[User]
Question A1
  a. Suppose you drop a block of mass m vertically onto a fixed ramp with angle θ with coefficient
     of static and kinetic friction µ. The block is dropped in such a way that it does not rotate
     after colliding with the ramp. Throughout this problem, assume the time of the collision is
     negligible.

       i. Suppose the block’s speed just before it hits the ramp is v and the block slides down the
          ramp immediately after impact. What is the speed of the block right after the collision?
      ii. What is the minimum µ such that the speed of the block right after the collision is 0?

  b. Now suppose you drop a sphere with mass m, radius R and moment of inertia βmR2 vertically
     onto the same fixed ramp such that it reaches the ramp with speed v.

       i. Suppose the sphere immediately begins to roll without slipping. What is the new speed
          of the sphere in this case?
      ii. What is the minimum coefficient of friction such that the sphere rolls without slipping
          immediately after the collision?

                  Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                 Part A                                           4

Solution:
Question A1
  a. Suppose you drop a block of mass m vertically onto a fixed ramp with angle θ with coefficient
     of static and kinetic friction µ. The block is dropped in such a way that it does not rotate
     after colliding with the ramp. Throughout this problem, assume the time of the collision is
     negligible.

       i. Suppose the block’s speed just before it hits the ramp is v and the block slides down the
          ramp immediately after impact. What is the speed of the block right after the collision?

                                                 Solution
          During the collision, the block receives impulses from the normal force, friction, and
          gravity. Since the collision is very short, the impulse due to gravity is negligible. Let pN
          and pF be the magnitudes of the impulses from the normal force and friction force.
          Since the block stays on the ramp after the collision, its final momentum is parallel to the
          ramp. Then the normal force must completely eliminate the block’s initial momentum
          perpendicular to the ramp, so
                                                 pN = mv cos θ.
          Since the block still moves after the collision,

                                                  pF = µpN .

          The block’s initial momentum parallel to the ramp is mv sin θ, so

                                             mv sin θ − pF = mu,

          where u is the final speed of the block. Solving for u gives

                                             u = v(sin θ − µ cos θ).

      ii. What is the minimum µ such that the speed of the block right after the collision is 0?

                                                 Solution
          We set u = 0 to obtain
                                                   µ = tan θ.
          Note that this is simply the no-slip condition for a block resting on an inclined plane!
          This is because in both cases, equality is achieved when the normal force and maximal
          friction force sum to a purely vertical force.

  b. Now suppose you drop a sphere with mass m, radius R and moment of inertia βmR2 vertically
     onto the same fixed ramp such that it reaches the ramp with speed v.

                  Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam              Part A                                               4

       i. Suppose the sphere immediately begins to roll without slipping. What is the new speed
          of the sphere in this case?

                                                Solution
         If the sphere immediately begins to roll without slipping, we can calculate the frictional
         impulse independently of the normal impulse. We have

                                           mv sin θ − pF = mu.

         The frictional impulse is responsible for the sphere’s rotation, so its angular momentum
         about its center of mass is L = pF R. But we also know that

                                          L = βmR2 ω = βmRu.

         Then
                                                pF = βmu.
         Substituting into the previous expression gives
                                                                      v sin θ
                                mv sin θ = (1 + β)mu       ⇒     u=           .
                                                                      1+β

      ii. What is the minimum coefficient of friction such that the sphere rolls without slipping
          immediately after the collision?

                                                Solution
         As in part (a), the normal impulse is pN = mv cos θ and the maximal frictional impulse
         is pF = µpN . From the previous part, we need
                                                     βmv sin θ
                                              pF =
                                                      1+β
         and equating these expressions gives
                                                     β tan θ
                                                µ=           .
                                                     1+β

                 Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam               Part A                                             5

USAPhO 2018 Question A1


[User]
Two moles of a monatomic ideal gas are taken through the following cycle.

        • The gas begins at point A with pressure P0 and volume V0 .

        • The gas is heated at constant volume until it doubles its pressure, reaching point B.

        • The gas is expanded at constant pressure until it doubles its volume, reaching point C.

      • The gas is cooled at constant volume until it halves its pressure, reaching point D.

      • The gas is compressed at constant pressure until it halves its volume, returning to point A.
    Assume that all processes are quasistatic and reversible.

      (a) Draw the process on a P V diagram.

      (b) Calculate the net work done by the gas during the cycle.

      (c) Calculate the efficiency of the cycle.

      (d) Calculate the change in entropy of the gas as the system goes from state A to state D.

Solution:
Two moles of a monatomic ideal gas are taken through the following cycle.

        • The gas begins at point A with pressure P0 and volume V0 .

        • The gas is heated at constant volume until it doubles its pressure, reaching point B.

        • The gas is expanded at constant pressure until it doubles its volume, reaching point C.

        • The gas is cooled at constant volume until it halves its pressure, reaching point D.

        • The gas is compressed at constant pressure until it halves its volume, returning to point A.

    Assume that all processes are quasistatic and reversible.

     (a) Draw the process on a P V diagram.

     (b) Calculate the net work done by the gas during the cycle.

        (c) Calculate the efficiency of the cycle.

     (d) Calculate the change in entropy of the gas as the system goes from state A to state D.

    Answer. (b) P0 V0 , (c) 2/13, (d) 5R log 2

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User] Ühendatud silindrilistesse anumatesse diameetritega $d_1$ ja $d_2$ on valatud vedelik tihedusega $\rho$. Kui palju tõuseb vedeliku tase anumates, kui ühte anumasse pannakse ujuma vedeliku tihedusest väiksema tihedusega keha massiga $m$?

Solution:
Kuna anumad on ühendatud ning vedelikes antakse rõhk igas suunas võrdselt edasi, on mõlemas anumas rõhud samal kõrgusel võrdsed. Järelikult võib antud anumaid vaadelda ühe anumana, mille pindala on
\[
S=\pi\left(\frac{d_{1}}{2}\right)^{2}+\pi\left(\frac{d_{2}}{2}\right)^{2}=\frac{\pi\left(d_{1}^{2}+d_{2}^{2}\right)}{4}.
\]
Väljasurutud vedeliku mass on võrdne lisatava ujuva keha massiga. Seega on lisanduva vedeliku ruumala $\Delta V = m/\rho$ ning kõrguse muut avaldub kui
\[
\Delta h=\frac{\Delta V}{S}=\frac{4 m}{\pi \rho\left(d_{1}^{2}+d_{2}^{2}\right)}.
\]

Vedelik

Ühendatud silindrilistesse anumatesse diameetritega $d_1$ ja $d_2$ on valatud vedelik tihedusega $\rho$. Kui palju tõuseb vedeliku tase anumates, kui ühte anumasse pannakse ujuma vedeliku tihedusest väiksema tihedusega keha massiga $m$?

Kuna anumad on ühendatud, on rõhud mõlemas anumas samal kõrgusel samad ja anumate vedelikutasemed võrdsed.


[User]
7. Passive air-cooling (9 points)

Consider a passive cooling system depicted in
figure. Cold air (at normal conditions: p0 =
105
Pa, T0 = 293K) flows over the heat sink of a
chip ofpower dissipation P = 100W,into a ver-
tical pipe of length L = 1m and cross-sectional
area S = 25cm2
. After passing the pipe, air
enters the ambient room. Assume that the air
inside the pipe becomes well mixed; neglect the
viscousandturbulentfrictionofairinsidethepi-
pe and heat sink. Air can be considered as an
ideal gas with adiabatic exponent γ = 1.4 and
molar mass μ = 29g/mol.
1)Expressheatcapacitanceatconstantpressure
cp via quantites γ and R (1 pt).
2)Findarelationshipbetweentheoutflowingair
density ρ and temperature T (the relationship
may contain also the parameters defined above)
(2 pts).
3) Find a relationship between the air flow ve-
locity in the pipe v and outflowing air density ρ
(the relationship may contain also the parame-
ters defined above) (2 pts).
4) Express the power disspation P in terms of
the air flow velocity v, the outflowing air tem-
perature T, and density ρ (the relationship may
contain also the parameters defined above) (2
pts).
5) What is the temperature T of the outflowing
air? In your calculations, you may use approxi-
mation T − T0 ≪ T0(2 pts)?

Solution:
7. Passive air-cooling

1) Using γ = cp/cV and cp = cV + R we arrive
at cp = γ
γ−1R.
2) From the ideal gas equation, p0 = ρ
μRT (the
process is by constant pressure, otherwise the-
re would be huge acceleration due to pressure
drop).
3) Different air densities inside and outside the
pipe give rise to small residual (as compared to
the static pressure distribution inside the pipe)
pressuredifference betweentheopenendsofthe
pipe, ∆p = −∆ρgL. This pressure difference is
responsible for the acceleration of the air, from
zero,up to the velocity of the air flow v. The mo-
mentum balance for small time interval τ yields
S∆pτ = ρ(Svτ)v, hence (ρ0 − ρ)gL = ρv2
. .
Here, the cold air density ρ0 = p0μ/RT0. Final-
ly, ( p0μ
RT0
− ρ)gL = ρv2
.
4) Heat flux: P = γ
γ−1R(T − T0)Svρ/μ.
5) From the result of question 2, we obtain
∆ρ
ρ = −∆T
T . From the result of question
3, ∆ρ
ρ = −v2
gL. Substituting these values in-
to the equation obtained for question 4, P =
γ
γ−1R v3
gLTSρ/μ. Using the gas equation, this
simplifies to v3
= γ
γ−1
gL
S
P
p0
. So, T = T0[1 +
( γ
γ−1
gL
S
P
p0
)2/3
/gL] ≈ 322K.

Passive air-cooling


[User] 2009. aasta veebruaris põrkasid Siberi kohal 780 kilomeetri kõrgusel kokku USA ja Venemaa satelliidid. Pidades silmas, et ümber Maa tiirleb juba tuhandeid satelliite ning nende kõigi orbiite pole seetõttu võimalik omavahel koordineerida, hinnake, mitme aasta tagant keskeltläbi niisugused juhuslikud kokkupõrked aset leiavad. Oma lahenduses kasutage järgmisi hinnanguid ja lähendusi: maalähedaste satelliitide arv $N=\num{2500}$; orbiidid jäävad maapinnast kõrguste vahemikku $h_1=\SI{200}{km}$ kuni $h_2=\SI{2000}{km}$ ning satelliidid on jaotunud selles kihis ühtlase ruumtihedusega; tüüpilise satelliidi ristlõikepindala $S=\SI{10}{m^2}$. Maa raadius $R=\SI{6400}{km}$, raskuskiirendus maapinnal $g=\SI{10}{m/s^2}$.

