[User]
Uurime lihtsustatud mudeli abil kergelt kaldus pinnal asetseva mobiiltelefoni liikumist värinaalarmi töötamise ajal.
Kujutagu lauale asetatud mobiil risttahukat massiga $M$, mille sees liigub
üles-alla väike keha massiga $m$.
Liikugu see keha ajahetkedel $t=0, 2\tau, 4\tau, \ldots$ vahemaa $x$ võrra
hetkeliselt üles ning ajahetkedel $t=\tau, 3\tau, 5\tau, \ldots$ algasendisse
tagasi.
Olgu mobiiltelefoni ja laua vaheline hõõrdetegur $\mu$ ning laua kaldenurk
$\alpha \ll 1$. Mobiiltelefoni ja laua vahelised põrked lugege absoluutselt
plastseiks.
Millise keskmise kiirusega $u$ hakkab mobiiltelefon mööda lauda liikuma?

Solution:
Kuivõrd väike keha liigub alla hetkeliselt, siis süsteemi mobiil+keha masskese püsib paigal, mistõttu mobiil kerkib lauapinnast kõrgusele $h=x\frac mM$.
Edasi hakkab mobiil raskusjõu toimel langema; lauapinnale jõudmiseks kuluv aeg $t=\sqrt{2h/g}$. Hõõrdejõud peatab mobiili ilma libisemata, kui $\mu>\tan\alpha$; et
$\alpha \ll 1$, siis võime eeldada, et see nii ka juhtub. Kui $t=\sqrt{2h/g}<\tau$, siis jõuab mobiil liikuda langemise jooksul lauapinna sihis vahemaa $\delta=h\sin\alpha\approx h\alpha$,
mis annab keskmiseks kiiruseks $$u=\frac{xm\alpha}{2\tau M}.$$ Kui $\sqrt{2h/g}<\tau$, siis ei jõua mobiil lõpuni langeda, vaid väikese keha kerkimine surub mobiili ennatlikult vastu lauda tagasi.
Mobiil jõuab langeda vahemaa $H=g\tau^2/2$ võrra, mis annab keskmiseks kiiruseks
$$u=\frac{H\alpha}{2\tau}=\frac 14 g\tau\alpha.$$
Kokkuvõtlikult võib vastuse anda kujul
$$u=\min\left(\frac{xm\alpha}{2\tau M}, \frac 14 g\tau\alpha\right).$$

Värinaalarm

Uurime lihtsustatud mudeli abil kergelt kaldus pinnal asetseva mobiiltelefoni liikumist värinaalarmi töötamise ajal.
Kujutagu lauale asetatud mobiil risttahukat massiga $M$, mille sees liigub
üles-alla väike keha massiga $m$.
Liikugu see keha ajahetkedel $t=0, 2\tau, 4\tau, \ldots$ vahemaa $x$ võrra
hetkeliselt üles ning ajahetkedel $t=\tau, 3\tau, 5\tau, \ldots$ algasendisse
tagasi.
Olgu mobiiltelefoni ja laua vaheline hõõrdetegur $\mu$ ning laua kaldenurk
$\alpha \ll 1$. Mobiiltelefoni ja laua vahelised põrked lugege absoluutselt
plastseiks.
Millise keskmise kiirusega $u$ hakkab mobiiltelefon mööda lauda liikuma?

Väikse keha hetkelise liikumise käigus püsib mobiil+keha massikese paigal. Seega liigub mobiil iga $\tau$ tagant sarnaselt kehaga hetkeliselt üles või alla. Iga kord kui mobiil üles liigub, nihkub ta gravitatsiooni tõttu ka veidike laua sihis edasi.


[User] Two football players tried a trick move, where two balls would collide in the air. The players stood from each other at a distance $d = \SI{20}{m}$ and hit their balls at the same time, both giving the ball an initial speed of $v = \SI{15}{m/s}$. In what area could the balls collide during the flight? To answer, make a figure from top view where the locations of the players are marked and the area of all the possible collision points. Also give the dimensions of that area. You do not have to find at what height from the ground the possible collision points are. Gravitational acceleration is $g = \SI{9,8}{m/s^2}$.

