difference relative to that of the light from the slit at the edge of the grating.
Context answer:
\boxed{证明题}


Context question:
(iv) Determine the intensities of the principal intensity maxima.
Context answer:
\boxed{$N^{2} a^{2}$}


Question: (v) Show that the number of principal maxima cannot exceed

$$
\left(\frac{2 d}{\lambda}+1\right)
$$

Solution:
(v) Adjacent max. estimate $I_{l}$ :

$$
\sin ^{2} N \beta=1, \quad \beta=2 \pi p \mp \frac{3 \pi}{2 N} \text { i.e } \beta= \pm \frac{3 \pi}{2 N}
$$

$$
\begin{gathered}
{\left[\beta=\pi p \pm \frac{\pi}{2 N}\right] \text { does not give a maximum as can be observed from the graph. }} \\
\qquad I_{1}=a^{2} \frac{1}{\frac{3 \pi^{2}}{2 n}}=\frac{a^{2} N^{2}}{23} \text { for } N \gg>1
\end{gathered}
$$

Adjacent zero intensity occurs for $\beta=\pi \rho \pm \frac{\pi}{N}$ i.e. $\delta= \pm \frac{\pi}{N}$

For phase differences much greater than $\delta, \quad \mathrm{I}=\mathrm{a}^{2}\left(\frac{\sin N \beta}{\sin \beta}\right)=a^{2}$
Subfield: Mechanics


Question: (i) If each mass is displaced from equilibrium by small displacements $u_{1}, u_{2}$ and $u_{3}$ respectively, write down the equation of motion for each mass.

Solution:
Equations of motion:

$$
\begin{aligned}
& m \frac{d^{2} u_{1}}{d t^{2}}=k\left(u_{2}-u_{1}\right)+k\left(u_{3}-u_{1}\right) \\
& m \frac{d^{2} u_{2}}{d t^{2}}=k\left(u_{3}-u_{2}\right)+k\left(u_{1}-u_{2}\right) \\
& m \frac{d^{2} u_{3}}{d t^{2}}=k\left(u_{1}-u_{3}\right)+k\left(u_{2}-u_{3}\right)
\end{aligned}
$$

Substituting $u_{n}(t)=u_{n}(0) \cos \omega t$ and $\omega_{o}^{2}=\frac{k}{m}$ :

$$
\begin{aligned}
\left(2 \omega_{o}^{2}-\omega^{2}\right) u_{1}(0)-\omega_{o}{ }^{2} u_{2}(0)-\omega_{o}^{2} u_{3}(0) & =0 \\
-\omega_{o}{ }^{2} u_{1}(0)+\left(2 \omega_{o}^{2}-\omega^{2}\right) u_{2}(0)-\omega_{o}{ }^{2} u_{3}(0) & =0 \\
-\omega_{o}{ }^{2} u_{1}(0)-\omega_{o}{ }^{2} u_{2}(0)+\left(2 \omega_{o}{ }^{2}-\omega^{2}\right) u_{3}(0) & =0
\end{aligned}
$$

Solving for $u_{1}(0)$ and $u_{2}(0)$ in terms of $u_{3}(0)$ using (a) and (b) and substituting into (c) gives the equation equivalent to

$$
\begin{gathered}
\left(3 \omega_{o}{ }^{2}-\omega^{2}\right)^{2} \omega^{2}=0 \\
\omega^{2}=3 \omega_{o}{ }^{2}, 3 \omega_{o}{ }^{2} \text { and } 0 \\
\omega=\sqrt{3} \omega_{o}, \sqrt{3} \omega_{o} \text { and } 0
\end{gathered}
$$

Final answer: $$
\begin{aligned}
& m \frac{d^{2} u_{1}}{d t^{2}}=k\left(u_{2}-u_{1}\right)+k\left(u_{3}-u_{1}\right) \\
& m \frac{d^{2} u_{2}}{d t^{2}}=k\left(u_{3}-u_{2}\right)+k\left(u_{1}-u_{2}\right) \\
& m \frac{d^{2} u_{3}}{d t^{2}}=k\left(u_{1}-u_{3}\right)+k\left(u_{2}-u_{3}\right)
\end{aligned}
$$
Subfield: Mechanics

Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1.