Solution:
Lähtume analoogiast molekulaarfüüsikaga, kus ühe molekuli vaba tee hindamisel arvestatakse, et molekul liigub ilma põrgeteta tüüpiliselt aja jooksul, mil tema kokkupõrke-ristlõige on katnud ruumala, milles asub tüüpiliselt üks osake (see ruumala avaldub kui anuma ruumala jagatud osakeste arvuga). Kokkupõrke-ristlõige pole päris identne osakese enda ristlõikega -- vaatleme näiteks kera-kujulisi osakesi raadiusega $r$, osakesed põrkuvad, kui nende tsentrid ei ole teineteisest kaugemal kui $2r$, niisiis on ühe osakese kokkupõrke-ristlõige neli korda suurem tema ristlõikest.

Satelliidid liiguvad ruumiosas ruumalaga
\[ V=\frac{4\pi}{3}\left[(R+h_2)^3-(R+h_1)^3\right]\approx \SI{1.2e12}{km^3}.\]
Liikumisruum ühe satelliidi kohta on seega $V/N$ (niisuguse ruumalaga suvaliselt valitud ruumiosast leiame tüüpiliselt ühe satelliidi).

Aja $t$ jooksul katab ühe satelliidi kokkupõrke-ristlõige ruumala
\[V_t=4Svt,\]
kus $v$ on tüüpiline satelliidi liikumise kiirus. Me ei tee suurt viga, võttes $v$ väärtuseks esimese kosmilise kiiruse (kiirus sõltub raadiuse ruutjuurest ning suhteline viga oleks ainult \mbox{$\sqrt{\frac{6400+2000}{6400}}\approx\num{1.15}$}):
\[\frac{v^2}{R}=\frac{GM}{R^2}=g.\]
Niisiis,
\[V_t=\sqrt{gR}4St.\]

Eelneva arutluse kohaselt arvestame, et ühel satelliidil tuleb kokkupõrget oodata niisugune ajavahemik $t$, et $V_t=V/N$. Et meil on aga $N$ satelliiti, siis esimese niisuguse kokkupõrkeni kulub $N$ korda vähem aega. Seega,
\[
\Delta t=\frac{V}{N^24S\sqrt{gR}}=\frac{\num{1.2e12}\cdot 10^9}{\num{4}\cdot\num{2.5}^2\cdot 10^6\cdot \num{10}\cdot \sqrt{10 \cdot \num{64} \cdot 10^2\cdot 10^3}}s=\SI{6e8}{s},
\]
ehk
\[
\Delta t=\frac{6\cdot10^{8}}{3600\cdot24\cdot365}\approx \SI{19}{a}.
\]

Satelliidid

2009. aasta veebruaris põrkasid Siberi kohal 780 kilomeetri kõrgusel kokku USA ja Venemaa satelliidid. Pidades silmas, et ümber Maa tiirleb juba tuhandeid satelliite ning nende kõigi orbiite pole seetõttu võimalik omavahel koordineerida, hinnake, mitme aasta tagant keskeltläbi niisugused juhuslikud kokkupõrked aset leiavad. Oma lahenduses kasutage järgmisi hinnanguid ja lähendusi: maalähedaste satelliitide arv $N=\num{2500}$; orbiidid jäävad maapinnast kõrguste vahemikku $h_1=\SI{200}{km}$ kuni $h_2=\SI{2000}{km}$ ning satelliidid on jaotunud selles kihis ühtlase ruumtihedusega; tüüpilise satelliidi ristlõikepindala $S=\SI{10}{m^2}$. Maa raadius $R=\SI{6400}{km}$, raskuskiirendus maapinnal $g=\SI{10}{m/s^2}$.

Satelliidi tüüpiline kiirus on võrreldav esimese paokiirusega, sest satelliitide orbitaalraadiused ei erine üksteisest märkimisväärselt. Ülesande eelduste kohaselt liiguvad satelliidid sarnaselt molekulidega gaasis. Gaasis molekuli vaba tee hindamisel arvestatakse, et molekul liigub ilma põrgeteta ligikaudu aja jooksul, mil tema kokkupõrke ristlõige on katnud ruumala, milles asub tüüpiliselt üks osake.


[User]
8. Magnetic pulse (7 points)

Consider an electric circuit
consisting of a coil of negligibly small inductance, consisting of
N = 10 turns and with the surface area of a single loop S =
10cm2
), resistors R1 = R2 = 3Ω, capacitor C = 0.2F,
and an inductance L = 1H, connected as shown in Fig. At the
moment of time t = 0, a magnetic field, parallel to the axis of
thecoilisswitchedon. Theinductanceofthemagneticfieldstarts
growing linearly, starting from B = 0 until the maximal value
B = 1T is achieved at t = 10ms. Further, the inductance of
the magnetic fieldremains constant (andequalto 1T).
i)FindthecurrentthroughtheresistorsR1 andR2 atthemoment
oftime t1 = 5ms.
ii) Find the current through the resistors R1 and R2 at the
moment oftimet2 = 15ms.
iii) Whatis the net chargepassing through the resistorR2?

Solution:
8. Magneticpulse

Sincewecanneglecttheinductance
of the coil, it performes as a voltage source, which outputs U =
NSB/τ = 1V during the time period between t = 0ms and
10ms, and 0V otherwise. The characteristic time scales of the
RC andLC cirquitsareτ1 = R1C = 0.6sandτ2 = L/R2 ≈
0.3s. So, for both cirquits, the processes are very fast, i.e. the
capacitor is effectively short-circuited, and almost all the voltage
fallson the inductance.
i) According to the considerations given above, I1 = U/R1 ≈
0.33A. As for I2, it starts growing from 0A at t = 0 at a rate,
given by LdI
dt
= U,i.e.I2 = Ut1/L = 5mA.
ii) When the voltage U is switched off (at t = τ = 10ms),
the capacitor will (almost completely, because t2 − τ ≪ R1C)
retain the charge it has accumulated, Q = I1τ. All the voltage
of the capacitor (Q/C) will fall on the resistor R1, so that I′
1 =
Q/R1C = Uτ/R2
1C ≈ 5.6mA. As forthe inductance, it will
retain(almostcompletely,becauset2 −τ ≪ L/R2)thecurrent
it has acquiredduring the first 10ms,I′
2 = Uτ/L = 10mA.
iii) Since the current in R2 will decay very slowly, as compared
to its growth during the first 10ms, we can neglect the charge
passing thorugh it during t < τ. Then we can write the Kirhoff’s
lawintheformLdI
dt
+R2
dq
dt
= 0,fromwhereLdI+R2dq = 0,
and L∆I = −R2∆q. Since ∆I = −I′
2, we obtain ∆q =
LI′
2/R2 = 3.3mC.

Magnetic pulse


[User] A boy who is sitting on a tree branch wants to drop onto a horse that happens to be galloping by right under the tree. The horse is moving at a speed of $v = \SI{10}{m/s}$ and the branch is $h = \SI{3}{m}$ higher than the saddle. What does the horizontal distance between the saddle and the branch have to be at the moment when the boy lets go of the branch?

Solution:
Vaba langemise aeg
\[
t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \cdot 3}{\num{9,81}}} \approx \SI{0,78}{s}
\]
Seega otsitav kaugus on $s = vt = \SI{10}{m/s} \cdot \SI{0,78}{s} = \SI{7,8}{m}$.

Hobune

Puuoksal istub poiss, kes soovib hüpata puu alt mööda galopeeriva hobuse selga. Hobuse kiirus on $v = \SI{10}{m/s}$ ja puuoksa kõrgus sadula suhtes $h = \SI{3}{m}$. Kui suur peab olema horisontaalsihiline distants sadula ja puuoksa vahel sel hetkel, kui poiss oksast lahti laseb?

Langemise aeg on avaldatav valemi $s = \frac{at^2}{2}$ kaudu.


[User]
Planck’s law can be used to understand atomic physics, as Einstein showed and as
    you will now show. Suppose we have a collection of two-state atoms encased inside a cavity, whose
    walls are blackbodies with temperature T . In equilibrium, there will be N1 atoms in the ground
    state and N2 atoms in the excited state, bathed in a photon gas of temperature T . Three physical
    processes can occur:

       1. Atoms in the excited state can each spontaneously decay with rate A.

       2. Atoms in the ground state can absorb a photon. For each atom, the rate of this process is
          Bρ(f ) where ρ(f ) is the energy density of radiation at frequency f .

       3. Atoms in the excited state can undergo stimulated emission when interacting with a photon,
          in the process e + γ → g + γ + γ. For each atom, the rate of this process is B ′ ρ(f ).

    These three “Einstein coefficients” depend on the frequency, but not on the temperature.

     (a) Argue that ρ(f ) is proportional to I(f ) in Planck’s law. In fact,
                                                               c
                                                     I(f ) =     ρ(f )

            though you do not have to show this.

     (b) Write down an expression for dN2 /dt.

        (c) In the steady state, for any temperature, N2 /N1 must be given by the Boltzmann distribution.
            Using this, find A and B ′ in terms of B.

    This is an extraordinary result: one might have expected a difficult calculation to find each of
    A, B, and B ′ , but if you know one, then by thermodynamics you know them all. In particular,
    thermodynamics requires spontaneous emission to be possible, A ̸= 0.

Solution:
(a) Here ρ(f ) refers to the distribution of radiation energy inside the blackbody, while
         I(f ) refers to the intensity of radiation escaping the blackbody if there were a hole. The
         reason these are proportional is that all electromagnetic radiation travels at the same speed;
         hence light of all frequencies will escape a blackbody at the same rate. (Note that this would
         not be true for an ideal gas, where higher-energy particles would escape at a higher rate.)

     (b) Adding up the contributions from the three processes,

                                     dN2
                                         = −AN2 + Bρ(f )N1 − B ′ ρ(f )N2 .
                                      dt

      (c) We know that in thermal equilibrium,

                                                   N2
                                                      = e−hf /kB T .
                                                   N1
         Plugging this in above and setting dN2 /dt = 0, we have

                                                A
                                                     = Behf /kB T − B ′ .
                                               ρ(f )

         Using the given expression for ρ(f ),

                                          c3
                                    A          (ehf /kB T − 1) = Behf /kB T − B ′ .
                                        8πhf 3
         Now, we see the left-hand side diverges in the limit T → ∞, which is physically reasonable.
         It is only possible for the right-hand side to do the same thing if

                                                       B ′ = B.

