weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: In an old coal factory, a conveyor belt will move at a constant velocity of $20.3 \mathrm{~m} / \mathrm{s}$ and can deliver a maximum power of $15 \mathrm{MW}$. Each wheel in the conveyor belt has a diameter of $2 \mathrm{~m}$. However a changing demand has pushed the coal factory to fill their coal hoppers with a different material with a certain constant specific density. These "coal" hoppers have been modified to deliver a constant $18 \mathrm{~m}^{3} \mathrm{~s}^{-1}$ of the new material to the conveyor belt. Assume that the kinetic and static friction are the same and that there is no slippage. What is the maximum density of the material?

Solution: The maximal force the convey belt can provide to a particle is:
$$
F=\frac{P}{v}
$$

The conveyor belt must provide an impulse to the particles to have a momentum of $p=m v$, where $m$ is the mass of the particle and $v$ is the velocity.

$$
F=\frac{d p}{d t}
$$

where $\frac{d p}{d t}$ is:

$$
\rho \dot{V} v
$$

Solving for for the maximum density we get:

$$
\begin{gathered}
\rho=\frac{P}{\dot{V} v^{2}} \\
\rho=2022.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}}
\end{gathered}
$$

Answer: $2022.2$
Problem: Neutrinos are extremely light particles and rarely interact with matter. The Sun emits neutrinos, each with an energy of $8 \times 10^{-14} \mathrm{~J}$ and reaches a flux density of $10^{11}$ neutrinos $/\left(\mathrm{s} \mathrm{cm}^{2}\right)$ at Earth's surface.

In the movie 2012, neutrinos have mutated and now are completely absorbed by the Earth's inner core, heating it up. Model the inner core as a sphere of radius $1200 \mathrm{~km}$, density $12.8 \mathrm{~g} / \mathrm{cm}^{3}$, and a specific heat of $0.400 \mathrm{~J} / \mathrm{g} \mathrm{K}$. The time scale, in seconds, that it will take to heat up the inner core by $1^{\circ} \mathrm{C}$ is $t=1 \times 10^{N}$ where $N$ is an integer. What is the value of $N$ ?

Solution: The cross sectional area is $\pi r^{2}$, so the incoming power generated by the neutrinos is:
$$
P=\pi r^{2} E \Phi
$$

where $E$ is the energy of each neutrino and $\Phi$ is the flux density. We want to cause a change in energy of:

$$
\Delta Q=m c \Delta T=\rho \frac{4}{3} \pi r^{3} c \Delta T
$$

which can be accomplished in a time:

$$
P t=\Delta Q \Longrightarrow t=\frac{\rho\left(4 \pi r^{3}\right) c \Delta T}{3 \pi r^{2} E \Phi}=\frac{4 \rho r c \Delta T}{3 E \Phi}=1 \times 10^{14} \mathrm{~s}
$$

Answer: $1 \times 10^{14}$
Problem: Eddie is experimenting with his sister's violin. Allow the "A" string of his sister's violin have an ultimate tensile strength $\sigma_{1}$. He tunes a string up to its highest possible frequency $f_{1}$ before it breaks. He then builds an exact copy of the violin, where all lengths have been increased by a factor of $\sqrt{2}$ and tunes the same string again to its highest possible frequency $f_{2}$. What is $f_{2} / f_{1}$ ? The density of the string does not change.

Note: The ultimate tensile strength is maximum amount of stress an object can endure without breaking. Stress is defined as $\frac{F}{A}$, or force per unit area.

