weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: Adithya is launching a package from New York City $\left(40^{\circ} 43^{\prime} \mathrm{N}\right.$ and $\left.73^{\circ} 56^{\prime} \mathrm{W}\right)$ to Guam $\left(13^{\circ} 27^{\prime} \mathrm{N}\right.$ and $\left.144^{\circ} 48^{\prime} \mathrm{E}\right)$. Find the minimal launch velocity $v_{0}$ from New York City to Guam. Ignore the rotation of the earth, effects due to the atmosphere, and the gravitational force from the sun. Additionally, assume the Earth is a perfect sphere with radius $R_{\oplus}=6.37 \times 10^{6} \mathrm{~m}$ and mass $M_{\oplus}=5.97 \times 10^{24} \mathrm{~kg}$.

Solution: We first want to find the angular distance between New York City and Guam. Let this be $\theta$. Let New York City be point $A$ and Guam be point $B$. Consider the north pole $P$ and the spherical triangle $P A B$. By the spherical law of cosines,
$$
\cos \theta=\cos \left(90^{\circ}-\phi_{A}\right) \cos \left(90^{\circ}-\phi_{B}\right)+\sin \left(90^{\circ}-\phi_{A}\right) \sin \left(90^{\circ}-\phi_{B}\right) \cos \left(\ell_{B}-\ell_{A}\right) .
\tag{23}
$$



Equation 23 simplifies to

$$
\cos \theta=\sin \phi_{A} \sin \phi_{B}+\cos \phi_{A} \cos \phi_{B} \cos \left(\ell_{B}-\ell_{A}\right)
\tag{24}
$$

from which we find $\theta=115.05^{\circ}$.

Now that we have determined the angular distance, we will proceed with the orbital mechanics problem. By the vis-viva equation, the speed at the launch point is

$$
v_{0}=\sqrt{G M_{\oplus}\left(\frac{2}{R_{\oplus}}-\frac{1}{a}\right)}
\tag{25}
$$

where $a$ is the semimajor axis for the orbit. It is clear that in order to minimize $v_{0}$, we must minimize $a$. The orbit is an ellipse with focii $F_{1}$ and $F_{2}$, where $F_{1}$ is the center of the earth. By the definition of an ellipse,

$$
A F_{1}+A F_{2}=2 a
\tag{26}
$$

Since $A F_{1}=R_{\oplus}$, is suffices to minimize $A F_{2}$. By symmetry, line $F_{1} F_{2}$ is the perpendicular bisector of $A B$. Since $F_{2}$ is on a fixed line, to minimize $F_{2}$, we place it at the foot of the perpendicular from $A$ to this line. Then, we obtain $A F_{2}=R_{\oplus} \sin \left(\frac{\theta}{2}\right)$. Thus,

$$
a=\frac{1+\sin \left(\frac{\theta}{2}\right)}{2} R_{\oplus}
$$

and from Equation 25 we find

$$
v_{0}=\sqrt{\frac{G M_{\oplus}}{R_{\oplus}} \frac{2 \sin \frac{\theta}{2}}{1+\sin \frac{\theta}{2}}} .
$$

Answer: $7564$
Problem: Consider a container filled with argon, with molar mass $39.9 \mathrm{~g} \mathrm{~mol}^{-1}$ whose pressure is much smaller than that of atmospheric pressure. Suppose there is a plate of area $A=10 \mathrm{~mm}^{2}$ moving with a speed $v$ perpendicular to its plane. If the gas has density $\rho=4.8 \times 10^{-7} \mathrm{~g} \mathrm{~cm}^{-3}$, and temperature $T=100 \mathrm{~K}$, find an approximate value for the drag force acting on the plate. Suppose that the speed of the plate is $v=100 \mathrm{~m} \mathrm{~s}^{-1}$.