Solution:
Vaatame pallide algkiirusi komponentide kaupa. Olgu 1. palli algkiiruse vertikaalkomponent $v_{1\mathrm{vert}}$ ning horisontaalkomponent pikki jalgpallureid ühendavat sirget $v_{1\mathrm{piki}}$ ja risti selle sirgega $v_{1\mathrm{risti}}$. Teise palli jaoks on vastavad komponendid $v\idx{2\mathrm{vert}}$, $v_{2\mathrm{piki}}$ ja $v_{2\mathrm{risti}}$. Kokkupõrke hetkel peavad mõlemad pallid olema läbinud ristsuunas võrdse vahemaa. Kuna pallide ristsuunaline kiirus ei muutu, siis peab kehtima $v_{1\mathrm{risti}}=v_{2\mathrm{risti}}$.\\
Kokkupõrkeks peavad pallid olema jõudnud ka samale kõrgusele. See tingimus on ilmselt täidetud, kui pallide algsed vertikaalkiirused on võrdsed, mille korral on pallid igal ajahetkel samal kõrgusel. Oletame, et pallid võiksid teatud hetkel samale kõrgusele jõuda ka juhul, kui näiteks esimese palli algne vertikaalkiirus on väiksem. Pallide tõusuaeg on arvutatav valemist $t=v_{\mathrm{vert}}/g$. Seega pöördub esimene väiksema vertikaalsuunalise algkiirusega pall varem tagasi. Kui teine pall lõpuks laskuma hakkab, siis liigub esimene pall juba maa poole. Kuna mõlemad pallid kiirenevad võrdselt raskuskiirendusega, siis on laskumisel esimese palli kiirus alati suurem ja teine pall ei jõua selle pallide õhusoleku ajal järele. Seega pole kokkupõrge erinevate vertikaalkiiruste korral võimalik ja peab kehtima $v_{1\mathrm{vert}}=v_{2\mathrm{vert}}$.\\
Kuna algkiirused olid moodulilt võrdsed ja $v^2=v_{\mathrm{vert}}^2+v_{\mathrm{piki}}^2+v_{\mathrm{risti}}^2$, siis peavad ka algkiiruste pikikomponendid olema absoluutväärtuselt võrdsed. Võimalikud kokkupõrkekohad on mõlemast jalgpallurist võrdsel kaugusel ja nende punktihulk moodustab maapinna tasandile projekteerituna sirge, mille otspunktide kaugus mõlemast jalgpallurist on määratud palli maksimaalse lennukaugusega. Selle leidmiseks avaldame näiteks lennuaja, mis on võrdne kahekordse ajaga, mis kulub raskusjõu mõjul vertikaalkiiruse kahanemiseks nullini
\[
t=2\frac{v_{\mathrm{vert}}}{g}.
\]
Selle ajaga liigutakse horisontaalsihis vahemaa
\[
s=v\idx{hor}t=2\frac{v\idx{hor}v\idx{vert}}{g}.
\]
Suurim lennukaugus saavutatakse, kui $v\idx{vert}=v\idx{hor}=v / \sqrt{2}$ ja seega $s\idx{max}=v^2 / g$.\\
Täisnurksest kolmnurgast leiame kokkupõrkepunktide piirkonna pikkuse
\[
l=2\sqrt{{s\idx{max}}^2-(d/2)^2}=2\sqrt{\frac{v^4}{g^2}-\frac{d^2}{4}},
\]
mille arvväärtus on \SI{41}{m}.