Figure 3.1

<img_4463>
Context question:
(i) If each mass is displaced from equilibrium by small displacements $u_{1}, u_{2}$ and $u_{3}$ respectively, write down the equation of motion for each mass.
Context answer:
$$
\begin{aligned}
& m \frac{d^{2} u_{1}}{d t^{2}}=k\left(u_{2}-u_{1}\right)+k\left(u_{3}-u_{1}\right) \\
& m \frac{d^{2} u_{2}}{d t^{2}}=k\left(u_{3}-u_{2}\right)+k\left(u_{1}-u_{2}\right) \\
& m \frac{d^{2} u_{3}}{d t^{2}}=k\left(u_{1}-u_{3}\right)+k\left(u_{2}-u_{3}\right)
\end{aligned}
$$


Question: (ii) Verify that the system has simple harmonic solutions of the form

$$
u_{n}=a_{n} \cos \omega t
$$

with accelerations, $\left(-\omega^{2} u_{n}\right)$ where $a_{n}(n=1,2,3)$ are constant amplitudes, and $\omega$, the angular frequency, can have 3 possible values,

$$
\omega_{o} \sqrt{3}, \omega_{o} \sqrt{3} \text { and } 0 \text {. where } \omega_{o}^{2}=\frac{k}{m} \text {. }
$$

Solution:
(ii) Equation of motion of the n'th particle:

$$
\begin{aligned}
& m \frac{d^{2} u_{n}}{d t^{2}}=k\left(u_{1+n}-u_{n}\right)+k\left(u_{n-1}-u_{n}\right) \\
& \frac{d^{2} u_{n}}{d t^{2}}=k\left(u_{1+n}-u_{n}\right)+\omega_{o}^{2}\left(u_{n-1}-u_{n}\right)
\end{aligned}
$$

Substituting $u_{n}(t)=u_{n}(0) \sin \left(2 n s \frac{\pi}{N}\right) \cos \omega_{s} t$

$$
\begin{aligned}
& -\omega_{s}^{2}\left(\sin \left(2 n s \frac{\pi}{N}\right)\right)=\omega_{o}^{2}\left[\sin \left(2(n+1) s \frac{\pi}{N}\right)-2 \sin \left(2 n s \frac{\pi}{N}\right)+\sin \left(2(n-1) s \frac{\pi}{N}\right)\right] \\
& -\omega_{s}^{2}\left(\sin \left(2 n s \frac{\pi}{N}\right)\right)=2 \omega_{o}^{2}\left[\frac{1}{2} \sin \left(2(n+1) s \frac{\pi}{N}\right)+\sin \left(2 n s \frac{\pi}{N}\right)-\frac{1}{2} \sin \left(2(n-1) s \frac{\pi}{N}\right)\right] \\
& -\omega_{s}^{2}\left(\sin \left(2 n s \frac{\pi}{N}\right)\right)=2 \omega_{o}^{2}\left[\sin \left(2 n s \frac{\pi}{N}\right) \cos \left(2 s \frac{\pi}{N}\right)-\sin \left(2 n s \frac{\pi}{N}\right)\right] \\
& \therefore \omega_{s}^{2}=2 \omega_{o}^{2}\left[1-\cos \left(2 s \frac{\pi}{N}\right)\right]: \quad(s=1,2, \ldots . . N)
\end{aligned}
$$

As $2 \sin ^{2} \theta=1-\cos 2 \theta$

This gives

$$
\omega_{s}=2 \omega_{o} \sin \left(\frac{s \pi}{N}\right) \quad(s=1,2, \ldots N)
$$

$\omega_{s}$ can have values from 0 to $2 \omega_{o}=2 \sqrt{\frac{k}{m}}$ when $N \rightarrow \infty$; corresponding to range $s=1$ to $\frac{N}{2}$.
Subfield: Mechanics

Problem T1. Focus on sketches

Part A. Ballistics

A ball, thrown with an initial speed $v_{0}$, moves in a homogeneous gravitational field in the $x-z$ plane, where the $x$-axis is horizontal, and the $z$-axis is vertical and antiparallel to the free fall acceleration $g$. Neglect the effect of air drag.