Solution: We note from a simple dimensional analysis that the angular frequency of the string $\omega$ will consist


of the tension $T$, the length of the string $L$ and the mass of the string $m$.

$$
\begin{aligned}
T & =\left[M L T^{-2}\right] \\
L & =[L] \\
m & =[M] \\
\omega & =\left[T^{-1}\right]
\end{aligned}
$$

Therefore, by rearranging, we find that

$$
\begin{aligned}
\omega & =T^{\alpha} L^{\beta} m^{\gamma} \\
{\left[T^{-1}\right] } & =\left[M L T^{-2}\right]^{\alpha}[L]^{\beta}[M]^{\gamma}
\end{aligned}
$$

Distributing the exponents, and rearranging gives us

$$
T^{-1}=M^{\alpha+\gamma} L^{\alpha+\beta} T^{-2 \alpha}
$$

We now have three equations

$$
\begin{aligned}
\alpha+\gamma & =0 \\
\alpha+\beta & =0 \\
-2 \alpha & =-1
\end{aligned}
$$

From here, we find that $\alpha=1 / 2$. Substituting this into the first equation gives us

$$
1 / 2+\gamma=0 \Longrightarrow \gamma=-1 / 2
$$

then substituting $\alpha$ into the second equation gives us

$$
1 / 2+\beta=0 \Longrightarrow \beta=-1 / 2
$$

We now find that the angular frequency is given by

$$
\omega=A \sqrt{\frac{T}{L m}}
$$

where $A$ is an arbritary constant. Noting that $\omega=2 \pi f$, we find that

$$
f=\frac{A}{2 \pi} \sqrt{\frac{T}{L m}}
$$

From this analysis, we can then see that $f_{2} / f_{1}=\sqrt{2} / 2 \approx 0.707$.

Answer: $\frac{\sqrt{2}}{2}$
Problem: A one horsepower propeller powered by a battery and is used to propel a small boat initially at rest. You have two options:

1. Put the propeller on top of the boat and push on the air with an initial force $F_{1}$
2. Put the propeller underwater and push on the water with an initial force $F_{2}$.

The density of water is $997 \mathrm{~kg} / \mathrm{m}^{3}$ while the density of air is $1.23 \mathrm{~kg} / \mathrm{m}^{3}$. Assume that the force is both cases is dependent upon only the density of the medium, the surface area of the propeller, and the power delivered by the battery. What is $F_{2} / F_{1}$ ? You may assume (unrealistically) the efficiency of the propeller does not change. Round to the nearest tenths.

Solution: The force exerted on the fluid is roughly proportional to the change in momentum with respect to time:
$$
F=\frac{d p}{d t}=v \frac{d m}{d t}=v \frac{d}{d t}(\rho A x)=\rho A v^{2}
$$

It is kept at a constant power $P=F v$, which can allow us to solve for the speed $v$ of the propellers.

$$
P=\rho A v^{3} \Longrightarrow v=\left(\frac{P}{\rho A}\right)^{1 / 3}
$$

so the force is given by:

$$
F=\rho A\left(\frac{P}{\rho A}\right)^{2 / 3} \Longrightarrow F \propto \rho^{1 / 3}
$$

Therefore:

$$
F_{2} / F_{1}=(997 / 1.23)^{1 / 3}=9.26 \text { times }
$$

Answer: 9.26
Problem: A professional pastry chef is making a sweet which consists of 3 sheets of chocolate. The chef leaves a gap with width $d_{1}=0.1 \mathrm{~m}$ between the top and middle layers and fills it with a chocolate syrup with uniform viscosity $\eta_{1}=10 \mathrm{~Pa} \cdot \mathrm{s}$ and a gap with width $d_{2}=0.2 \mathrm{~m}$ between the middle and bottom sheet and fills it with caramel with uniform viscosity $\eta_{2}=15 \mathrm{~Pa} \cdot \mathrm{s}$. If the chef pulls the top sheet with a velocity $2 \mathrm{~m} / \mathrm{s}$ horizontally, at what speed must he push the bottom sheet horizontally such that the middle sheet remains stationary initially? Ignore the weight of the pastry sheets throughout the problem and the assume the sheets are equally sized.

Note: Shear stress is governed by the equation $\tau=\eta \times$ rate of strain.