Solution: Let $N$ be the number of particles colliding with the plate in a unit time, and let all the molecules arrive at the plate with the same speed $u$. The force on the plate will be the momentum imparted over a unit time, i.e. $F=2 N m(v-u) / t$ where the coefficient off two reflects on the fact that the average speed of the molecules hitting the wall is the same as the molecules departing. Note that Maxwells distribution dictates that the average speed of particles in all direction $\langle u\rangle=0$ which means that the average force acting on the plate is simply $\langle F\rangle=2 N m v / t$. During a time period $t$, these molecules arrive a wall that is of thickness $v_{R} t$ where $v_{R}$ is the relative velocity between the molecules and the plate (which is not negligible as the mean free path and velocity of the plate are comparable.) This means the number of collisions found in this layer will be $N=\frac{1}{2} n V \approx A v_{R} t$ where the factor of $1 / 2$ reflects on how half the molecules go to the plate and the other half go the other direction. Thus the force acting on the plate will become $\left\langle F_{D}\right\rangle=2 m v N / t=2 n m v \cdot v_{R} A$. To find the relative velocity, consider the vector diagram:


<img_4277>

By law of cosines, the magnitude of $v_{R}$ will simply be

$$
v_{R}^{2}=v^{2}+u^{2}-2 \boldsymbol{v} \cdot \boldsymbol{u}
$$

The direction of $\boldsymbol{u}$ points homogeneously in all directions as the orientation of molecules changes with each individual one. Therefore, $2 \boldsymbol{v} \cdot \boldsymbol{u}$ will point in all directions averaging to 0 . Thus, the magnitude of $v_{R}$ will simply be $\left\langle v_{R}\right\rangle=\sqrt{v^{2}+\langle u\rangle^{2}}$ where $\langle u\rangle^{2}$ is the average thermal velocity of molecules. In terms of density, we can then express the drag force to be $2 \rho A v \sqrt{v^{2}+\langle u\rangle^{2}}$. $2.41 \times 10^{-4} \mathrm{~N}$.

Answer: $2.41\times 10^{-4}$
Problem: Consider a $1 \mathrm{~cm}$ long slit with negligible height. First, we divide the slit into thirds and cover the middle third. Then, we perform the same steps on the two shorter slits. Again, we perform the same steps on the four even shorter slits and continue for a very long time.
Then, we shine a monochromatic, coherent light source of wavelength $500 \mathrm{~nm}$ on our slits, which creates an interference pattern on a wall 10 meters away. On the wall, what is the distance between the central maximum and the first side maximum? Assume the distance to the wall is much greater than the width of the slit. Answer in millimeters.

Solution: This problem is essentially an infinite convolution. In detail, consider two separate amplitude functions $f(x)$ and $g(x)$ that correspond to two different interference patterns. One can think of a convolution of $f$ and $g$ as taking $f$ and placing the same function in every position that exists in $g$. To be more specific, the amplitude of one interference pattern can be thought of as the product of the amplitude patterns of two separate amplitude functions. For example, the diffraction pattern of two slits is the same as the product of the amplitudes for one slit and two light sources (this will be given as an exercise to prove). To make more sense of this, let us consider that very example and let $f$ pertain to the amplitude function of a single slit and let $g$ pertain to the amplitude function of two light sources. One can then write in phase space with a generalized coordinate $x^{\prime}$ that $\{f * g\}(x)=\int_{-\infty}^{\infty} f\left(x-x^{\prime}\right) g\left(x^{\prime}\right) \mathrm{d} x^{\prime}$ according to the convolution theorem. The integral then runs through all values and places a copy of $f\left(x-x^{\prime}\right)$ at the peaks of $g\left(x^{\prime}\right)$.
With this in hand, we can use the convolution theorem to our advantage. Let us designate the amplitude function of the entire Cantor slit to be $F(\theta)$. Since $\theta$ is small, we can designate the phase angle as $\phi=k x \theta$ where $x$ is a variable moving through all of the slits. If we consider a single slit that is cut into a third, the angle produced will be a third of its original as well, or the new amplitude function will simply be a function of $\theta / 3$ or $F(\theta / 3)$. The distance between the midpoints of the two slits will simply be $d=2 / 3 \mathrm{~cm}$ which then becomes $2 / 9$, then $2 / 27$ and so on. In terms of the original function, we can decompose it into the function of $F(\theta / 3)$ and the amplitude function of a single light source. In other words, $F(\theta)=F(\theta / 3) \cos (k d \theta / 2)$ where $k=\frac{2 \pi}{\lambda}$. To the limit of infinity, this can be otherwise written as

$$
F(\theta)=\prod_{i=0}^{\infty} \cos \left(\frac{k d \theta}{2 \cdot 3^{i}}\right)=\cos \left(\frac{k d \theta}{2}\right) \cos \left(\frac{k d \theta}{6}\right) \cos \left(\frac{k d \theta}{18}\right) \ldots
$$

Finding an exact mathematical solution would be difficult, thus it is simple enough to just multiply the first 4 or 5 terms to achieve the final answer.