         Plugging this back in, we have
                                                         8πhf 3
                                                    A=          B.
                                                           c3


[User]
Consider a string with the following shape.

      (a) If this is a traveling wave moving to the right with velocity v, carefully draw the velocity and
          acceleration of every point on the string.

      (b) Now suppose the string is held in place, with zero velocity. If it is suddenly relased, sketch
          the subsequent behavior of the string.

Solution:
(a) You can figure out the velocity in two different ways. First, since the wave is
         proportional to f (x − vt), the vertical velocity is
                                                ∂y               ∂y
                                                   = −vf ′ = −v
                                                ∂t               ∂x
          so it is proportional to the slope of the string. Alternatively, you can think about how the
          string has to move so that a moment later, its shape is the same but translated to the right.
          The result is shown below:

          To derive the acceleration, you can think about how the velocity profile has to change as the
          string moves, or you can think about how it comes about from the tension in the string. In
          general, the net force depends on the concavity ∂ 2 y/∂x2 of the string. In this case, it’s only
          nonzero at the three kinks.

      (b) To keep the string in that position, we must hold it at three points. Since information can’t
          travel faster than the speed of waves, only the bits of string near those three points can move
          right after release, because they’re the only ones that know about the release. The direction
          of motion can be found with the wave equation (the middle goes down, the ends go up).
          Now, for the general solution, note that a solution to the wave equation with zero initial
          velocity may be written in the form f (x + vt) + f (x − vt). Here, the function f has the same
          shape as the wave, but half the height. Evidently, two traveling waves split off in opposite
          directions.


[User]
Läbipaistvast materjalist poolsilindri raadius on $R$ ja aine murdumisnäitaja $n$. Laserkiir langeb risti poolsilindri tasapinnalisele küljele (vt. joonist). Leida maksimaalne kaugus $d_{\max}$, mille korral laserkiir veel väljub silindrilise pinna kaudu. Millistes piirides varieerub kaugus $l$ (punkti $O$ ning laserkiire ja optilise telje lōikepunkti $A$ vahemaa), kui $d_{\max }>d>0$.

\emph{Märkus.} Väikeste nurkade korral võib lugeda $\sin \alpha \approx \tan \alpha \approx \alpha$. Eeldatakse, et $l$ on monotoonne funktsioon $d$-st.
\begin{center}
	\includegraphics[width=0.8\linewidth]{2004-v3g-07-yl.pdf}
\end{center}

Solution:
$d$ maksimaalse väärtuse määrab piirjuht, kus laserkiir langeb tagumisele pinnale täieliku sisepeegelduse piirnurga all, s.t. $\sin \alpha=1 / n$. Teiselt poolt, $\sin \alpha=d_{\max } / R$, seega $d_{\max }=R / n$.
\begin{center}
	\includegraphics[width=0.8\linewidth]{2004-v3g-07-lah.pdf}
\end{center}
Kui $d=d_{\max}$, siis väljuv laserkiir on silinderpinnale puutujaks. Seega
$$
\begin{aligned}
	l &=R \cos \alpha+d_{\max } \tan \alpha=R\left[\sqrt{1-\sin ^{2} \alpha}+\frac{1}{n} \frac{\sin \alpha}{\sqrt{1-\sin ^{2} \alpha}}\right] \\
	&=R\left[\sqrt{1-1 / n^{2}}+\frac{1}{n} \frac{1 / n}{\sqrt{1-1 / n^{2}}}\right]=\frac{n R}{\sqrt{n^{2}-1}}
\end{aligned}
$$
Kui $d \rightarrow 0$, siis laserkiir langeb tagumisele pinnale nurga all $\alpha \approx d / R$. Murdumisnurk on $\beta=n \alpha=n d / R.$ Pinnanormaali kaldenurk on samuti $\alpha$, seega väljunud kiire kaldenurk on $\gamma=\beta-\alpha=(n-1) d / R$. Nüüd saame
$$
l=R+\frac{d}{\gamma}=R+\frac{R}{n-1}=\frac{n R}{n-1} .
$$

Laser

Läbipaistvast materjalist poolsilindri raadius on $R$ ja aine murdumisnäitaja $n$. Laserkiir langeb risti poolsilindri tasapinnalisele küljele (vt. joonist). Leida maksimaalne kaugus $d_{\max}$, mille korral laserkiir veel väljub silindrilise pinna kaudu. Millistes piirides varieerub kaugus $l$ (punkti $O$ ning laserkiire ja optilise telje lōikepunkti $A$ vahemaa), kui $d_{\max }>d>0$.

\emph{Märkus.} Väikeste nurkade korral võib lugeda $\sin \alpha \approx \tan \alpha \approx \alpha$. Eeldatakse, et $l$ on monotoonne funktsioon $d$-st.
\begin{center}
	\includegraphics[width=0.8\linewidth]{2004-v3g-07-yl.pdf}
\end{center}


[User] A boy is measuring the speed of wind with respect to himself while riding on a bicycle: if he is riding along the road at one direction with a speed 10 km/h then he gets the result 20 km/h and if he rides the opposite direction with a speed 20 km/h then he also gets the result 20 km/h. How fast with respect to the ground is the wind blowing?

Solution:
Jalgratturi mõõdetav tuul $\vec w'$ on tuule kiirusvektori $\vec w$ ja jalgratturi kiirusvektori $\vec v$ vahe $\vec w'=\vec w - \vec v$. Olgu jalgratturi kiirusvektorid $\vec v_1$ ja $\vec v_2$ ning tuule kiirusvektorid $\vec w'_1$ ja $\vec w'_2$ vastavalt ühele ja teisele poole sõites.

Antud juhul teame ainult kiirusi, mitte nende suundi. Teades eelnevalt, et mõõdetud tuule kiirus on sama suur mõlemas suunas liikudes, saab kiirusvektorid esitada võrdhaarse kolmnurgana. (Kolmnurga mõlemad pooled vastavad erinevas suunas sõitmisele ning vastavad ülal mainitud valemile. Kuna tuule vektor on mõlemal juhul sama ja kiirused paralleelsed, saab selle konstrueerida nagu joonisel.)

\begin{center}
\includegraphics[width=0.4\textwidth]{2013-v3g-03-jalgrattur}\\
\end{center}

Koosinusteoreemist:

$$
\begin{cases}
|\vec w|^2 = |\vec w'_1|^2 + |\vec v_1|^2 - 2 |\vec w'_2| |\vec v_1|\cos \alpha \\
|\vec w|^2 = |\vec w'_2|^2 + |\vec v_2|^2 - 2 |\vec w'_2| |\vec v_2|\cos \alpha.
\end{cases}
$$
Teades, et $ |\vec v_2|=2 |\vec v_1|$ ja et $ |\vec w'_1|=2 |\vec w'_2|$, saame esimese võrrandi korrutada kahega ja teise sellest lahutada.

$$|\vec w|^2 = |\vec w'_1|^2 - 2|\vec v_1|^2. $$

Ehk tuule tegelik kiirus on:
$$|\vec w|=\sqrt{|\vec w'_1|^2-2|\vec v_1|^2} \approx \SI{14}{km \per h}.$$

Jalgrattur

Poiss mõõdab jalgrattaga sõites tuule kiirust enda suhtes: kui ta sõidab piki teed ühes
suunas kiirusega \SI{10}{km/h}, saab ta tulemuseks \SI{20}{km/h}, ning kui ta sõidab
vastassuunas kiirusega \SI{20}{km/h}, siis saab ta tulemuseks samuti
\SI{20}{km/h}. Kui kiiresti maa suhtes puhub tuul?

Vaadeldes tuule ja jalgratturi kiirusi vektoritena, on võimalik geomeetriliselt konstrueerida vastavad tekstis toodud tingimused ning geomeetria põhjal tuule kiirus välja arvutada.


[User]
3. A small angle prism of crown glass and a small angle prism of flint glass are arranged
    to deviate a light beam in opposite directions. A white light beam is incident on
    the crown glass prism of the arrangement. The refracting angle of the crown prism
    is 10.00 and that of the flint prism is such that the net deviation of the light beam
    from both the prisms for the C and F spectral lines are equal. Assume minimum
    deviation condition for both the prisms. See Table 1 below.

                                         E
      Table 1: Values of n (refractive indices) for three different spectral lines.
                                       C       D
                             Flint 1.644 1.650 1.665
                            Crown 1.517 1.520 1.527
                                                       F

                                                                     [ 3 + 2 + 5 = 10 ]

    (a) State the refracting angle (AF ) of the flint prism in degrees.

                   CS
          AF =

    (b) State the net deviation (∆D ) for the D line in degrees.
          ∆D =

     (c) Draw a schematic ray diagram of the D line for this arrangement indicating the
         numerical values of the following important angles: (i) incidence on each prism;
         (ii) deviation from each prism and (iii) angle between the adjacent surfaces of
         the two prisms. You may use a pencil for this diagram.

HB
                                        Page 5

INPhO - 2010                                           Roll Number:P 1 0 0 0 0

Solution:
3. (a) AF = 4.76 0
   (b) ∆D = −2.1 0
                                    3R
                                     2
                                       ln
                                          585
                                          32
                                              + R ln
                                                     E
                                                     65
                                                     27

                         CS
   (c) See Fig. (3). Note that all angles are expressed in degrees.

                             Crown                      Flint

                           7.6                 5.2

                                              3.9               3.1

HB
                                                        3.7

                                             Figure 3:

             mgR

INPhO 2010 Problem 3


[User]
Suletud balloon ruumalaga $V = \SI{10}{l}$ oli täidetud veega temperatuuril $t_0 = \SI{0}{\degreeCelsius}$. Samal temperatuuril külmutati vesi jääks, mille tulemusena ballooni kest venis välja ja vesi avaldas kogu jäätumise protsessi käigus balloonile rõhku $p = \SI{5e7}{Pa}$.