\begin{center}
\includegraphics{2013-v3g-09-jalgpallurid}
\end{center}

Jalgpallurid

Kaks jalgpallurit proovisid trikilööki, kus kaks palli õhus kokku põrkavad.
Jalgpallurid seisid teineteisest $d = \SI{20}{m}$ kaugusel ja andsid samal ajahetkel
sooritatud löögiga kumbki oma pallile algkiiruse $v = \SI{15}{m/s}$. Mis piirkonnas võisid pallid lennul kokku põrgata?
Vastuseks tehke pealtvaates joonis, kuhu on kantud jalgpallurite
asukohad ja kõikvõimalike kokkupõrkepunktide piirkond. Esitage ka selle
piirkonna mõõdud. Võimalike kokkupõrkepunktide kõrgust maapinnast pole
vaja eraldi välja arvutada ega joonisele kanda. Raskuskiirendus
on $g = \SI{9,8}{m/s^2}$.

Lihtsuse huvides tasub pallide kiirusi komponentide järgi vaadelda. Vertikaalne komponent $v_z$, horisontaalkomponent pikki jalgpallureid ühendavat sirget $v_y$ ning risti selle sirgega $v_x$. Kokkupõrke hetkel peab risti jalgpallureid ühendava sirgega läbitud vahemaa mõlemal pallil sama olema.


[User]
Leida laengud $q_1$, $q_2$ ja $q_3$ kõikidel skeemil toodud kondensaatoritel.

\begin{center}
	\includegraphics[width=0.7\linewidth]{2006-lahg-02-yl}
\end{center}

Solution: Kondensaatori $C_1$ plaadid on ühendatud läbi takistite $R_1$ ja $R_2$. Seepärast laeng selle kondensaatori plaatidel on $q_1 = 0$ (pärast seda, kui on lõppenud kondensaatorite $C_2$ ja $C_3$ laadimine). Kuna pärast kondensaatorite laadimist voolud skeemis ei kulge, pinged kondensaatoritel $C_2$ ja $C_3$ on võrdsed $\mathcal E$. Järelikult, $q_2 = C_2\mathcal E$ ja $q_3 = C_3\mathcal E$.

Elektriskeem

Leida laengud $q_1$, $q_2$ ja $q_3$ kõikidel skeemil toodud kondensaatoritel.

\begin{center}
	\includegraphics[width=0.7\linewidth]{2006-lahg-02-yl}
\end{center}

Kuna süsteemi stabiilses olekus on kondensaatorite pinge konstantne, ei läbi neid ka vool.


[User]
A student measures the latent heat of vaporization per unit mass of a substance X. He begins by
putting approximately 200 mL of liquid X on a combined hotplate and scale. He inserts a ther-
mometer, turns on the hotplate at time t = 0 min, and records the liquid’s temperature as well as
the combined mass. After 24 minutes, no more liquid remains in the beaker. The measured results
as shown below.

Plots of the temperature as a function of time, and the rate of change of temperature as a function
of temperature are given below.

  1. Explain why the second graph has a sharp increase, then a gradual decrease, then a sharp
     decrease.

  2. The specific heat capacity of liquid X is known to be 2.19 kJ kg−1 K−1 . Using this information
     and the given data, find the net power that goes into vaporizing the liquid, when it is boiling.

  3. Find the latent heat of vaporization of liquid X, in units of kJ kg−1 .

Part B

Solution:
A student measures the latent heat of vaporization per unit mass of a substance X. He begins by
putting approximately 200 mL of liquid X on a combined hotplate and scale. He inserts a ther-
mometer, turns on the hotplate at time t = 0 min, and records the liquid’s temperature as well as
the combined mass. After 24 minutes, no more liquid remains in the beaker. The measured results
as shown below.

Plots of the temperature as a function of time, and the rate of change of temperature as a function
of temperature are given below.