Question: i. By adjusting the launching angle for a ball thrown with a fixed initial speed $v_{0}$ from the origin, targets can be hit within the region given by

$$
z \leq z_{0}-k x^{2}
$$

You can use this fact without proving it. Find the constants $z_{0}$ and $k$.

Solution:
When the stone is thrown vertically upwards, it can reach the point $x=0, z=v_{0}^{2} / 2 g$ (as it follows from the energy conservation law). Comparing this with the inequality $z \leq z_{0}-k x^{2}$ we conclude that

$$
z_{0}=v_{0}^{2} / 2 g
$$

Let us consider the asymptotics $z \rightarrow-\infty$; the trajectory of the stone is a parabola, and at this limit, the horizontal displacement (for the given $z$ ) is very sensitive with respect to the curvature of the parabola: the flatter the parabola, the larger the displacement. The parabola has the flattest shape when the stone is thrown horizontally, $x=v_{0} t$ and $z=-g t^{2} / 2$, i.e. its trajectory is given by $z=-g x^{2} / 2 v_{0}^{2}$. Now, let us recall that $z \leq z_{0}-k x^{2}$, i.e. $-g x^{2} / 2 v_{0}^{2} \leq z_{0}-k x^{2} \Rightarrow k \leq g / 2 v_{0}^{2}$. Note that $k<g / 2 v_{0}^{2}$ would imply that there is a gap between the parabolic region $z \leq z_{0}-k x^{2}$ and the given trajectory $z=-g x^{2} / 2 v_{0}^{2}$. This trajectory is supposed to be optimal for hitting targets far below $(z \rightarrow-\infty)$, so there should be no such a gap, and hence, we can exclude the option $k<g / 2 v_{0}^{2}$. This leaves us with

$$
k=g / 2 v_{0}^{2}
$$

Final answer: $\frac{v_{0}^{2}}{ 2 g}$ , $\frac{g}{2 v_{0}^{2}}$
Subfield: Mechanics


Question: iii. What is the minimal launching speed $v_{\min }$ needed to hit the topmost point of a spherical building of radius $R$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_72b956b534088c26e196g-1.jpg?height=923&width=968&top_left_y=1680&top_left_x=59)

La Geode, Parc de la Villette, Paris. Photo: katchooo/flickr.com

Solution:
The brute force approach would be writing down the condition that the optimal trajectory intersects with the building at two points and touches at one. This would be described by a fourth order algebraic equation and therefore, it is not realistic to accomplish such a solution within a reasonable time frame.

Note that the interior of the building needs to lie inside the region where the targets can be hit with a stone thrown from the top with initial speed $v_{\min }$. Indeed, if we can throw over the building, we can hit anything inside by lowering the throwing angle. On the other hand, the boundary of the targetable region needs to touch the building. Indeed, if there were a gap, it would be possible to hit a target just above the point where the optimal trajectory touches the building; the trajectory through that target wouldn't touch the building anywhere, hence we arrive at a contradiction.

So, with $v_{0}$ corresponding to the optimal trajectory, the targetable region touches the building; due to symmetry, overall there are two touching points (for smaller speeds, there would be four, and for larger speeds, there would be none). With the origin at the top of the building, the intersection points are defined by the following system of equations:

$$
x^{2}+z^{2}+2 z R=0, \quad z=\frac{v_{0}^{2}}{2 g}-\frac{g x^{2}}{2 v_{0}^{2}}
$$