Leida balloonis olnud vee ($\mathrm{H_2O}$) siseenergia muut koos märgiga. Jää tihedus $\rho_j = \SI{900}{kg/m^3}$ ja sulamissoojuseks antud rõhul $\lambda = \SI{317}{kJ/kg}$. Jää ja vee kokkusurutavust mitte arvestada.

Solution:
Termodünaamika I seaduse kohaselt $\Delta U = Q - A$, kus $Q$ on süsteemi antud soojushulk ning $A$ on välisjõudude vastu tehtud töö. Antud ülesande kontekstis on $Q$ negatiivne ja $A$ positiivne ning $A = p\Delta V$ ja $Q = -\lambda \rho_vV$. Niisiis,
\[
\Delta U=-\lambda \rho_{v} V-p\left(\frac{V \rho_{v}}{\rho_{j}}-V\right)=-\rho_{v} V\left[\lambda+p\left(\rho_{j}^{-1}-\rho_{v}^{-1}\right)\right].
\]
Paneme tähele, et avaldis nurksulgudes peaks kujutama endast sulamissoojust normaaltingimustel (sest vee võib viia samasse lõppolekusse ka teisel viisil --- muutes ta jääks normaaltingimustel ning seejärel viies rõhu etteantud väärtuseni; et jää loeme kokkusurumatuks, siis rõhu tõstmisel tööd ei tehta). Paistab, et tegemist pole siiski päris hariliku veega, sest
\[
\lambda+p\left(\rho_{j}^{-1}-\rho_{v}^{-1}\right)= \SI{323}{kJ/kg} \neq \lambda_{0}=\SI{334}{kJ/kg}.
\]
Arvandmete asendamisel leiame $\Delta U = \SI{-3,23}{MJ}$.

Balloon

Suletud balloon ruumalaga $V = \SI{10}{l}$ oli täidetud veega temperatuuril $t_0 = \SI{0}{\degreeCelsius}$. Samal temperatuuril külmutati vesi jääks, mille tulemusena ballooni kest venis välja ja vesi avaldas kogu jäätumise protsessi käigus balloonile rõhku $p = \SI{5e7}{Pa}$.

Leida balloonis olnud vee ($\mathrm{H_2O}$) siseenergia muut koos märgiga. Jää tihedus $\rho_j = \SI{900}{kg/m^3}$ ja sulamissoojuseks antud rõhul $\lambda = \SI{317}{kJ/kg}$. Jää ja vee kokkusurutavust mitte arvestada.

Termodünaamika I seaduse kohaselt $\Delta U = Q - A$, kus $Q$ on süsteemi antud soojushulk ning $A$ on välisjõudude vastu tehtud töö. Antud üesande kontekstis on $Q$ negatiivne ja $A$ positiivne.


[User]
Fikseeritud telgedega hammasrattad raadiustega $r_1$ ja $r_2$ hambuvad ja on
ühendatud venimatu nööriga,
mis on mõlemale puutujaks. Esimest ratast pööratakse
jõumomendiga $M$. Kui suur on nööri pinge $T$?

\begin{center}
	\includegraphics{2010-v3g-04-joonis_hr_ipe.pdf}
\end{center}

Solution:
Kasutame virtuaalse nihke meetodit: oletame, et nöör pole siiski päris venimatu ning
saame esimest ratast pöörata väikese nurga $\alpha$ võrra. Hõõre puudub,
mistõttu salvestub kogu välise jõumomendi töö nööri elastsusjõu potentsiaalseks
energiaks. Välisjõumomendi töö on $M \alpha$ (kui jõumomenti avaldab üks jõud
õlaga $\tilde o$ ja suurusega $M / \tilde o$, siis nihkub ta rakenduspunkt $\alpha
\tilde o$ võrra ja töö on $\alpha \tilde o M / \tilde o = M \alpha$). Väikesel
nihkel ei jõua $T$ oluliselt muutuda, seega peab nööri venitamise töö olema $T
s$, kus $s$ on nööri pikenemine. Hambumusse jäävate hammasrataste pinnapunktide
läbitavad teepikkused on võrdsed --- mõlemal $\alpha r_1$, järelikult $s = 2
\alpha r_1$ ja $M \alpha = 2 \alpha r_1 T$, kust $T = \frac{M}{2 r_1}$.

\vspace{0.5\baselineskip}
\textit{Alternatiivne lahendus}\\
Ratastele mõjuvad jõud ja jõumomendid on tasakaalus. Lihtsaim on kirjutada
jõumomentide tasakaalud rataste tsentrite suhtes, kuna siis on võllide poolt
avaldatava tundmatute jõudude õlad nullid. (Muidu saame lahenduse, kui avaldame
need jõud jõudude tasakaaluvõrranditest.) Rattad mõjutavad teineteist
puutujasihilise jõuga; kui teine ratas avaldab esimesele jõudu $\vec F$, siis
avaldab Newtoni III seaduse järgi esimene teisele $-\vec F$. Jõumomentide
tasakaal esimesele rattale on nii $M = (F+T) r_1$ ning teisele $T=F$, sestap
$T = \frac{M}{2 r_1}$.

Hammasrattad

Fikseeritud telgedega hammasrattad raadiustega $r_1$ ja $r_2$ hambuvad ja on
ühendatud venimatu nööriga,
mis on mõlemale puutujaks. Esimest ratast pööratakse
jõumomendiga $M$. Kui suur on nööri pinge $T$?

\begin{center}
	\includegraphics{2010-v3g-04-joonis_hr_ipe.pdf}
\end{center}

Kuna nöör on venimatu, on mõlemad hammasrattad paigal ja seega tasakaalus. Kehtib nii jõudude kui ka jõumomentide tasakaal. Antud juhul on kõige mugavam vaadelda jõumomentide tasakaalu hammasrataste keskpunktide suhtes.


[User]
Theoretical competition. Tuesday, 15 July 2014                                      1/2
                     Problem3. Simplest model of gas discharge (10 points)
       An electric current flowing through a gas is called a gas discharge. There are many types of gas
discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark
discharge that occurs between the clouds and the earth in the form of lightning.
                      Part А. Non-self-sustained gas discharge (4.8points)
        In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a
permanent operationan external ionizer is needed, which creates Zext pairs of singly ionized ions and free
electrons per unit volume and per unit time uniformly in the volume.
        When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited
increase in the number densities of electrons and ions in the gas is prevented by the recombination process in
which a free electron recombineswith an ion to form a neutral atom. The number of recombining
eventsZrec that occurs in the gas per unit volume and per unit time is given by
                                                 Zrec = rne ni ,
where r is a constant called the recombination coefficient, and ne ,ni denotethe electron and ion number
densities, respectively.
        Suppose that at time t = 0the external ionizer is switched on and the initial number densities of
electrons and ions in the gas are both equal to zero. Then, the electron number density ne (t) depends on
timetas follows:
                                            ne t = n0 + a tanh bt,
where n0 , a and bare some constants, and tanh xstands for the hyperbolic tangent.
 A1    Findn0 , a, band express them in terms ofZext andr.(1.8points)
       Assume that there are two external ionizers available. When the first one is switched on, the electron
number density in the gas reaches its equilibrium value of be ne1 = 12 ∙ 1010 cm−3 . When the second
external ionizer is switched on, the electron number density reaches its equilibrium value of ne2 = 16 ∙
1010 cm−3 .
 A2    Find the electron number densityne at equilibriumwhen both external ionizers are switched on
       simultaneously.(0.6 points)
        Attention!In what follows it is assumed that the external ionizer is switched on for quite long period
of time such that all processes have become stationary and do not depend on time. Completely neglect the
electric field due the charge carriers.
        Assume that the gas fills in the tube between the two parallel conductive plates of area S separated by
the distance L ≪ S from each other. The voltage U is applied across the plates to create an electric field
between them. Assume that the number densities of both kinds of charge carriers remain almost constant
along the tube.
        Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti)
acquire the same ordered speedvdue to the electric field strengthE found as
                                                   v = βE,
where bisa constant called charge mobility.
 A3    Express the electric currentIin the tube in terms of U, β, L, S, Zext , r and ewhichis the elementary
       charge.(1.7 points)
 A4    Find the resistivityρgas of the gas at sufficiently small values of the voltage applied and express it in
       terms ofβ, L, Zext , r and e.(0.7 points)

                      Theoretical competition. Tuesday, 15 July 2014                                      2/2
                          PartB. Self-sustained gas discharge (5.2 points)
        In this part of the problemthe ignition of the self-sustained gas discharge is consideredto show how
the electric current in the tube becomes self-maintaining.
        Attention!In the sequel assume that the external ionizer continues to operatewith the same Zext rate,
neglect the electric field due to the charge carriers such that the electric field is uniform along the tube, and
the recombination can be completely ignored.
        For the self-sustained gas discharge there are two
important processes not considered above. The first process
isasecondary electron emission, and the second one isa
formation of electron avalanche. The secondary electron
emission occurs when ions hit on the negative electrode,
called a cathode, and the electrons are knocked out of it to
move towards the positive electrode, called an anode. The
ratio of the number of the knocked electronsNe per unit
timeto the number of ionsNi hitting the cathode per unit
time is called the coefficient of the secondary electron
emission,γ = Ne /Ni .The formation of the electron
avalanche isexplained as follows. The electric field accelerates free electrons which acquire enough kinetic
energy to ionize the atoms in the gas by hitting them. As a result the number of free electrons moving
towards the anodesignificantly increases. This process is described by the Townsend coefficient α, which
characterizes an increase in the number of electronsdNe due to moving Ne electrons that have passed the
distancedl, i.e.
                                                   dNe
                                                       = αNe .
                                                    dl
        The total currentI in any cross section of the gas tube consists of the ion Ii (x)and the electronIe (x)
currents which, in the steady state, depend on the coordinate x, shown in the figureabove. The electron
current Ie (x) varies along the x-axis according to the formula
                                             Ie x = C1 e A1 x + A2 ,
where A1 , A2 , C1 are some constants.
 B1    FindA1 , A2 and express them in terms ofZext , α, e, L, S.(2 points)
      The ion current Ii (x) varies along the x-axis according to the formula
                                            Ii x = C2 + B1 e B2 x ,
where B1 , B2 , C2 are some constants.
 B2    FindB1 , B2 and express them in terms ofZext , α, e, L, S, C1 .(0.6 points)
 B3    Write down the condition for Ii (x)atx = L.(0.3 points)
 B4    Write down the condition forIi (x)andIe (x)at x = 0.(0.6 points)
 B5    Find the total currentIand express it in terms ofZext , α, γ, e, L, S.Assume that it remains finite
       (1.2 points)
         Let the Townsend coefficient αbe constant. When the length of the tube turns out greater than some
critical value, i.e. L > Lcr , the external ionizer can be turned off and the discharge becomes self-sustained.
 B6    FindLcr and express it in terms ofZext , α, γ, e, L, S.(0.5 points)