  1. Explain why the second graph has a sharp increase, then a gradual decrease, then a sharp
     decrease.
  2. The specific heat capacity of liquid X is known to be 2.19 kJ kg−1 K−1 . Using this information
     and the given data, find the net power that goes into vaporizing the liquid, when it is boiling.
  3. Find the latent heat of vaporization of liquid X, in units of kJ kg−1 .
Solution. This is AuPhO 2009, problem 14, whose official solution is here. (It was later copy-pasted
to become INPhO 2015, problem 5.) The answers are:
  1. First, the hot plate needs to warm up. Once it fully warms up, it can start heating the liquid.
     As the liquid gets hotter, it starts to lose more heat to its environment, so its rate of heating
     decreases. Finally, the liquid starts to boil, which happens at constant temperature.

  2. We need to account for the fact that the liquid will always be losing heat to its environment,
     so we should look at dT /dt just before it starts boiling. Referring to the second graph, it is
     9 ◦ C/min. Then we have

                             dT
                    P = mc      = (310 g − 69 g)(2.19 kJ/(kg · K))(9 K/min) = 80 W
                             dt
    where we remembered to subtract off the mass of the beaker and thermometer.
    Or, if you want to do a bit better, you could fit a line to the part of the dT /dt versus T curve,
    between roughly 60 ◦ C and 150 ◦ C, as this region should be linear by Newton’s law of cooling,
    and extrapolating it to 160 ◦ C. This will give an answer lower by about 10%, and either is
    good enough for the purposes of this question.

  3. We plot the mass versus time, in the regime where the liquid has started to boil, to get a
     boiling rate of 26 g/min. This implies L = 180 kJ/kg.

Part B

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


[User]
Question A2
Death Metal
A droplet of liquid metal has constant mass density and constant surface tension σ, which causes it
to form into a sphere of radius R. (Throughout this problem, you may neglect gravity.) A thin wire
is inserted into the droplet and connected to an electric current source which slowly charges the
droplet. There is a critical value of the charge, Q0 , that causes the droplet to split in half. Each
half takes half the total charge, Q0 /2, and half the mass of the original droplet. The “ejected half”
is repelled far away from the other half, which remains in contact with the wire.

a. For simplicity, assume the droplet splits as soon as the final state (after the droplet has split in
   half and the two halves are well separated) has a lower total energy than that of the initial, single
   droplet. What is the value of Q0 ? Give your answer in terms of a dimensionless constant A
   multiplied by a product of powers of σ, R, and the vacuum permittivity ϵ0 , and give the numeric
   value of A to three significant figures.

b. As more charge is added to the droplet by the current source, it continues to split in half
   repeatedly. What is the charge qn on the nth ejected droplet? Give your answer in terms of Q0 .

c. In the limit where all of the initial mass of the droplet has been ejected, what is the total work
   done by the current source? Give your answer in terms of a dimensionless constant B multiplied
   by a product of powers of σ, R, and ϵ0 , and give the numeric value of B to three significant
   figures.

                   Copyright ©2022 American Association of Physics Teachers

2022 USAPhO                                     Part A                                                 6

Solution:
Question A2
Death Metal
A droplet of liquid metal has constant mass density and constant surface tension σ, which causes it
to form into a sphere of radius R. (Throughout this problem, you may neglect gravity.) A thin wire
is inserted into the droplet and connected to an electric current source which slowly charges the
droplet. There is a critical value of the charge, Q0 , that causes the droplet to split in half. Each
half takes half the total charge, Q0 /2, and half the mass of the original droplet. The “ejected half”
is repelled far away from the other half, which remains in contact with the wire.
a. For simplicity, assume the droplet splits as soon as the final state (after the droplet has split in
   half and the two halves are well separated) has a lower total energy than that of the initial, single
   droplet. What is the value of Q0 ? Give your answer in terms of a dimensionless constant A
   multiplied by a product of powers of σ, R, and the vacuum permittivity ϵ0 , and give the numeric
   value of A to three significant figures.

                                               Solution
     Note that σ has units of energy/length2 , R has units of length, and ϵ0 has units of
     charge2 /(energy × length). There is only one combination with dimensions of charge,
     so we must have                           p
                                       Q0 = A ϵ0 σR3 .