Answer: $0.647$
Problem: A certain planet with radius $R=$ $3 \times 10^{4} \mathrm{~km}$ is made of a liquid with constant density $\rho=1.5 \mathrm{~g} / \mathrm{cm}^{3}$ with the exception of a homogeneous solid core of radius $r=10 \mathrm{~km}$ and mass $m=2.4 \times 10^{16} \mathrm{~kg}$. Normally, the core is situated at the geometric center of the planet. However, a small disturbance has moved the center of the core $x=1 \mathrm{~km}$ away from the geometric center of the planet. The core is released from rest, and the fluid is inviscid and incompressible.

Calculate the magnitude of the force due to gravity that now acts on the core. Work under the assumption that $R \gg r$.

Solution: We solve by simplifying the configuration into progressively simpler but mathematically equivalent formulations of the same problem.
Because the core is spherical and homogeneous, all the gravitational forces on the core are equivalent to the gravitational forces on a point mass located at the center of the core. Thus, the problem is now the force of gravity on a point mass of mass $m$ located $x$ distance away from the center of planet, with a sphere of radius $r$ evacuated around the point mass.

But if we filled this evacuated sphere, would the force on the point mass be any different? No! If we filled up the sphere, all the added liquid (of the same density as the rest of the planet) would add no additional force on the point mass because it is symmetrically distributed around the point mass. Thus, the problem is now the force of gravity on a point mass of mass $m$ located $x$ distance away from the center of the planet.

By shell theorem, we can ignore all the fluid that is more than a distance $x$ away from the center of the planet. Thus, the problem is now the force of gravity on a point mass $m$ situated on the surface of a liquid planet of radius $x$. This force is not difficult to calculate at all:

$$
\begin{aligned}
F & =\frac{G M m}{x^{2}} \\
& =\frac{G m}{x^{2}}\left(\frac{4}{3} \pi x^{3} \rho\right) \\
& =\frac{4}{3} \pi G m \rho x
\end{aligned}
$$

$1.0058 \times 10^{13} \mathrm{~N}$

Answer: $1.0058 \times 10^{13}$
Problem: Suppose a ping pong ball of radius $R$, thickness $t$, made out of a material with density $\rho_{b}$, and Young's modulus $Y$, is hit so that it resonates in mid-air with small amplitude oscillations. Assume $t \ll R$. The density of air around (and inside) the ball is $\rho_{a}$, and the air pressure is $p$, where $\rho_{a} \ll \rho_{b} \frac{t}{R}$ and $p \ll Y \frac{t^{3}}{R^{3}}$.



An estimate for the resonance frequency is $\omega \sim R^{a} t^{b} \rho_{b}^{c} Y^{d}$. Find the value of $4 a^{2}+3 b^{2}+2 c^{2}+d^{2}$.
Hint: The surface of the ball will oscillate by "bending" instead of "stretching", since the former takes much less energy than the latter.

Solution: Throughout the problem, we will work to order of magnitude and ignore prefactors.
The hint says the surface of the ball will bend instead of stretch, so we need to develop a theory of bending. First, we consider the simplified scenario of a long beam with thickness $t$, width $w$, length $L$, made out of a material with Young's modulus $Y$ and density $\rho$. When the beam bends, the top part of the beam will be in tension and the bottom part will be in compression. Thus, this is how the potential energy is stored in the beam. Suppose the beam is bent with curvature $\kappa$. Then the top part of the beam will stretch by $L t \kappa$, and the bottom part will compress by the same amount. Using Hooke's law, we can approximate the total potential energy stored in the beam as $U \sim \frac{Y t w}{L}(L t \kappa)^{2} \sim Y t^{3} w L \kappa^{2}$. Note that if the relaxed state of the beam was already curved, we simply replace $\kappa$ with the change in curvature.

To find the oscillation frequency of the beam, we need to find the kinetic energy in terms of $\dot{\kappa}$. Since curvature is on the order of second derivative of displacement, we can multiply $\kappa$ by $L^{2}$ to get an estimate for displacement. Then $\dot{\kappa} L^{2}$ gives an estimate for speed, so the kinetic energy is $K \sim \rho t w L\left(\dot{\kappa} L^{2}\right)^{2} \sim \rho t w L^{5} \dot{\kappa}^{2}$. Thus, the frequency of oscillations is $\omega \sim \sqrt{\frac{Y t^{2}}{\rho L^{4}}}$. Again, if the beam



was already curved, we can replace $\kappa$ everywhere with the change in curvature.