Solution:
Theoretical competition. Tuesday, 15 July 2014                                        1/3
                        Problem 3. Simplest model of gas discharge
                                                 Solution
                                 Part А. Non-self-sustained gas discharge
A1.Let us derive an equation describing the change of the electron number density with time. It is
determined by the two processes; the generation of ion pairs by external ionizer and the recombination of
electrons with ions. At ionization process electrons and ions are generated in pairs, and at recombination
processthey disappear in pairs as well.Thus, their concentrations are alwaysequal at any given time, i.e.
                        n t = ne t = ni (t)                                                      (A1.1).
       Then the equation describing the numberdensityevolution of electrons and ions in time can be written
as
                        dn (t)
                               = Zext − rn(t)2                                                   (A1.2).
                         dt
       It is easy to show that att → 0 the function tanh bt → 0, therefore, by virtue of the initial condition
n 0 = 0,one finds
                        n0 = 0                                                                   (A1.3).
       Substituting ne t = a tanh bt in (A1.2) and separating it in the independent functions (hyperbolic,
or 1 ande x ), one gets
                                Zext
                      a=                                                                         (A1.4),
                                  r
                       b = rZext                                                              (A1.5).
A2.According to equation (A1.4) the number density of electronsat steady-state is expressed in terms of the
external ionizer activity as
                                   Zext 1
                      ne1 =                                                                      (A2.1),
                                       r
                                   Zext 2
                      ne2 =                                                                      (A2.2),
                                       r
                                  Zext 1 +Zext 2
                      ne =                                                                       (A2.3).
                                           r
       Thus,the following analogue of the Pythagorean theorem is obtained as
                                2     2
                       ne = ne1    + ne2  = 20.0 ∙ 1010 cm−3 .                                    (A2.4)
A3.In the steady state, the balance equations of electrons and ions in the tube volume take the form
                                             I
                       Zext SL = rne ni SL + e                                                    (A3.1),
                                                       e
                                                       Ii
                      Zext SL = rne ni SL + e                                                     (A3.2).
       It follows from equations (A3.1) and (A3.2) that the ion and electron currents are equal, i.e.
                      Ie = Ii                                                                     (A3.3).
       At the same time the total current in each tube section is the sum of the electron and ion currents
                      I = Ie + Ii                                                                 (A3.4).
       By definition ofthe current density the following relations hold
                            I
                      Ie = 2 = ene vS = eβne ES                                                   (A3.5),
                            I
                      Ii = 2 = eni vS = eβni ES                                           (A3.6).
        Substituting (A3.5) and (A3.6) into (A3.1) and (A3.2), the following quadratic equation for the
current is derived
                                                   I   2      I
                      Zext SL = rSL 2eβES                   + 2e                                 (A3.7).
       The electric field strength in the gas is equal to
                            U
                       E=L                                                                       (A3.8).
and solution to the quadratic equation (A3.7) takes the form
                           eβ 2 U 2 S                       4rZext L4
                      I=                   −1 ± 1 +                                              (A3.9).
                                rL 3                          β 2U2

                      Theoretical competition. Tuesday, 15 July 2014                                       2/3
       It is obvious that only positive root does make sense, i.e.
                           eβ 2 U 2 S                   4rZext L4
                      I=                        1+                  −1                           (A3.10).
                             rL 3                        β 2U2
A4.At low voltages (A3.10) simplifies and gives the following expression
                                        Zext S
                      I = 2Ueβ                      .                                            (A4.1)
                                            r   L
which is actually the Ohm law.
       Using the well-known relation
                           U
                       R=I                                                                       (A4.2)
together with
                             L
                       R = ρS                                                                    (A4.3),
one gets
                             1          r
                      ρ = 2eβ                                                                    (A4.4).
                                    Zext

                                    Part B. Self-sustained gas discharge
B1.Consider a gas layer located between xandx + dx.The rate of change in the electron number inside the
layer due to the electric current is givenfor a small time interval dt by
                                 I x +dx −I (x)                     1 dI x
                       dNeI = e             e
                                               dt = e dx e
                                                             dxdt.                                (B1.1).
                                        e
       This change is due to the effect of the external ionization and the electron avalanche formation.
       The external ionizer creates the following number of electrons in the volumeSdx
                       dNeext = Zext Sdxdt                                                        (B1.2).
whereas the electron avalanche produces the number of electrons found as
                                                            I (x)
                       dNea = αNe dl = ne Sdxvdt = α e e dxdt                                     (B1.3).
       The balance equationfor the number of electrons is written as
                       dNeI = dNeext + dNea                                                       (B1.4),
whichresults in the following differential equation for the electron current
                       dIe (x)
                               = eZext S + αIe (x)                                                (B1.5).
                         dx
                                       A1 x
       On substitutingIe x = C1 e           + A2 ,one derives
                       A1 = α                                                                     (B1.6),
                                 eZext S
                       A2 = −                                                                     (B1.7).
                                   α
B2.Given the fact that the ions flow in the direction opposite to the electron motion,the balance equationfor
the number of ionsis written as
                       dNiI = dNiext + dNia                                                       (B2.1),
where
                                 I x −I (x+dx )                      1 dI x
                        dNiI = i         i               i
                                               dt = − e dx   dxdt                              (B2.2).
                                         e
                            ext
                        dNi = Zext Sdxdt                                                       (B2.3).
                            a       Ie (x)
                        dNi = α e dxdt                                                         (B2.4).
         Hence, the following differential equation for the ion current is obtained
                           dIi (x)
                        − dx       = eZext S + αIe (x).                                        (B2.5)
         Onsubstituting the previouslyfound electron current together with the ion current,Ii x = C2 +
    B2 x
B1 e ,yields
                        B1 = −C1                                                               (B2.6),
                        B2 = α                                                                 (B2.7).
B3.Sincetheionsstartstomovefrom the anode located atx = L, the following condition holds
                        Ii L = 0                                                               (B3.1).
B4.By definition of secondary electron emission coefficient the following condition should be imposed
                        Ie 0 = γIi 0                                                           (B4.1).

                        Theoretical competition. Tuesday, 15 July 2014                                  3/3
B5.Total current in each tube section is the sum of the electron and ion currents:
                                             eZ S
                       I = Ie + Ii = C2 − ext                                                 (B5.1).
                                               α
Aftersubstituting the boundary conditions (B3.1) and (B4.1):
                       C2 − C1 e αL = 0                                                       (B5.2)
and
                            eZ S
                       C1 − ext    = γ(C2 − C1 )                                              (B5.3).
                               α
Solving (B5.2) and (B5.3) one can obtain:
                             eZ S          1
                       C2 = ext       −αL
                                                                                (B5.4).
                               α    e     (1+γ)−γ
So the total current:
                             eZext S        1
                        I=                             −1                                     (B5.5).
                               α       e −αL (1+γ)−γ
B6.When the discharge gap length is increased, the denominator in formula (B5.1) decreases. At that
moment, when it turns zero, the electric current in the gas becomes self-sustaining and external ionizer can
be turned off. Thus,
                           1          1
                     Lcr = α ln 1 + γ                                                          (B6.1).

An electric current flowing through a gas is called a gas discharge. There are m


[User]
Põrandale asetatakse kõrvuti kaks ühesugust silindrilist toru --- paralleelselt ja küljetsi üksteist puutuvana. Kolmas samasugune toru asetatakse nende peale --- samuti paralleelselt, nõnda et see toetub kahele alumisele.
Milliseid tingimusi peavad rahuldama hõõrdetegur $\mu$ toru ja põranda vahel ning hõõrdetegur $k$ kahe toru vahel
selleks, et pealmine toru kahte alumist üksteisest eemale ei vajutaks?

\begin{center}
	\includegraphics[width=0.3\linewidth]{2010-v2g-10-torud.pdf}
\end{center}

Solution:
Kõigepalt paneme tähele, et põhimõtteliselt võiks antud süsteemis toimida rõhumisjõud kahe alumise silindri vahel, kuid see
kaob niipea, kui alumised silindrid natukenegi üksteisest eemalduvad; niisiis võime sellega mitte arvestada.

\begin{center}
	\includegraphics[width=0.6\textwidth]{2010-v2g-10-torudlah.pdf}
\end{center}

Esmalt oletame, et $\mu$ on piisavalt suur, nii et vastu põrandat toetuvad torud pigem veerevad kui libisevad (kui $k$ pole piisavalt suur). Vaatleme vastu põrandat toetuvale torule mõjuvate jõumomentide tasakaalu tingimust toru ja põranda kontaktpunkti $P$ suhtes. Põranda rõhumis- ja hõõrdeju õlg on null; ka raskusjõud $mg$ õlg on null. Vaadeldavale torule mõjub veel vaid üksainus jõud --- ülemise toru põhjustatud hõõrde- ja rõhumisjõu resultant, mis on rakendatud puutepunkti $Q$ (vt joonist) ja kui tegemist on libisemise piirjuhuga (st veidigi väiksem hõõrdetegur $k$ viiks libisemisele), siis moodustab see vektor pinnanormaaliga nurga $\arctan k$ (sest antud vektor moodustub üksteisega risti olevate rõhumisjõu $N$ ja hõõrdejõu $F_h$ vektorite resultandina ning nurga tangens on $F_h/N=k$). Et ülejäänud jõudude moment oli null, siis peab ka selle jõu moment olema null,
st jõu vektor peab olema suunatud punkti $P$. Et kolmnurk $OQP$ on võrdhaarne (vt joonist),
siis
\[
k\ge \tan 15^\circ\approx \num{0,27}.
\]
Nüüd oletame, et $k \ge \tan 15^\circ$ ning vaatleme libisemise piirjuhtu punktis $P$. Selleks vaatleme jõumomentide tasakaalu punkti $Q$ suhtes.
Silindrile mõjuv raskusjõud $mg$ ning punktis $P$ toimiv rõhumisjõud $\frac 32 mg$ (mis kompenseerib poolteise silindri raskusjõu) on rakendatud sirge $OP$ sihis ning nende summaarne jõumoment $\frac 14mgR$ (kus $R$ on silindri raadius) tasakaalustab hõõrdejõu momendi
\[
\frac 32 mg\mu \left(R+R\sin 60^\circ\right).
\]
Siinjuures arvestasime, et punktis $P$ toimiv hõõrdejõud on $\mu$-kordne rõhumisjõud $\frac 32 mg$ ning on horisontaalne ja omab seetõttu õlga $R+R\sin 60^\circ$. Seega,
\[
\mu \ge \frac{1}{6\left(1+\sin 60^\circ\right)}=\frac{1}{6\left(1+\frac{\sqrt 3}{2}\right)}\approx \num{0,09}.
\]