Solution: The plates are equal sizes so all we have to do is simply balance the shear stresses which act in opposing directions on the middle plate:
$$
\begin{gathered}
\tau_{1}=\tau_{2} \\
\eta_{1} \cdot \frac{v_{1}}{d_{1}}=\eta_{2} \cdot \frac{v_{2}}{d_{2}} \\
10 \cdot \frac{2}{0.1}=15 \frac{v}{0.2} \\
v=2.667 \mathrm{~m} / \mathrm{s}
\end{gathered}
$$

Answer: $2.667$
Problem: A magnetic field is located within a region enclosed by an elliptical island with semi-minor axis of $a=100 \mathrm{~m}$ and semi-major axis of $b=200 \mathrm{~m}$. A car carrying charge $+Q=1.5 \mathrm{C}$ drives on the boundary of the island at a constant speed of $v=5 \mathrm{~m} / \mathrm{s}$ and has mass $m=2000 \mathrm{~kg}$. Any dimensions of the car can be assumed to be much smaller than the dimensions of the island. Ignore any contributions to the magnetic field from the moving car and assume that the car has enough traction to continue driving in its elliptical path.

Let the center of the island be located at the point $(0,0)$ while the semi major and semi minor axes lie on the $x$ and $y$-axes, respectively.

On this island, the magnetic field varies as a function of $x$ and $y: B(x, y)=k_{b} e^{c_{b} x y} \hat{z}$ (pointing in the upward direction, perpendicular to the island plane in the positive $z$-direction). The constant $c_{b}=10^{-4} \mathrm{~m}^{-2}$ and the constant $k_{b}=2.1 \mu \mathrm{T}$

At what point on the island is the force from the magnetic field a maximum? Write the distance of this point from the $x$-axis in metres.

Solution: To find a minimum or maximum, the gradient of the constraint function $f(x, y)=\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}-1$ and the gradient of the $B$ field function should be scalar multiples of each other.

$$
\begin{aligned}
& \frac{2 x}{b^{2}} \mu=c_{b} y e^{x y} \\
& \frac{2 y}{a^{2}} \mu=c_{b} x e^{x y}
\end{aligned}
$$

Solving the two equations, we get that a maximum point $(x, y)$ is of the form $\left(\frac{b}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right)$ or $\left(-\frac{b}{\sqrt{2}},-\frac{a}{\sqrt{2}}\right)$. The distance from the $\mathrm{y}$-axis is thus $\frac{a}{\sqrt{2}}=70.7 \mathrm{~m}$.

Answer: 70.7
Problem: Inside a laboratory at room temperature, a steel tuning fork in the shape of a $\mathrm{U}$ is struck and begins to vibrate at $f=426 \mathrm{~Hz}$. The tuning fork is then brought outside where it is $10^{\circ} \mathrm{C}$ hotter and the experiment is performed again. What is the change in frequency, $\Delta f$ of the tuning fork? (A positive value will indicate an increase in frequency, and a negative value will indicate a decrease.)

Note: The linear thermal coefficient of expansion for steel is $\alpha=1.5 \times 10^{-5} \mathrm{~K}^{-1}$ and you may assume the expansion is isotropic and linear. When the steel bends, there is a restoring torque $\tau=-\kappa \theta$ such that $\kappa \equiv G J$ where $G=77 \mathrm{GPa}$ is constant and $J$ depends on the geometry and dimensions of the cross-sectional area.

Solution: Note that $\kappa$ has units of torque so dimensionally, $J$ must be proportional to $L^{3}$. Therefore, we have:
$$
\beta M L^{2} \alpha \propto-L^{3} \theta \Longrightarrow f \propto \sqrt{L}
$$