We can model the ping pong ball as a "beam" of length order $R$, width order $R$, and thickness $t$. This is a very crude approximation, but will give a dimensionally correct answer (since we are ignoring prefactors). The angular frequency is thus $\omega \sim \frac{t}{R^{2}} \sqrt{\frac{Y}{\rho_{b}}}$.

Answer: 19.75
Problem: A player throws two tennis balls on a level ground at $v=20 \mathrm{~m} / \mathrm{s}$ in the same direction, once at an angle of $\alpha=35^{\circ}$ and once at an angle $\beta=55^{\circ}$ to the horizontal. The distance between the landing spots of the two balls is $d$. Find $d$ in meters.
Assume the height of the player is negligble and ignore air resistance.

Solution: The range of a projectile is proportional as $R \propto \sin 2 \theta$, or $R \propto \cos \theta \sin \theta$. As $\cos (90-\theta)=\sin \theta$, and $\alpha+\beta=90$, the distance travelled by both projectiles are the same.
$0 \mathrm{~m}$

Answer: 0
Problem: For this problem, assume the Earth moves in a perfect circle around the sun in the $x y$ plane, with a radius of $r=1.496 \times 10^{11} \mathrm{~m}$, and the Earth has a mass $m=5.972 \times 10^{24} \mathrm{~kg}$. An alien stands far away from our solar system on the $x$ axis such that it appears the Earth is moving along a one dimensional line, as if there was a zero-length spring connecting the Earth and the Sun.
For the alien at this location, it is impossible to tell just from the motion if it's 2D motion via gravity or 1D motion via a spring. Let $U_{g}$ be the gravitational potential energy ignoring its self energy if Earth moves via gravity, taking potential energy at infinity to be 0 and $U_{s}$ be the maximum spring potential energy if Earth moves in $1 \mathrm{D}$ via a spring. Compute $U_{g} / U_{s}$.

Solution: One naive idea is to directly compute $U_{g}$ and $U_{s}$, but we can use the fact that their frequencies are the same, or:
$$
\omega^{2}=\frac{k}{m}=\frac{G M}{r^{3}} \Longrightarrow k r^{2}=\frac{G M}{r}
$$

Then,

$$
U_{g}=-\frac{G M m}{r}=-k r^{2}
$$

and

$$
U_{s}=\frac{1}{2} k r^{2}
$$

Therefore,

$$
U_{g} / U_{s}=-2
$$

Answer: -2
Problem: Battle ropes can be used as a full body workout (see photo). It consists of a long piece of thick rope (ranging from $35 \mathrm{~mm}$ to $50 \mathrm{~mm}$ in diameter), wrapped around a stationary pole. The athlete grabs on to both ends, leans back, and moves their arms up and down in order to create waves, as shown in the photo.

The athlete wishes to upgrade from using a $35 \mathrm{~mm}$ diameter rope to a $50 \mathrm{~mm}$ diameter rope, while keeping everything else the same (rope material, rope tension, amplitude, and speed at which her arms move back and forth). By doing so, the power she needs to exert changes from $P_{0}$ to $P_{1}$. Compute $P_{1} / P_{0}$.

Solution: The power transmitted by a wave is given by
$$
P=\frac{1}{2} \mu \omega^{2} A^{2} v
$$

where $\mu=\frac{m}{L}$ is the linear mass density, $A$ is the amplitude, and $v$ is the speed of the wave. The speed of a wave on a rope is given by

$$
v=\sqrt{\frac{T}{\mu}}
\tag{2}
$$

where $T$ is the tension. Note that $\omega, A, T$ will all remain constant when changing the radius. Thus, $P \propto \sqrt{\mu} \propto \sqrt{m}$. As we increase the radius by a factor of $f=\frac{50}{35}$, we change the mass by $f^{2}$, so the power changes by a factor of $f$, giving us

$$
P_{1} / P_{0}=f=1.43
$$

Answer: 1.43
Problem: Given vertically polarized light, you're given the task of changing it to horizontally polarized light by passing it through a series of $N=5$ linear polarizers. What is the maximum possible efficiency of this process? (Here, efficiency is defined as the ratio between output light intensity and input light intensity.)