Torud

Põrandale asetatakse kõrvuti kaks ühesugust silindrilist toru --- paralleelselt ja küljetsi üksteist puutuvana. Kolmas samasugune toru asetatakse nende peale --- samuti paralleelselt, nõnda et see toetub kahele alumisele.
Milliseid tingimusi peavad rahuldama hõõrdetegur $\mu$ toru ja põranda vahel ning hõõrdetegur $k$ kahe toru vahel
selleks, et pealmine toru kahte alumist üksteisest eemale ei vajutaks?

\begin{center}
	\includegraphics[width=0.3\linewidth]{2010-v2g-10-torud.pdf}
\end{center}

Tasakaalu korral peab iga toru jaoks kehtima jõudude ning jõumomentide tasakaal. Alumiste torude jaoks on jõumomentide tasakaalu kõige mugavam vaadelda maapinna puutepunkti suhtes, sest sellisel juhul on maapinna hõõrdejõu panus \num{0}. Lisaks paneme tähele, et kahe alumise silindri vahel rõhumisjõudu ei ole, sest see kaob niipea, kui alumised silindrid natukenegi üksteisest eemalduvad.


[User]
A submarine of unknown nationality is traveling near the bottom of the Baltic sea, at the depth
of h = 300 m. Its interior is one big room of volume V = 10 m3 filled with air (M = 29 g/mol) at
pressure p0 = 100 kPa and temperature t0 = 20 ◦ C. Suddenly it hits a rock and a large hole of area
A = 20 cm2 is formed at the bottom of the submarine. As a result, the submarine sinks to the
bottom and most of it is filled fast with water, leaving a bubble of air at increased pressure (no
air escapes the submarine). The density of water is ρ = 1000 kg/m3 and free fall acceleration is
g = 9.81 m/s2 . Treat the air as a diatomic ideal gas.

  1. What is the volume rate (in m3 /s) at which the water flows into the submarine immediately
     after the formation of the hole?

  2. The flow rate is so large that the submarine is filled with water so fast that heat exchange
     between the gas and the water can be neglected, in both this part and the next part. What
     is the volume of the air bubble once water flow has stopped?

  3. At this point, what is the change in internal energy of the air?

  4. The water stream rushing into the submarine creates a turbulent flow, which ultimately causes
     energy to be dissipated as heat. How much total energy is dissipated to heat within the water,
     once the inflow has stopped due to equalized pressures?

Solution:
A submarine of unknown nationality is traveling near the bottom of the Baltic sea, at the depth
of h = 300 m. Its interior is one big room of volume V = 10 m3 filled with air (M = 29 g/mol) at
pressure p0 = 100 kPa and temperature t0 = 20 ◦ C. Suddenly it hits a rock and a large hole of area
A = 20 cm2 is formed at the bottom of the submarine. As a result, the submarine sinks to the
bottom and most of it is filled fast with water, leaving a bubble of air at increased pressure (no
air escapes the submarine). The density of water is ρ = 1000 kg/m3 and free fall acceleration is
g = 9.81 m/s2 . Treat the air as a diatomic ideal gas.

  1. What is the volume rate (in m3 /s) at which the water flows into the submarine immediately
     after the formation of the hole?

  2. The flow rate is so large that the submarine is filled with water so fast that heat exchange
     between the gas and the water can be neglected, in both this part and the next part. What
     is the volume of the air bubble once water flow has stopped?

  3. At this point, what is the change in internal energy of the air?

  4. The water stream rushing into the submarine creates a turbulent flow, which ultimately causes
     energy to be dissipated as heat. How much total energy is dissipated to heat within the water,
     once the inflow has stopped due to equalized pressures?

Solution. This is NBPhO 2018, problem 8. The official solutions are here, an outline is:
  1. (5) According to Bernoulli’s principle, the flow rate through the hole is
                                        s
                                          2∆P       p
                                   v=            = 2gh = 76.7 m/s.
                                             ρ
     The rate of volume flow is
                                         Q = Av = 0.153 m3 /s.

  2. (5) The air in the submarine is adiabatically compressed. The initial pressure is p0 , and the
     flow stops once its pressure is equal to the outside water pressure, pf ≈ ρgh = 3.0 × 106 Pa.
     Since γ = 7/5 for a diatomic gas,
                                                5/7
                                                 p0
                                       Vf = Vi          = 0.9 m3 .
                                                 pf

  3. (5) The change in internal energy is

                                          ∆Uair = nR∆T.

     Using the ideal gas law, we have n = 410. The change in temperature can be found using
                                       1−1/γ         2/7
                                       pf              pf
                              Tf = Ti            = Ti          = 770 K.
                                       p0              p0
     Plugging in the numbers,
                                            ∆Uair = 4 × 106 J.

  4. (5) We use energy conservation. The work done on the submarine by the external water is

                                W = pf ∆V = pf (Vi − Vf ) = 2.7 × 107 J.

     If this isn’t totally clear, you can also note that the total decrease in gravitational potential
     energy of the water is ρgh∆V . But since h is large, we have ρgh ≈ pf in hydrostatic equilibrium,
     recovering the result above.
     This energy goes into heating up the water in the submarine, and heating the air. But we
     only want the heat in the water, which is

                                     Q = W − ∆Uair = 2.3 × 107 J.

     Any answer within 10% of this is acceptable.


[User]
Consider a layer of the atmosphere with density ρ, pressure P , adiabatic index γ, and
    density and pressure gradients dρ/dz and dP/dz. Suppose that a small parcel of air in this layer
    acquires a small upward velocity. Under certain conditions, the parcel of air will begin oscillating
    in height, performing simple harmonic motion. Neglecting drag and heat transfer between the
    parcel of air and its surroundings, find the angular frequency ω of this motion. This is called the

                  01^
    Brunt–Vaisala or buoyancy frequency.

                        01^

Solution:
Let the parcel have fixed mass m. At its initial position, the parcel has volume V and
    density ρp , which is equal to the local atmospheric density ρ. After the parcel moves up by ∆z, it
    has volume V ′ and density ρ′p , and the local atmospheric density is ρ′ . Then the net force on the
    parcel is the buoyant force minus the weight,

                                              F = (ρ′ − ρ′p )gV ′

    and we have m = ρ′ V ′ . Thus, the acceleration is
                                                       ρ′p

                                              a = 1 − ′ g.
                                                       ρ
    For small displacements, we have
                                              dρ                         dρp
                                   ρ′ ≈ ρ +      ∆z,       ρ′p ≈ ρ +         ∆z
                                              dz                         dz
    so that at first order in ∆z, the acceleration is

                                                g dρ dρp
                                           a=         −      ∆z.
                                                ρ dz    dz

    To calculate dρp /dz, note that since the parcel expands and contracts adiabatically, Pp ∝ ργp .
    Moreover, the pressure of the parcel is always equal to that of the surrounding atmosphere, so

                                                   P ∝ ργp

    which for small displacements means
                                               1 dP    1 dρp
                                                    =γ       .
                                               P dz    ρ dz
    Plugging this into the acceleration equation gives

                                                1 dρ   1 dP
                                        a=g          −        ∆z
                                                ρ dz γP dz
    which implies                  s                              r
                                                                       d log(P 1/γ /ρ)

                                          1 dP   1 dρ
                              ω=     g         −               =   g                   .
                                         γP dz   ρ dz                        dz
    When the argument of the square root becomes negative, there is no frequency of oscillation; instead,
    the parcel just keeps on going. That describes an atmosphere unstable against convection. As a
    check, marginal stability occurs for an adiabatic atmosphere, where P ∝ ργ . In realistic atmospheres,
    you wouldn’t expect to see these kinds of oscillations because of damping effects, but they provide
    a characteristic timescale for dynamic processes.

                  01^
                        01^


[User]
Ühesugustest kondensaatoritest mahtuvusega $C$ on koostatud joonisel näidatud lõpmatu ahel. Leidke ahela kogumahtuvus $C_k$.

\begin{center}
	\includegraphics[width=0.3\linewidth]{2007-v2g-08-yl}
\end{center}

Solution:
Lõpmatust ahelast ühe lüli eemaldamisega mahtuvus ei muutu. Seetõttu võime tervet ahelat vaadelda kui jadaühendust $C$-st ning $C$ ja $C_k$ paralleelühendusest. Seega saame, kasutades veel asjaolu, et jadaühenduses liituvad mahtuvuse pöördväärtused ning rööpühenduses mahtuvused ise, võrrandi:
\[
C_{k}=\frac{1}{1 / C+1 /\left(C+C_{k}\right)}.
\]
Teisendades jõuame ruutvõrrandini:
\[
C_{k}^{2}+C C_{k}-C^{2}=0.
\]
Seda lahendades saame:
\[
C_{k}=\frac{-1 \pm \sqrt{5}}{2} C \approx \SI{0,6}{C}.
\]
Negatiivse lahendi heitsime kõrvale.

Kondensaatoriredel

Ühesugustest kondensaatoritest mahtuvusega $C$ on koostatud joonisel näidatud lõpmatu ahel. Leidke ahela kogumahtuvus $C_k$.

\begin{center}
	\includegraphics[width=0.3\linewidth]{2007-v2g-08-yl}
\end{center}

Lõpmatust ahelast ühe lüli eemaldamisega summaarne mahtuvus ei muutu.


[User]
4. Questions consist of sub-questions. Write your detailed answer in the blank space
   provided below the sub-question and final answer to the sub-question in the smaller
   box which follows the blank space.