     To find the value of A, note that a droplet of radius R and charge Q0 has a surface tension
     energy 4πR2 σ and an electrostatic energy Q20 /2C, where C = 4πϵ0 R is the capacitance of a
     conducting sphere relative to infinity. Thus, the total energy of the drop is

                                                          Q20
                                         E0 = 4πσR2 +           .
                                                         8πϵ0 R

     When the droplet splits, the new halves both have charge Q0 /2 and a radius R1 = R/21/3 ,
     since the total volume stays the same. The total energy of each of the halves is thus

                                                         (Q0 /2)2
                                        E1 = 4πσR12 +             .
                                                         8πϵ0 R1
     Setting E0 = 2E1 and solving for Q0 gives
                                           s
                                                  21/3 − 1
                                       A = 8π              ≈ 14.9.
                                                  2 − 21/3

     This model of the instability of a charged liquid drop is not exactly accurate, and was chosen
     to keep the problem simple. In reality, it isn’t enough for the final state to have lower energy;
     the repulsive electrostatic force needs to exceed the surface tension force, or else the bubble
     can’t get to this state of lower energy. Accounting for this requires a different calculation,
     first performed by Lord Rayleigh in 1882, which gives the “Rayleigh limit” of A = 8π ≈ 25.1.
     Students who successfully derived this result also received full credit for part (a).

b. As more charge is added to the droplet by the current source, it continues to split in half
   repeatedly. What is the charge qn on the nth ejected droplet? Give your answer in terms of Q0 .

                   Copyright ©2022 American Association of Physics Teachers

2022 USAPhO                                       Part A                                                 8

                                               Solution
     Consider the situation just before the nth split. At this point, the initial droplet has split in
     half n − 1 times, so it has a volume that is 2n−1 times smaller than the original volume, and
     thus a radius Rn−1 = R/2(n−1)/3 . By part (a), the charge required to split this drop is

                                                               Q0
                                              q
                                     Qn = A         3
                                               ϵ0 σRn−1 =            .
                                                            2(n−1)/2
     Once the split happens, the nth ejected droplet takes half of this charge, so
                                                  Qn     Q0
                                          qn =       = (n+1)/2 .
                                                   2  2

     Also note that the radius of this droplet is Rn = R/2n/3 .

c. In the limit where all of the initial mass of the droplet has been ejected, what is the total work
   done by the current source? Give your answer in terms of a dimensionless constant B multiplied
   by a product of powers of σ, R, and ϵ0 , and give the numeric value of B to three significant
   figures.

                                               Solution
     By dimensional analysis, the total work W done by the voltage source must be

                                               W = BσR2 .

     The work W is equal to the difference between the energy Ef of the final state, where there
     are many small droplets that have been dispersed to far away, and the energy Ei = 4πσR2 of
     the initial, uncharged droplet. The final energy includes the surface tension and electrostatic
     energy of all the small droplets, which have radius Rn and charge qn found in the previous
     problem. This means that the final energy is
                                ∞
                                                        !
                               X                  q 2
                                                    n
                          Ef =       4πσRn2 +
                                                8πϵ0 Rn
                               n=1
                                          ∞                    ∞
                                      2
                                          X   1        Q20   X 1
                             = 4πσR              +
                                      n=1
                                          22n/3 16πR0 ϵ0 n=1 22n/3
                                                   ! ∞
                                             A 2     X 1
                             = 4πσR2 1 +
                                            64π 2        22n/3
                                                     n=1
                                                   !
                                    2        A2          1
                             = 4πσR 1 +          2    2/3
                                                               = 4πσR2 × 2.30.
                                            64π      2 −1

     Here we have inserted the results found in part (b), then used the result of part (a) and
     summed the geometric series. Thus,

                             W = Ef − Ei ≈ (4πσR2 ) × 1.30 = 16.3 σR2 .