Therefore, we have:

$$
\frac{\Delta f}{f}=\frac{\Delta L}{2 L}
$$

Since $\frac{\Delta L}{L}=\alpha \Delta T$, this gives us:

$$
\Delta f=\frac{1}{2} f \alpha \Delta T=0.0320 \mathrm{~Hz}
$$

Answer: 0.0320
Problem: A large metal conducting sphere with radius $10 \mathrm{~m}$ at an initial potential of 0 and an infinite supply of smaller conducting spheres of radius $1 \mathrm{~m}$ and potential $10 \mathrm{~V}$ are placed into contact in such a way: the large metal conducting sphere is contacted with each smaller sphere one at a time. You may also assume the spheres are touched using a thin conducting wire that places the two spheres sufficiently far away from each other such that their own spherical charge symmetry is maintained. What is the least number of smaller spheres required to be touched with the larger sphere such that the potential of the larger sphere reaches $9 \mathrm{~V}$ ? Assume that the charges distribute slowly and that the point of contact between the rod and the spheres is not a sharp point.

Solution: Let each sphere with radius $1 \mathrm{~m}$ have charge $q$. Note that each time the large metal conducting sphere is contacted with each of the smaller spheres, the potential is equalized between the two objects. The potential on a sphere is proportional to $\frac{q}{r}$, so the large conducting sphere must retain $\frac{10}{11}$ of the total charge after it is contacted with a smaller sphere. Furthermore, to reach $9 \mathrm{~V}$, the required end charge on the sphere of radius $10 \mathrm{~m}$ is at least $9 q$. Thus, we get a recursion for the charge of the large square $Q$ in terms of the number of small spheres touched $n$.
$$
Q(n+1)=(Q(n)+q) \cdot \frac{10}{11}
$$

Inductively applying this recursion, we obtain

$$
Q(n)=q\left[\left(\frac{10}{11}\right)^{n}+\cdots+\frac{10}{11}\right]
$$

We can now sum this geometric series:

$$
Q(n)=q\left(\frac{10}{11} \cdot \frac{\left(\frac{10}{11}\right)^{n}-1}{\frac{10}{11}-1}\right) .
$$

Thus, using $Q(n) \geq 9 q$, we find that $\left(\frac{10}{11}\right)^{n} \leq 0.1$, which provides $n \geq 24.1588$, or $n=25$ as the answer.

Answer: 25
Problem: During high speed motion in a strong electric field, a charged particle can ionize air molecules it collides with.

A charged particle of mass $m=0.1 \mathrm{~kg}$ and charge $q=0.5 \mu \mathrm{C}$ is located in the center of a cubical box. Each vertex of the box is fixed in space and has a charge of $Q=-4 \mu \mathrm{C}$. If the side length of the box is $l=1.5 \mathrm{~m}$ what minimum speed (parallel to an edge) should be given to the particle for it to exit the box (even if it's just momentarily)? Let the energy loss from Corona discharge and other radiation effects be $E=0.00250 \mathrm{~J}$.

Solution: Conservation of energy gives:
$$
T_{i}+U_{i}=T_{f}+U_{f}+E
$$

Solving for the initial potential energy gives

$$
U_{i}=-8 \frac{k q Q}{l \sqrt{3} / 2}=-\frac{16 k q Q}{\sqrt{3} l}
$$

And since the final kinetic energy is zero, the final potential energy is

$$
U_{f}=-4 \frac{k q Q}{l / \sqrt{2}}-4 \frac{k q Q}{l \sqrt{\frac{3}{2}}}=-\left(4 \sqrt{2}+4 \sqrt{\frac{2}{3}}\right) \frac{k q Q}{l}
$$

and thus solving for the initial kinetic energy:

$$
\frac{1}{2} m v^{2}=\left(-4 \sqrt{2}-4 \sqrt{\frac{2}{3}}+\frac{16}{\sqrt{3}}\right) \frac{k q Q}{l}+E
$$

The final answer is $v=0.354 \mathrm{~m} / \mathrm{s}$.

Answer: 0.354
Problem: Max finds himself trapped in the center of a mirror walled equilateral triangular room. What minimum beam angle must his flashlight have so that any point of illumination in the room can be traced back to his flashlight with at most 1 bounce? (Answer in degrees.) Since the room is large, assume the person is a point does not block light. Visualize the questions in a 2D setup. The floor/ceiling is irrelevant.