Upon eliminating $z$, this becomes a biquadratic equation for $x$ :

$$
x^{4}\left(\frac{g}{2 v_{0}^{2}}\right)^{2}+x^{2}\left(\frac{1}{2}-\frac{g R}{v_{0}^{2}}\right)+\left(\frac{v_{0}^{2}}{4 g}+R\right) \frac{v_{0}^{2}}{g}=0
$$

Hence the speed by which the real-valued solutions disappear can be found from the condition that the discriminant vanishes:

$$
\left(\frac{1}{2}-\frac{g R}{v_{0}^{2}}\right)^{2}=\frac{1}{4}+\frac{g R}{v_{0}^{2}} \Longrightarrow \frac{g R}{v_{0}^{2}}=2
$$

Bearing in mind that due to the energy conservation law, at the ground level the squared speed is increased by $4 g R$. Thus we finally obtain

$$
v_{\min }=\sqrt{v_{0}^{2}+4 g R}=3 \sqrt{\frac{g R}{2}} .
$$

Final answer: $v_{\min }=3 \sqrt{\frac{g R}{2}}$
Subfield: Electromagnetism

Part C. Magnetic straws

Consider a cylindrical tube made of a superconducting material. The length of the tube is $l$ and the inner radius is $r$ with $l \gg r$. The centre of the tube coincides with the origin, and its axis coincides with the $z$-axis.

<img_4512>

There is a magnetic flux $\Phi$ through the central cross-section of the tube, $z=0, x^{2}+y^{2}<r^{2}$. A superconductor is a material which expels any magnetic field (the field is zero inside the material).
Context question:
i. Sketch five such magnetic field lines, which pass through the five red dots marked on the axial cross-section of the tube, on the designated diagram on the answer sheet.
Context answer:
<img_4360>


Question: ii. Find the tension force $T$ along the $z$-axis in the middle of the tube (i.e. the force by which two halves of the tube, $z>0$ and $z<0$, interact with each other).

Solution:
Let us consider the change of the magnetic energy when the tube is stretched (virtually) by a small amount $\Delta l$. Note that the magnetic flux trough the tube is conserved: any change of flux would imply a non-zero electromotive force $\frac{d \Phi}{d t}$, and for a zero resistivity, an infinite current. So, the induction $B=\frac{\Phi}{\pi r^{2}}$. The energy density of the magnetic field is $\frac{B^{2}}{2 \mu_{0}}$. Thus, the change of the magnetic energy is calculated as

$$
\Delta W=\frac{B^{2}}{2 \mu_{0}} \pi r^{2} \Delta l=\frac{\Phi^{2}}{2 \mu_{0} \pi r^{2}} \Delta l
$$

This energy increase is achieved owing to the work done by the stretching force, $\Delta W=T \Delta l$. Hence, the force

$$
T=\frac{\Phi^{2}}{2 \mu_{0} \pi r^{2}}
$$

Final answer: $T=\frac{\Phi^{2}}{2 \mu_{0} \pi r^{2}}$
Subfield: Thermodynamics

Problem T2. Kelvin water dropper

The following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\sigma S$, where $S$ is the surface area of the interface and $\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\sigma l$, where $l$ is the length of a straight line separating the fragments.

<img_4427>

A long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\sigma$ and its density is $\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \ll r$.

Part A. Single pipe

Question: i. Find the radius $r_{\max }$ of a drop just before it separates from the nozzle.

Solution:
Let us write the force balance for the droplet. Since $d \ll r$, we can neglect the force $\frac{\pi}{4} \Delta p d^{2}$ due to the excess pressure $\Delta p$ inside the tube. So, the gravity force $\frac{4}{3} \pi r_{\max }^{3} \rho g$ is balanced by the capillary force. When the droplet separates from the tube, the water surface forms in the vicinity of the nozzle a "neck", which has vertical tangent. In the horizontal cross-section of that "neck", the capillary force is vertical and can be calculated as $\pi \sigma d$. So,

$$
r_{\max }=\sqrt[3]{\frac{3 \sigma d}{4 \rho g}}
$$

Final answer: $r_{\max }=\sqrt[3]{\frac{3 \sigma d}{4 \rho g}}$
Subfield: Thermodynamics

Problem T2. Kelvin water dropper

The following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\sigma S$, where $S$ is the surface area of the interface and $\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\sigma l$, where $l$ is the length of a straight line separating the fragments.