                   Copyright ©2022 American Association of Physics Teachers

2022 USAPhO                                      Part A                                                      9

USAPhO 2022 Question A2


[User]
Q T-3
                                                                                Page 1 of 2

                                                The Design of a Nuclear Reactor1                          (Total Marks: 10)
                                                                                235
        Uranium occurs in nature as UO2 with only 0.720% of the uranium atoms being U. Neutron induced
        fission occurs readily in 235U with the emission of 2-3 fission neutrons having high kinetic energy. This
        fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the
        kinetic energy of the fission neutrons, one can induce a chain of fissions in other 235U nuclei. This forms the
        basis of the power generating nuclear reactor (NR).
        A typical NR consists of a cylindrical tank of height H and radius R filled with a material called moderator.
        Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural UO2 in
        solid form of height H, are kept axially in a square array. Fission neutrons, coming outward from a fuel
        channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough
        energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid
        flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin,
        (B) Moderator and (C) NR of cylindrical geometry.

          Fig-I                 Fig-II                      Fig-III                    Schematic sketch of the
                                                                                Nuclear Reactor (NR)
                                                                                Fig-I: Enlarged view of a fuel
                                                                                channel (1-Fuel Pins)
                                                                                Fig-II: A view of the NR
                                                                                (2-Fuel Channels)
                                                                                Fig-III: Top view of NR
                                                                                (3-Square Arrangement of
                                                                                Fuel Channels and 4-Typical
                                                                                Neutron Paths).
                                                                                Only components relevant to
                                                                                the problem are shown (e.g.
                                                                                control rods and coolant are
                                                                                not shown).

    A   Fuel Pin
            Data     1. Molecular weight Mw = 0.270 kg mol-1          2. Density ρ = 1.060×104 kg m-3
          for UO2    3. Melting point Tm = 3.138×103 K                4. Thermal conductivity λ = 3.280 W m-1 K-1
        Consider the following fission reaction of a stationary 235U after it absorbs a neutron of negligible kinetic
        energy.
                                          235
                                              U + 1n → 94Zr + 140Ce + 2 1n +
A1                                                                                0.8
        Estimate       (in MeV) the total fission energy released. The nuclear masses are: m(235U) = 235.044 u;
        m( Zr) = 93.9063 u; m(140Ce) = 139.905 u; m(1n) = 1.00867 u and 1 u = 931.502 MeV c-2. Ignore charge
           94

        imbalance.
A2      Estimate N the number of 235U atoms per unit volume in natural UO2.                                               0.5
        Assume that the neutron flux density, φ = 2.000×1018 m-2 s-1 on the fuel is uniform. The fission cross-
        section (effective area of the target nucleus) of a 235U nucleus is σf = 5.400×10-26 m2. If 80.00% of the
A3                                                                                1.2
        fission energy is available as heat, estimate Q (in W m-3), the rate of heat production in the pin per unit
        volume. 1MeV = 1.602×10-13 J
        The steady-state temperature difference between the center (Tc) and the surface (Ts) of the pin can be
A4      expressed as Tc−Ts = k F(Q,a,λ), where k = 1 ∕ 4 is a dimensionless constant and a is the radius of the pin.      0.5
        Obtain F(Q,a,λ) by dimensional analysis. Note that λ is the thermal conductivity of UO2.

1
  Joseph Amal Nathan (BARC) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the
principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and
the International Board are gratefully acknowledged.

                                                         Q T-3
                                                                                Page 2 of 2