Solution: Let $\theta_{0}=0$ be the original direction of polarization and $\theta_{5}=\pi / 2$ the final direction of polarization. The 5 polarizers are directed along $\theta_{1}, \theta_{2}, \ldots, \theta_{5}$. Let $\delta_{k}=\theta_{k}-\theta_{k-1}$, so that the efficiency is

$$
\eta=\prod_{k=1}^{5} \cos ^{2} \delta_{k}
$$

We wish to maximize $\eta$ subject to the constraint that $\sum_{k} \delta_{k}=\pi / 2$. Clearly, the $\delta_{k}^{\prime} s$ should be non-negative, implying that $0 \leq \delta_{k} \leq \pi / 2$ and thus $\cos \delta_{k} \geq 0$ for all $k$.



We claim that the maximum is achieved when all $\delta_{k}$ are equal. If not, let $\delta_{i} \neq \delta_{i+1}$. Then

$$
\begin{aligned}
\cos \delta_{i} \cos \delta_{i+1} & =\frac{1}{2}\left[\cos \left(\delta_{i}+\delta_{i+1}\right)+\cos \left(\delta_{i}-\delta_{i+1}\right)\right] \\
& <\frac{1}{2}\left[\cos \left(\delta_{i}+\delta_{i+1}\right)+1\right] \\
& =\frac{1}{2}\left[\cos \left(\delta_{i}^{\prime}+\delta_{i+1}^{\prime}\right)+\cos \left(\delta_{i}^{\prime}-\delta_{i+1}^{\prime}\right)\right]
\end{aligned}
$$

where $\delta_{i}^{\prime}=\delta_{i+1}^{\prime}=\frac{\delta_{i}+\delta_{i+1}}{2}$. So replacing $\delta_{i}, \delta_{i+1}$ with $\delta_{i}^{\prime}, \delta_{i+1}^{\prime}$ increases $\eta$.

So $\eta$ is maximized when all $\delta_{k}$ are equal, i.e., $\delta_{k}^{*}=\frac{\pi}{10}$ for all $k$. Then

$$
\eta^{*}=\cos ^{10}\left(\frac{\pi}{10}\right) \approx 0.6054
$$

Answer: $\cos ^{10}(\frac{\pi}{10})$
Problem: In this problem, we explore how fast an iceberg can melt, through the dominant mode of forced convection. For simplicity, consider a very thin iceberg in the form of a square with side lengths $L=100 \mathrm{~m}$ and a height of $1 \mathrm{~m}$, moving in the arctic ocean at a speed of $0.2 \mathrm{~m} / \mathrm{s}$ with one pair of edges parallel to the direction of motion (Other than the height, these numbers are typical of an average iceberg). The temperature of the surrounding water and air is $2^{\circ} \mathrm{C}$, and the temperature of the iceberg is $0^{\circ} \mathrm{C}$. The density of ice is $917 \mathrm{~kg} / \mathrm{m}^{3}$ and the latent heat of melting is $L_{w}=334 \times 10^{3} \mathrm{~J} / \mathrm{kg}$.
The heat transfer rate $\dot{Q}$ between a surface and the surrounding fluid is dependent on the heat transfer coefficient $h$, the surface area in contact with the fluid $A$, and the temperature difference between the surface and the fluid $\Delta T$, via $\dot{Q}=h A \Delta T$.

In heat transfer, three useful quantities are the Reynold's number, the Nusselt number, and the Prandtl number. Assume they are constant through and given by (assuming laminar flow):

$$
\operatorname{Re}=\frac{\rho v_{\infty} L}{\mu}, \quad \mathrm{Nu}=\frac{h L}{k}, \quad \operatorname{Pr}=\frac{c_{p} \mu}{k}
$$

where:

- $\rho$ : density of the fluid
- $v_{\infty}$ : speed of the fluid with respect to the object (at a very far distance)
- $L$ : length of the object in the direction of motion



- $\mu$ : dynamic viscosity of the fluid
- $k$ : thermal conductivity of the fluid
- $c_{p}$ : the specific heat capacity of the fluid