Solution:
4. (a) Magnetic field =        l
                           0               ρ>r

       Value of magnetic field =

       where ρ is the radial distance.

   (b) L =
           µ0 N 2 πr2
                l
       Value of L = 1.97 × 10−2 H
   (c) E = 3.95 J
           e
                                      1.26 × 10−2 T
                                      0

                                                       E    ρ<r
                                                            ρ>r

                         CS
   (d) i = (1 − e−Rt/L ) + i0 e−Rt/L       if i0 6= 0
           R
                    di     Lv
   (e) e = iR + L − i
                    dt   l + vt
       where L = µ0 N 2 πr2 /(l + vt)
                           µ0 N i0 ωρ
                        
                        
                                     sin(ωt)            ρ<r
                        
                              2l
   (f) Electric field =
                          µ N i ωr2
                         0 0
                        
                                      sin(ωt)          ρ>r
                               2ρl
   (g) The plot of E with radial distance:

                              Electric field

HB
                       0.04 V/m

                                               r        radial distance

        Lines of forces: Note, the lines of force are dense upto ρ = r and increasingly sparse
        thereafter.

                                                   r

                                                   2

INPhO 2012 Problem 4


[User]
Theory Question 3
                                                                             Page 1 of 8

Theory Question 3
This question consists of five independent parts. Each of them asks for an estimate of an
order of magnitude only, not for a precise answer. Enter all your answers into the
Answer Script.

Digital Camera         Consider a digital camera with a square CCD chip with linear
dimension L = 35 mm having Np = 5 Mpix (1 Mpix = 106 pixels). The lens of this
camera has a focal length of f = 38 mm. The well known sequence of numbers (2, 2.8, 4,
5.6, 8, 11, 16, 22) that appear on the lens refer to the so called F-number, which is
denoted by F # and defined as the ratio of the focal length and the diameter D of the
aperture of the lens, F # = f / D .

3.1   (1.0) Find the best possible spatial resolution Δxmin , at the chip, of the camera as
      limited by the lens. Express your result in terms of the wavelength λ and the F-
      number F # and give the numerical value for λ = 500 nm.
3.2   (0.5) Find the necessary number N of Mpix that the CCD chip should possess in
      order to match this optimal resolution.
3.3   (0.5) Sometimes, photographers try to use a camera at the smallest practical
      aperture. Suppose that we now have a camera of N 0 = 16 Mpix, with the chip size
      and focal length as given above. Which value is to be chosen for F # such that the
      image quality is not limited by the optics?
3.4   (0.5) Knowing that the human eye has an approximate angular resolution of
      φ = 2 arcmin and that a typical photo printer will print a minimum of 300 dpi
      (dots per inch), at what minimal distance z should you hold the printed page from
      your eyes so that you do not see the individual dots?

Data 1 inch = 25.4 mm
     1 arcmin = 2.91 × 10−4 rad

                                                                                Theory Question 3
                                                                                Page 2 of 8

Hard-boiled egg           An egg, taken directly from the fridge at temperature T0 = 4°C, is
dropped into a pot with water that is kept boiling at temperature T1 .

3.5     (0.5) How large is the amount of energy U that is needed to get the egg
        coagulated?
3.6     (0.5) How large is the heat flow J that is flowing into the egg?
3.7     (0.5) How large is the heat power P transferred to the egg?
3.8     (0.5) For how long do you need to cook the egg so that it is hard-boiled?

Hint    You may use the simplified form of Fourier’s Law J = κ ΔT / Δr , where ΔT is the
        temperature difference associated with Δr , the typical length scale of the problem.
        The heat flow J is in units of W m−2.
Data Mass density of the egg: μ = 103 kg m−3
     Specific heat capacity of the egg: C = 4.2 J K−1 g−1
     Radius of the egg: R = 2.5 cm
     Coagulation temperature of albumen (egg protein): Tc = 65°C
     Heat transport coefficient: κ = 0.64 W K−1 m−1 (assumed to be the same for liquid
     and solid albumen)

Lightning      An oversimplified model of lightning is presented. Lightning is caused by
the build-up of electrostatic charge in clouds. As a consequence, the bottom of the cloud
usually gets positively charged and the top gets negatively charged, and the ground below
the cloud gets negatively charged. When the corresponding electric field exceeds the
breakdown strength value of air, a disruptive discharge occurs: this is lightning.
                     current
                          .
                        .....
                           .....
                              ...
                                ...
                                  ...
                                    ...
          Imax = 100 kA               ...... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ....
                                       ...
                                         ...
                                                                                ....
                                                                                .... ....
                                           ...
                                             ...                                                                                . .. ... ........
                                                                                . . ....
                                                                                ... ..
                                               ...
                                                 ...                                                                                ....
                                                   ...
                                                     ..                                                                                .. .. ...                                              ....
                                                                                ....
                                                      ...
                                                                                ..                         ..                                        ....
                                                        ...
                                                          ...
                                                                                .. .                             ..                                            ....
                                                            ...
                                                                                .. .                                   .
                                                                                .                                                 ....
                                                              ...
                                                                                ..                                        ..                                                      ....
                                                                ...
                                                                                ..                                            ..                                                          ....
                                                                  ..
                                                                   ...
                                                                                . .                                                                                ....
                                                                     ...
                                                                                ..                                                    .
                                                                                .                                                                     ....
                                                                       ...
                                                                                ..                                                       .
                                                                                .                                                                          ....
                                                                         ...
                                                                                ..                                                          ..                                                                               ....
                                                                           ...
                                                                                ..                                                              ..                                                                                ....
                                                                             ...
                                                                               ..                                                                                ..                                                                  ..                                                                                ....
                                                                                ...                                                                                ..                                                                                ....
                                                                                ...
                                                                                . . .                                                                         .
                                                                                .                                                                                ....
                                                                                ...
                                                                                . .                                                                              ..                                                                                ....
                                                                                ...
                                                                                . .                                                                                .
                                                                                .                                                                                ....
                                                                                ...
                                                                                ...                                                                 . .                                                                                ..                                                                                ....
                                                                                ...
                                                                                . .                                                                                .                                                                                ....
                                                                                ..
                                                                                ..                                                                                .                                                                                ....
                                                                                ...
                                                                                . .                                                                                ..
                                                                                ....
                                                                                ...
                                                                                . .                                                                                ..                                                                                ....
                                                                                ...
                                                                                ...                                      . .                                                                                .
                                                                                .                                                                                ....
                                                                                ...
                                                                                .. .                                                                                .                                                                                ....
                                                                                ...
                                                                                . .
                                                                                .
                                                                                ..                                                                                ....
                                                                                ... ..
                                                                                ..                                                                                ....
                                                                                ... ..
                                                                                ..
                                                                                ....
                                                                                ... ..
                                                                                ..                                                                                ....
                                                                                ... ..                                                                                .                                                                                ...
                                                                                ................................................................................................................................................................................................................................................................................................................................................................................................
                                                                                .
                                                                                ....

                       0                                                                                time
                                                                                τ = 0.1 ms

       Idealized current pulse flowing between the cloud and the ground during
       lightning.

                                                                       Theory Question 3
                                                                             Page 3 of 8

Answer the following questions with the aid of this simplified curve for the current as a
function of time and these data:
   Distance between the bottom of the cloud and the ground: h = 1 km;
   Breakdown electric field of humid air: E0 = 300 kV m-1;
   Total number of lightning striking Earth per year: 32 × 106;
   Total human population: 6.5 × 109 people.

3.9   (0.5) What is the total charge Q released by lightning?

3.10 (0.5) What is the average current I flowing between the bottom of the cloud and
     the ground during lightning?
3.11 (1.0) Imagine that the energy of all storms of one year is collected and equally
     shared among all people. For how long could you continuously light up a 100 W
     light bulb for your share?

Capillary Vessels      Let us regard blood as an incompressible viscous fluid with mass
density μ similar to that of water and dynamic viscosity η = 4.5 g m−1 s−1. We model
blood vessels as circular straight pipes with radius r and length L and describe the blood
flow by Poiseuille’s law,

                                        Δp = R D ,

the Fluid Dynamics analog of Ohm’s law in Electricity. Here Δp is the pressure
difference between the entrance and the exit of the blood vessel, D = Sυ is the volume
flow through the cross-sectional area S of the blood vessel and υ is the blood velocity.
The hydraulic resistance R is given by

                                             8ηL
                                        R=        .
                                             π r4

For the systemic blood circulation (the one flowing from the left ventricle to the right
auricle of the heart), the blood flow is D ≈ 100 cm3s−1 for a man at rest. Answer the
following questions under the assumption that all capillary vessels are connected in
parallel and that each of them has radius r = 4 μm and length L = 1 mm and operates
under a pressure difference Δp = 1 kPa.

3.12 (1.0) How many capillary vessels are in the human body?
3.13 (0.5) How large is the velocity υ with which blood is flowing through a capillary
     vessel?

                                                                        Theory Question 3
                                                                              Page 4 of 8

Skyscraper        At the bottom of a 1000 m high skyscraper, the outside temperature is
Tbot= 30°C. The objective is to estimate the outside temperature Ttop at the top. Consider a
thin slab of air (ideal nitrogen gas with adiabatic coefficient γ = 7/5) rising slowly to
height z where the pressure is lower, and assume that this slab expands adiabatically so
that its temperature drops to the temperature of the surrounding air.

3.14 (0.5) How is the fractional change in temperature dT / T related to dp / p , the
     fractional change in pressure?
3.15 (0.5) Express the pressure difference dp in terms of dz , the change in height.

3.16 (1.0) What is the resulting temperature at the top of the building?