     The desired temperature of the coolant is 5.770×102 K. Estimate the upper limit au on the radius a of the
A5                                                                                1.0
     pin.
B    The Moderator
     Consider the two dimensional elastic collision between a neutron of mass 1 u and a moderator atom of mass
     A u. Before collision all the moderator atoms are considered at rest in the laboratory frame (LF). Let ⃗⃗⃗⃗ and
     ⃗⃗⃗⃗ be the velocities of the neutron before and after collision respectively in the LF. Let ⃗⃗⃗⃗⃗ be the velocity
     of the center of mass (CM) frame relative to LF and θ be the neutron scattering angle in the CM frame. All
     the particles involved in collisions are moving at nonrelativistic speeds.
     The collision in LF is shown schematically, where θL is the scattering angle (Fig-IV). Sketch the collision
     schematically in CM frame. Label the particle velocities for 1, 2 and 3 in terms of ⃗⃗⃗⃗ , ⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ . Indicate
     the scattering angle θ.
                                                 ⃗⃗⃗⃗
                           Fig-IV
                                                   2            Collision in the Laboratory Frame
B1                                                                                1.0
                                                              1-Neutron before collision
                  ⃗⃗⃗⃗                     θL                 2-Neutron after collision
                                                              3-Moderator Atom before collision
                    1                                         4-Moderator Atom after collision
                                    3
                                                    4

     Obtain v and V, the speeds of the neutron and moderator atom in the CM frame after collision, in terms of A
B2                                                                                1.0
     and .
     Derive an expression for G(α, θ) = Ea ∕ Eb , where Eb and Ea are the kinetic energies of the neutron, in the
B3                                                                                1.0
     LF, before and after the collision respectively and                           .
     Assume that the above expression holds for D2O molecule. Calculate the maximum possible fractional
B4   energy loss              of the neutron for the D2O (20 u) moderator.                                                  0.5

C    The Nuclear Reactor
     To operate the NR at any constant neutron flux ψ (steady state), the leakage of neutrons has to be
     compensated by an excess production of neutrons in the reactor. For a reactor in cylindrical geometry the
     leakage rate is k1 [(2.405 ∕ R)2 + (π ∕ H)2] ψ and the excess production rate is k2 ψ. The constants k1 and k2
     depend on the material properties of the NR.
     Consider a NR with k1 = 1.021×10-2 m and k2 = 8.787×10-3 m-1. Noting that for a fixed volume the leakage
C1                                                                                1.5
     rate is to be minimized for efficient fuel utilization, obtain the dimensions of the NR in the steady state.
     The fuel channels are in a square arrangement (Fig-III) with the nearest neighbour distance 0.286 m. The
C2   effective radius of a fuel channel (if it were solid) is 3.617×10-2 m. Estimate the number of fuel channels Fn         1.0
     in the reactor and the mass M of UO2 required to operate the NR in steady state.

Solution:
Theoretical Task 3 (T-3) : Solutions                                        1 of 9

                          The Design of a Nuclear Reactor1

Uranium occurs in nature as UO2 with only 0.720% of the uranium atoms being 235 U. Neutron
induced fission occurs readily in 235 U with the emission of 2-3 fission neutrons having high
kinetic energy. This fission probability will increase if the neutrons inducing fission have low
kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain
of fissions in other 235 U nuclei. This forms the basis of the power generating nuclear reactor
(NR).

A typical NR consists of a cylindrical tank of height H and radius R filled with a material called
moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel
pins of natural UO2 in solid form of height H, are kept axially in a square array. Fission
neutrons, coming outward from a fuel channel, collide with the moderator, losing energy, and
reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat
generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In
the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator
and (C) NR of cylindrical geometry.

      Fig-I               Fig-II                       Fig-III
                                                                            Schematic sketch of the
                                                                             Nuclear Reactor (NR)
                                                                          Fig-I: Enlarged view of a
                                                                          fuel channel (1-Fuel Pins)
                                                                          Fig-II: A view of the NR
                                                                          (2-Fuel Channels)
                                                                          Fig-III: Top view of NR
                                                                          (3-Square Arrangement of
                                                                          Fuel     Channels      and
                                                                          4-Typical Neutron Paths).
                                                                          Only components relevant
                                                                          to the problem are shown
                                                                          (e.g. control rods and
                                                                          coolant are not shown).