<img_4427>

A long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\sigma$ and its density is $\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \ll r$.

Part A. Single pipe
Context question:
i. Find the radius $r_{\max }$ of a drop just before it separates from the nozzle.
Context answer:
\boxed{$r_{\max }=\sqrt[3]{\frac{3 \sigma d}{4 \rho g}}$}


Question: ii. Relative to the far-away surroundings, the pipe's electrostatic potential is $\varphi$. Find the charge $Q$ of a drop when its radius is $r$.

Solution:
Since $d \ll r$, we can neglect the change of the droplet's capacitance due to the tube. On the one hand, the droplet's potential is $\varphi$; on the other hand, it is $\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}$. So,

$$
Q=4 \pi \varepsilon_{0} \varphi r
$$

Final answer: $Q=4 \pi \varepsilon_{0} \varphi r$
Subfield: Modern Physics

Problem T3. Protostar formation

Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant.

Question: i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.

Solution:
$$
\begin{aligned}
& T=\text { const } \Longrightarrow p V=\text { const } \\
& V \propto r^{3} \\
& \therefore p \propto r^{-3} \Longrightarrow \frac{p\left(r_{1}\right)}{p\left(r_{0}\right)}=2^{3}=8 .
\end{aligned}
$$

Final answer: $n=8$
Subfield: Modern Physics

Problem T3. Protostar formation

Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant.
Context question:
i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.
Context answer:
\boxed{$n=8$}


Question: ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle.

Solution:
During the period considered the pressure is negligible. Therefore the gas is in free fall. By Gauss' theorem and symmetry, the gravitational field at any point in the ball is equivalent to the one generated when all the mass closer to the center is compressed into the center. Moreover, while the ball has not yet shrunk much, the field strength on its surface does not change much either. The acceleration of the outermost layer stays approximately constant. Thus,

$$
t \approx \sqrt{\frac{2\left(r_{0}-r_{2}\right)}{g}}
$$

where

$$
\begin{aligned}
g & \approx \frac{G m}{r_{0}^{2}} \\
\therefore t & \approx \sqrt{\frac{2 r_{0}^{2}\left(r_{0}-r_{2}\right)}{G m}}=\sqrt{\frac{0.1 r_{0}^{3}}{G m}} .
\end{aligned}
$$

Final answer: $\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$
Subfield: Modern Physics

Problem T3. Protostar formation

Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant.
Context question:
i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.
Context answer:
\boxed{$n=8$}


Context question:
ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle.
Context answer:
\boxed{$\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$}


Context question:
iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws.
Context answer:
\boxed{$t_{r \rightarrow 0}=\pi \sqrt{\frac{r_{0}^{3}}{8 G m}}$}


Question: iv. At some radius $r_{3} \ll r_{0}$, the gas becomes dense enough to be opaque to the heat radiation. Calculate the amount of heat $Q$ radiated away during the collapse from the radius $r_{0}$ down to $r_{3}$.

Solution:
By Clapeyron-Mendeleyev law,

$$
p=\frac{m R T_{0}}{\mu V}
$$

Work done by gravity to compress the ball is

$$
W=-\int p d V=-\frac{m R T_{0}}{\mu} \int_{\frac{4}{3} \pi r_{0}^{3}}^{\frac{4}{3} \pi r_{3}^{3}} \frac{d V}{V}=\frac{3 m R T_{0}}{\mu} \ln \frac{r_{0}}{r_{3}}
$$

The temperature stays constant, so the internal energy does not change; hence, according to the $1^{\text {st }}$ law of thermodynamics, the compression work $W$ is the heat radiated.

Final answer: $Q=\frac{3 m R T_{0}}{\mu} \ln \frac{r_{0}}{r_{3}}$