Through experiments, the relationship between the three dimensionless numbers is, for a flat plate:

$$
\mathrm{Nu}=0.664 \operatorname{Re}^{1 / 2} \operatorname{Pr}^{1 / 3}
$$

Use the following values for calculations:

|  | Air | Water |
| :--- | :--- | :--- |
| $\rho\left(\mathrm{kg} / \mathrm{m}^{3}\right)$ | 1.29 | 1000 |
| $\mu(\mathrm{kg} /(\mathrm{m} \cdot \mathrm{s}))$ | $1.729 \times 10^{-5}$ | $1.792 \times 10^{-3}$ |
| $c_{p}(\mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K}))$ | 1004 | 4220 |
| $k(\mathrm{~W} /(\mathrm{m} \cdot \mathrm{K}))$ | 0.025 | 0.556 |

The initial rate of heat transfer is $\dot{Q}$. Assuming this rate is constant (this is not true, but will allow us to obtain an estimate), how long (in days) would it take for the ice to melt completely? Assume convection is only happening on the top and bottom faces. Round to the nearest day.

Solution: The heat transfer coefficient for water-ice and air-ice contact can be figured out with the relationship between the three dimensionless numbers
$$
\mathrm{Nu}=0.664 \operatorname{Re}^{1 / 2} \operatorname{Pr}^{1 / 3} \Longrightarrow h=0.664 \frac{k}{L}\left(\frac{\rho v_{\infty} L}{\mu}\right)^{1 / 2}\left(\frac{c_{p} \mu}{k}\right)^{1 / 3}
$$

As

$$
\frac{\mathrm{d} Q}{\mathrm{~d} t}=h A \Delta T
$$

we then have

$$
t\left(\dot{Q}_{a}+\dot{Q}_{w}\right)=\left(h_{w} A_{w}+h_{a} A_{a}\right) \Delta T \Longrightarrow t=\frac{\rho L^{2} H L_{w}}{\Delta T} \frac{1}{h_{w} L^{2}+h_{a} L^{2}}=59.84 \approx 60 \text { days. }
$$

Answer: 60
Problem: In a galaxy far, far away, there is a planet of mass $M=6 \cdot 10^{27} \mathrm{~kg}$ which is a sphere of radius $R$ and charge $Q=10^{3} \mathrm{C}$ uniformly distributed. Aliens on this planet have devised a device for transportation, which is an insulating rectangular plate with mass $m=1 \mathrm{~kg}$ and charge $q=10^{4} \mathrm{C}$. This transportation device moves in a circular orbit at a distance $r=8 \cdot 10^{6} \mathrm{~m}$ from the center of the planet. The aliens have designated this precise elevation for the device, and do not want the device to deviate at all. In order to maintain its orbit, the device contains a relatively small energy supply. Find the power (in Watts) that the energy supply must release in order to sustain this orbit.
The velocity of the device can be assumed to be much smaller than the speed of light, so that relativistic effects can be ignored. The device can also be assumed to be small in comparison to the size of the planet.

Solution: The centripetal force is given by
$$
\frac{m v^{2}}{r}=\frac{G M m}{r^{2}}-\frac{q Q}{4 \pi \epsilon_{0} r^{2}}
$$



which implies the centripetal acceleration is

$$
a=\frac{v^{2}}{r}=\frac{G M}{r^{2}}-\frac{q Q}{4 \pi \epsilon_{0} m r^{2}}
$$

Now, the device loses energy due to its acceleration, as given by the Larmor formula. The power needed to sustain motion is

$$
P=\frac{q^{2} a^{2}}{6 \pi \epsilon_{0} c^{3}}=0.522 \mathrm{~W}
$$

Answer: $0.522$
Problem: A raindrop of mass $M=0.035 \mathrm{~g}$ is at height $H=2 \mathrm{~km}$ above a large lake. The raindrop then falls down (without initial velocity), mixing and coming to equilibrium with the lake. Assume that the raindrop, lake, air, and surrounding environment are at the same temperature $T=300 \mathrm{~K}$. Determine the magnitude of entropy change associated with this process (in $J / K$ ).

Solution: The total heat gain is equal to the change in potential energy of the raindrop, which spreads through out the whole environment at thermally equilibrium temperature $T$ (the environment is very large so any change in $T$ is negligible). The entropy gain $\Delta S$ is thus generated by the dissipation of this potential energy $M g H$ to internal energy $\Delta U$ in the environment (given that the specific volume of water doesn't change much, $\Delta U \approx M g H$ ). Hence the entropy change associated with this process can be estimated by:
$$
S=\frac{\Delta U}{T} \approx \frac{M g H}{T} \approx 2.29 \times 10^{-3} \mathrm{~J} / \mathrm{K}
\tag{8}
$$

Answer: $2.29 \times 10^{-3}$