The point of illumination refers to any point in the room that is lit.

Solution: Each time light hits a mirror, we can reflect the entire equilateral triangle about that mirror and continue to trace the straight-line path of the light. For a maximum of 1 bounce, we can reflect our triangle about each of its initial sides. In order for the light to hit every part of the triangle (or an image of that part), by symmetry we require a $120^{\circ}$ angle. In this case, we would just shine the flashlight so that light directly reaches the entirety of one side, and reflections of light will fully reach the other two sides of the triangle.

Answer: $120$
Problem: Kushal finds himself trapped in a large room with mirrors as walls. Being scared of the dark, he has a powerful flashlight to light the room. All references to "percent" refer to area. Since the room is large, assume the person is a point does not block light. Visualize the questions in a 2D setup. The floor/ceiling is irrelevant. The point of illumination refers to any point in the room that is lit.

What percent of a large circular room can be lit up using a flashlight with a 20 degree beam angle if Kushal stands in the center?

Solution: Each ray emitted follows a straight line, even when it is reflected since it originates from the center of a circle. Thus, the light rays in total trace out two circular sectors with an angle of $\theta=20^{\circ}$ each. Thus, the total percent of the room illuminated is:

$$
f=\frac{2 \theta}{360}=11.1 \% \text {. }
$$

Answer: 11.1
Problem: Two identical neutron stars with mass $m=4 \times 10^{30} \mathrm{~kg}$ and radius $15 \mathrm{~km}$ are orbiting each other a distance $d=700 \mathrm{~km}$ away from each other ( $d$ refers to the initial distance between the cores of the neutron stars). Assume that they orbit as predicted by classical mechanics, except that they generate gravitational waves. The power dissipated through these waves is given by:

$$
P=\frac{32 G^{4}}{5}\left(\frac{m}{d c}\right)^{5}
$$

How long does it take for the two stars to collide? Answer in seconds. Note: $d$ is the distance between the cores of the stars.

Solution: Due to Virial theorem, we have:
$$
K=-\frac{1}{2} U
$$

so the total energy is:

$$
E=U-\frac{1}{2} U=-\frac{G m^{2}}{2 R}
$$

We know that the power dissipated gives the change in energy, or:

$$
P=\frac{32 G^{4}}{5}\left(\frac{m}{R c}\right)^{5}=\frac{d}{d t} \frac{G m^{2}}{2 R}
$$

or:

$$
\frac{32 G^{4}}{5}\left(\frac{m}{R c}\right)^{5} d t=-\frac{G m^{2}}{2 R^{2}} d R \Longrightarrow \int_{0}^{t} \frac{64 G^{3}}{5} \frac{m^{3}}{c^{5}} d t=\int_{d}^{2 r}-R^{3} d R
$$

Solving this leads us to:

$$
\frac{64 G^{3} m^{3}}{5 c^{5}} t=\frac{d^{4}-r^{4}}{4} \Longrightarrow t=\frac{5 c^{5}\left(d^{4}-16 r^{4}\right)}{256 G^{3} m^{3}}
$$

Plugging in the numbers gives:

$$
t=590 \mathrm{sec}
$$

Note that we can also assume that $d^{4} \gg(2 r)^{4}$ which will simplify calculations, and not introduce any noticeable error.