Data Boltzmann constant: k = 1.38 × 10−23 J K−1
     Mass of a nitrogen molecule: m = 4.65 × 10−26 kg
     Gravitational acceleration: g = 9.80 m s−2

Answer Script                                              Theory Question 3
                                                                 Page 5 of 8

       Country Code                     Student Code    Question Number

                                                              3

Answer Script                                                      For
                                                                   Examiners
                                       Digital Camera              Use
                                                                   Only
3.1   The best spatial resolution is
                                                                   0.7
       (formula:) Δxmin =

      which gives                                                  0.3
       (numerical:) Δxmin =

      for λ = 500 nm.
                                                                   0.5
3.2   The number of Mpix is
       N=

3.3   The best F-number value is                                   0.5

       F# =

                                                                   0.5
3.4   The minimal distance is
       z=

Answer Script                                              Theory Question 3
                                                                 Page 6 of 8

      Country Code                     Student Code     Question Number

                                                              3
                                      Hard-boiled egg
                                                                   For
3.5   The required energy is                                       Examiners
                                                                   Use
                                                                   Only
       U=
                                                                   0.5

3.6   The heat flow is                                             0.5

       J=

3.7   The heat power transferred is                                0.5

       P=

3.8   The time needed to hard-boil the egg is                      0.5

       τ=

Answer Script                                            Theory Question 3
                                                               Page 7 of 8

      Country Code                   Student Code     Question Number

                                                            3
                                      Lightning
                                                                 For
3.9   The total charge is                                        Examiners
                                                                 Use
                                                                 Only
       Q=
                                                                 0.5

3.10 The average current is                                      0.5

        I=

3.11 The light bulb would burn for the duration                  1.0

       t=

                                  Capillary Vessels

                                                                 1.0
3.12 There are

        N=

      capillary vessels in a human body.

3.13 The blood flows with velocity                               0.5

       υ=

Answer Script                                          Theory Question 3
                                                             Page 8 of 8

      Country Code                   Student Code   Question Number

                                                          3

                                     Skyscraper
                                                               For
                                                               Examiners
3.14 The fractional change in temperature is                   Use
                                                               Only

        dT
           =                                                   0.5
        T

3.15 The pressure difference is                                0.5

       dp =

3.16 The temperature at the top is                             1.0

       Ttop =

Solution:
SOLUTIONS to Theory Question 3
Digital Camera Two factors limit the resolution of the camera as a pho-
tographic tool: the diffraction by the aperture and the pixel size. For diffrac-
tion, the inherent angular resolution θR is the ratio of the wavelength λ of
the light and the aperture D of the camera,
                                           λ
                                 θR = 1.22   ,
                                          D
where the standard factor of 1.22 reflects the circular shape of the aperture.
When taking a picture, the object is generally sufficiently far away from the
photographer for the image to form in the focal plane of the camera where
the CCD chip should thus be placed. The Rayleigh diffraction criterion then
states that two image points can be resolved if they are separated by more
than

                               ∆x = f θR = 1.22λ F ] ,
3.1
       which gives
                                   ∆x = 1.22 μm

if we choose the largest possible aperture (or smallest value F ] = 2) and
assume λ = 500 nm for the typical wavelength of daylight
    The digital resolution is given by the distance ` between the center of two
neighboring pixels. For our 5 Mpix camera this distance is roughly
                                L
                            `= q    = 15.65 μm .
                                 Np
Ideally we should match the optical and the digital resolution so that neither
aspect is overspecified. Taking the given optical resolution in the expression
for the digital resolution, we obtain

                                        2
                                L

3.2                          N=              ≈ 823 Mpix .
                                ∆x

   Now looking for the unknown optimal aperture, we note that we should
have ` ≥ ∆x, that is: F ] ≤ F0 with
                                             s
                              L       N
                     F0 =      √   =2    = 14.34 .
                          1.22λ N0    N0

                                       1

Since this F ] value is not available, we choose the nearest value that has a
higher optical resolution,

3.3                                  F0 = 11 .

   When looking at a picture at distance z from the eye, the (small) sub-
tended angle between two neighboring dots is φ = `/z where, as above, ` is
the distance between neighboring dots. Accordingly,

                    `   2.54 × 10−2 /300 dpi
3.4            z=     =                      = 14.55 cm ≈ 15 cm .
                    φ     5.82 × 10−4 rad

Hard-boiled egg All of the egg has to reach coagulation temperature.
This means that the increase in temperature is

                  ∆T = Tc − T0 = 65 ◦ C − 4 ◦ C = 61 ◦ C .

Thus the minimum amount of energy that we need to get into the egg such
that all of it has coagulated is given by U = μV C∆T where V = 4πR3 /3 is
the egg volume. We thus find

3.5                           4πR3
                       U =μ        C(Tc − T0 ) = 16768 J .
                                3

The simplified equation for heat flow then allows us to calculate how much
energy has flown into the egg through the surface per unit time. To get an
approximate value for the time we assume that the center of the egg is at the
initial temperature T = 4 ◦ C. The typical length scale is ∆r = R, and the
temperature difference associated with it is ∆T = T1 − T0 where T1 = 100 ◦ C
(boiling water). We thus get

3.6                     J = κ(T1 − T0 )/R = 2458 W m−2 .

Heat is transferred from the boiling water to the egg through the surface of
the egg. This gives

                                     2

3.7                    P = 4πR2 J = 4πκR(T1 − T0 ) ≈ 19.3 W

for the amount of energy transferred to the egg per unit time. From this we
get an estimate for the time τ required for the necessary amount of heat to
flow into the egg all the way to the center:

                    U   μCR2 Tc − T0   16768
3.8            τ=     =              =       = 869 s ≈ 14.5 min .
                    P    3κ T1 − T0     19.3

Lightning The total charge Q is just the area under the curve of the
figure. Because of the triangular shape, we immediately get

                                          I0 τ
3.9                                  Q=        = 5C.
                                           2

The average current is

                                              I0
3.10                           I = Q/τ =         = 50 kA ,
                                              2

simply half the maximal value.
    Since the bottom of the cloud gets negatively charged and the ground
positively charged, the situation is essentially that of a giant parallel-plate ca-
pacitor. The amount of energy stored just before lightning occurs is QE0 h/2
where E0 h is the voltage difference between the bottom of the cloud and the
ground, and lightning releases this energy. Altogether we thus get for one
lightning the energy QE0 h/2 = 7.5 × 108 J. It follows that you could light
up the 100 W bulb for the duration

                              32 × 106    7.5 × 108 J
3.11                     t=             ×             ≈ 10 h .
                              6.5 × 109     100 W

Capillary Vessels        Considering all capillaries, one has
                                    ∆p
                           Rall =      = 107 Pa m−3 s .
                                    D
                                          3

All capillaries are assumed to be connected in parallel. The analogy between
Poiseuille’s and Ohm’s laws then gives the hydraulic resistance R of one
capillary as
                                    1      N
                                        =     .
                                  Rall     R
We thus get
                                          R
                                   N=
                                         Rall
for the number of capillary vessels in the human body. Now calculate R using
Poiseuille’s law,
                           8ηL
                       R=       ≈ 4.5 × 1016 kg m−4 s−1 ,
                           πr4
and arrive at

3.12                          4.5 × 1016
                           N≈       7
                                         = 4.5 × 109 .
                                  10

The volume flow is D = Sall v where Sall = N πr2 is the total cross-sectional
area associated with all capillary vessels. We then get

                               D       r2 ∆p
3.13                     v=        2
                                     =       = 0.44 mm s−1 ,
                              N πr      8ηL

where the second expression is found by alternatively considering one capil-
lary vessel by itself.

Skyscraper When the slab is at height z above the ground, the air in
the slab has pressure p(z) and temperature T (z) and the slab has volume
V (z) = Ah(z) where A is the cross-sectional area and h(z) is the thickness
of the slab. At any given height z, we combine the ideal gas law

          pV = N kT      (N is the number of molecules in the slab)

with the adiabatic law

                      pV γ = const or (pV )γ ∝ pγ−1

                                                                       dp   dT
to conclude that pγ−1 ∝ T γ . Upon differentiation this gives (γ −1)      =γ ,
                                                                       p    T
so that

                                       4

                                dT            dp
3.14                               = (1 − 1/γ) .
                                T             p

Since the slab is not accelerated, the weight must be balanced by the force
that results from the difference in pressure at the top and bottom of the slab.
Taking downward forces as positive, we have the net force
                                                 pV      V dp
             0 = N mg + A[p(z + h) − p(z)] =        mg +      h,
                                                 kT      h dz
          dp    mg p
so that      =−      or
          dz    k T

                                           mg p
3.15                              dp = −        dz .
                                           k T

Taken together, the two expressions say that
                                              mg
                           dT = −(1 − 1/γ)       dz
                                              k
and therefore we have
                                                  mgH
                        Ttop = Tbot − (1 − 1/γ)
                                                   k
for a building of height H, which gives

3.16                               Ttop = 20.6 ◦ C

for H = 1 km and Tbot = 30 ◦ C.

                                      5

Theory Question 3


[User]
01^
                         USAPhO 2022, problem B2. A nice problem on deriving the time dilation
    formula for an electrostatic “clock”.

Solution:
01^
                         USAPhO 2022, problem B2. A nice problem on deriving the time dilation
    formula for an electrostatic “clock”.


[User] Lääts tekitab esemest $d = \SI{24}{cm}$ kaugusele ekraanile kujutise, mis on esemest \num{3} korda suurem. Leidke läätse fookuskaugus.

Solution:
Kuna kujutis tekib ekraanile, siis on kujutis tegelik ning tegemist on koondava läätsega.

\begin{center}
	\includegraphics[width=0.9\linewidth]{2009-lahg-02-lah}
\end{center}

Olgu $a$ kaugus esemest läätseni, $k$ kaugus kujutisest läätseni ning $f$ läätse fookuskaugus. Et kujutis on esemest \num{3} korda suurem, siis sarnastest kolmnurkadest $ABO$ ja $A'B'O$
\[
\frac{k}{a}=3 \quad \Rightarrow \quad k=3 a.
\]
Kujutis tekib kugusele $d = \SI{24}{cm}$, seega
\[
a+k=4 a=\SI{24}{cm} \quad \Rightarrow \quad a=\SI{6}{cm}, k=\SI{18}{cm}.
\]
Nüüd läätse valemist
\[
\frac{1}{a}+\frac{1}{k}=\frac{1}{f}
\]
leiame, et $f = \SI{4,5}{cm}$.

\emph{Märkus:} läätse valemi asemel võib fookuskauguse leidmiseks kasutada sarnaseid kolmnurki $A'B'F$ ja $OKF$. Saame
\[
\frac{k-f}{f}=3 \quad \Rightarrow \quad f=\frac{k}{4}=\SI{4,5}{cm}.
\]

Lääts

Lääts tekitab esemest $d = \SI{24}{cm}$ kaugusele ekraanile kujutise, mis on esemest \num{3} korda suurem. Leidke läätse fookuskaugus.

Tasub teha selge joonis ning kasutada sarnaseid kolmnurki.