Answer: 590
Problem: In the cosmic galaxy, the Sun is a mainsequence star, generating its energy mainly by nuclear fusion of hydrogen nuclei into helium. In its core, the Sun fuses hydrogen to produce deuterium $(2 \mathrm{H})$ and tritium $(3 \mathrm{H})$, then makes about 600 million metric tons of helium (4He) per second. Of course, there are also some relatively smaller portions of fission reactions in the Sun's core, e.g. a nuclear fission reaction with Uranium-235 (235U). The Fusion reaction:

$$
{ }^{2} \mathrm{H}+{ }^{3} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+n+\text { Released Energy }
$$

The Fission reaction:

$$
{ }^{235} U+n+(\text { Initial Energy }) \rightarrow{ }^{144} \mathrm{Ba}+{ }^{90} \mathrm{Kr}+2 n+\text { Released Energy }
$$

Isotope Mass (at rest) 

| Isotope Names | Mass (at rest) $(\mathrm{u})$ |
| :--- | :--- |
| Deuterium $\left({ }^{2} \mathrm{H}\right)$ | 2.0141 |
| Tritium $\left({ }^{3} \mathrm{H}\right)$ | 3.0160 |
| Helium $\left({ }^{4} \mathrm{He}\right)$ | 4.0026 |
| Neutron $(\mathrm{n})$ | 1.0087 |
| Uranium-235 $\left({ }^{235} \mathrm{U}\right)$ | 235.1180 |
| Barium-144 $\left({ }^{144} \mathrm{Ba}\right)$ | 143.8812 |
| Krypton-90 $\left({ }^{90} \mathrm{Kr}\right)$ | 89.9471 |

Calculate the kinetic energy (in $\mathrm{MeV}$ ) released by the products in one fusion reaction.

Solution: Let the kinetic energy released be

$$
K E_{\text {released }}=-\Delta m c^{2}
$$

Let the mass of helium be $m_{h}$, deuterium be $m_{d}$, tritium $m_{t}$, and mass of neutron $m_{n}$ Therefore.

$$
-\Delta m=m_{d}+m_{t}-m_{h}-m_{n}=3.0160+2.0141-4.0026-1.0087=0.0188 \mathrm{u}
$$

which gives

$$
K E_{\text {released }}=(0.0188 \mathrm{u}) \cdot\left(931.494 \frac{\mathrm{MeV}}{\mathrm{u}}\right)=17.51 \mathrm{MeV}
$$

Answer: 17.51
Problem: A particle of rest mass $m$ moving at a speed $v=0.7 c$ decomposes into two photons which fly off at a separated angle $\theta$. What is the minimum value of the angle of separation assuming that the two photons have equal wavelength. (Answer in degrees)

Solution: Conservation of momentum and energy gives:
$$
\begin{gathered}
p_{m}=2 E_{\gamma} \cos (\theta / 2) \\
E_{m}=2 E_{\gamma}
\end{gathered}
$$

Relativity demands that:

$$
E_{m}^{2}=m^{2}+p_{m}^{2}
$$

Solving this system of three equations gives:

$$
\begin{aligned}
4 E_{\gamma}^{2} & =m^{2}+p_{m}^{2} \\
\frac{\gamma^{2} m^{2} v^{2}}{\cos ^{2}(\theta / 2)} & =m^{2}+\gamma^{2} m^{2} v^{2} \\
\gamma^{2} v^{2}\left(-1+\frac{1}{\cos ^{2}(\theta / 2)}\right) & =1 \\
\frac{1}{\cos ^{2}(\theta / 2)} & =\frac{1}{\gamma^{2} v^{2}}+1 \\
\cos (\theta / 2) & =v \\
\theta & =91.1^{\circ}
\end{aligned}
$$

Answer: $91.1$
Problem: At an amusement park, there is a ride with three "teacups" that are circular with identical dimensions. Three friends, Ethan, Rishab, and Kushal, all pick a teacup and sit at the edge. Each teacup rotates about its own axis clockwise at an angular speed $\omega=1 \mathrm{rad} / \mathrm{s}$ and can also move linearly at the same time.

The teacup Ethan is sitting on (as always) is malfunctional and can only rotate about its own axis. Rishab's teacup is moving linearly at a constant velocity $2 \mathrm{~m} / \mathrm{s}[\mathrm{N}]$ and Kushal's teacup is also moving linearly at a constant velocity of $4 \mathrm{~m} / \mathrm{s}\left[\mathrm{N} 60^{\circ} \mathrm{E}\right]$. All three teacups are rotating as described above. Interestingly, they observe that at some point, all three of them are moving at the same velocity. What is the radius of each teacup?

Note: $\left[\mathrm{N} 60^{\circ} \mathrm{E}\right]$ means $60^{\circ}$ clockwise from north e.g. $60^{\circ}$ east of north.

Solution: We can plot the motion on a $v_{y}-v_{z}$ graph instead of carrying out calculations. We have three points at locations $(0,0),(0,2)$, and $(2 \sqrt{3}, 2)$ which represent the velocity of the center of mass of the teacups. The velocity that they are moving at can be traced as a circle with radius $r \omega$, centered at these points.
The problem now becomes, at what value $r$ will the three circles intersect. Drawing a diagram, or carrying out trigonometric calculations gives $r=2 \mathrm{~m}$.

Answer: 2
Problem: Life on Earth would not exist as we know it without the atmosphere. There are many reasons for this, but one of which is temperature. Let's explore how the atmosphere affects the temperature on Earth. Assume that all thermal energy striking the earth uniformly and ideally distributes itself across the Earth's surface.

- Assume that the Earth is a perfect black body with no atmospheric effects. Let the equilibrium temperature of Earth be $T_{0}$. (The sun outputs around $3.846 \times 10^{2^{6}} \mathrm{~W}$, and is $1.496 \times 10^{8} \mathrm{~km}$ away.)
- Now assume the Earth's atmosphere is isothermal. The short wavelengths from the sun are nearly unaffected and pass straight through the atmosphere. However, they mostly convert into heat when they strike the ground. This generates longer wavelengths that do interact with the atmosphere. Assume that the albedo of the ground is 0.3 and $e$, the emissivity and absorptivity of the atmosphere, is 0.8 . Let the equilibrium average temperature of the planet now be $T_{1}$.

What is the percentage increase from $T_{0}$ to $T_{1}$ ?

Note: The emissivity is the degree to which an object can emit longer wavelengths (infrared) and the absorptivity is the degree to which an object can absorb energy. Specifically, the emissivity is the ratio between the energy emitted by an object and the energy emitted by a perfect black body at the same temperature. On the other hand, the absorptivity is the ratio of the amount of energy absorbed to the amount of incident energy.

Solution: Let us solve this problem in the case of the Earth being a graybody first and then substitute values for when it is a blackbody. The portion of energy that reaches the Earth is given by the ratio between the cross-sectional area of the satellite and the area of an imaginary sphere centered around the sun with a radius of $L$. Thus, the incoming radiation is multiplied by a factor of $\gamma=(R / 2 L)^{2}$. The energy from the sun that the surface absorbs is $\gamma(1-\alpha) E$, where $E$ is the energy output of the sun. Here $\gamma=1 / 4$ as the sphere encompassing will be 4 times the area of its intercept.
We can now write two systems of equations at the atmosphere and the ground of the Earth. At the top of the atmosphere, we require equilibrium meaning that zero net radiation leaves the atmosphere or:

$$
-\frac{1}{4} S_{0}(1-\alpha)+\varepsilon \sigma T_{a}^{4}+(1-\varepsilon) \sigma T_{s}^{4}=0
$$

Similarly, at the ground, we write another equilibrium equation of:

$$
\frac{1}{4} S_{0}(1-\alpha)+\varepsilon \sigma T_{a}^{4}-\sigma T_{s}^{4}=0
$$

Thus, solving the ground equilibrium equation yields us $T_{a}=2^{-1 / 4} T_{s}$ and plugging back into the atmosphere equilibrium equation tells us:

$$
\frac{1}{4} S_{0}(1-\alpha)=\left(1-\frac{\varepsilon}{2}\right) \sigma T_{s}^{4} \Longrightarrow T_{s}=289.601 \mathrm{~K}
$$

Answer: 289.601