A. Fuel Pin
                                                Data for UO2
                                              −1
1. Molecular weight Mw =0.270 kg mol                                2. Density ρ=1.060×104 kg m−3
3. Melting point Tm =3.138×103 K                        4. Thermal conductivity λ=3.280 W m−1 K−1

A1 Consider the following fission reaction of a stationary 235 U after it absorbs a neutron of
   negligible kinetic energy.
                                   235
                                         U +1 n −→94 Zr +140 Ce + 2 1 n + ∆E
  1
    Joseph Amal Nathan (BARC) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the
principal authors of this problem. The contributions of the Academic Committee, Academic Development Group
and the International Board are gratefully acknowledged.

                    Theoretical Task 3 (T-3) : Solutions                                  2 of 9

    Estimate ∆E (in MeV) the total fission energy released. The nuclear masses are: m(235 U)
    = 235.044 u; m(94 Zr) = 93.9063 u; m(140 Ce) = 139.905 u; m(1 n) = 1.00867 u and 1 u =
    931.502 MeV c−2 . Ignore charge imbalance.                                                     [0.8]

     Solution: ∆E = 208.684 MeV

     Detailed solution: The energy released during the transformation is

                   ∆E = [m(235 U) + m(1 n) − m(94 Zr) − m(140 Ce) − 2m(1 n)]c2

     Since the data is supplied in terms of unified atomic masses (u), we have

                        ∆E = [m(235 U) − m(94 Zr) − m(140 Ce) − m(1 n)]c2

                            = 208.684 MeV [Acceptable Range (208.000 to 209.000)]
     from the given data.

A2 Estimate N the number of 235 U atoms per unit volume in natural UO2 .                           [0.5]

     Solution: N = 1.702 × 1026 m−3

     Detailed solution: The number of UO2 molecules per m3 of the fuel N1 is given
     in the terms of its density ρ, the Avogadro number NA and the average molecular
     weight Mw as
           ρNA
     N1 =
           Mw
           10600 × 6.022 × 1023
         =                       = 2.364 × 1028 m−3
                  0.270
     Each molecule of UO2 contains one uranium atom. Since only 0.72% of these are
     235
       U,
     N = 0.0072× N1
       = 1.702 × 1026 m−3 [Acceptable Range (1.650 to 1.750)]

A3 Assume that the neutron flux φ = 2.000 × 1018 m−2 s−1 on the fuel is uniform. The fission
   cross-section (effective area of the target nucleus) of a 235 U nucleus is σf = 5.400 ×10−26
   m2 . If 80.00% of the fission energy is available as heat, estimate Q (in W m−3 ) the rate of
   heat production in the pin per unit volume. 1MeV = 1.602 ×10−13 J.                              [1.2]

     Solution: Q = 4.917 × 108 W/m3

     Detailed solution: It is given that 80% of the fission energy is available as heat
     thus the heat energy available per fission Ef is from a-(i)
     Ef = 0.8 × 208.7 MeV
        = 166.96 MeV
        = 2.675 × 10−11 J
     The total cross-section per unit volume is N × σf . Thus the heat produced per unit

                    Theoretical Task 3 (T-3) : Solutions                                  3 of 9

     volume per unit time Q is
     Q = N × σf × φ × Ef
       = (1.702 × 1026 ) × (5.4 × 10−26 ) × (2 × 1018 ) × (2.675 × 10−11 ) W/m3
       = 4.917 × 108 W/m3 [Acceptable Range (4.800 to 5.000)]

A4 The steady-state temperature difference between the center (Tc ) and the surface (Ts ) of the
   pin can be expressed as Tc − Ts = kF (Q, a, λ) where k = 1/4 is a dimensionless constant
   and a is the radius of the pin. Obtain F (Q, a, λ) by dimensional analysis.                     [0.5]

                            Qa2
     Solution: Tc − Ts =        .
                            4λ
     Detailed solution: The dimensions of Tc − Ts is temperature. We write this as
     Tc − Ts = [K]. Once can similarly write down the dimensions of Q, a and λ. Equating
     the temperature to powers of Q, a and λ, one could state the following dimensional
     equation:
     K = Qα aβ λγ
     = [M L−1 T −3 ] α [L] β [M L1 T −3 K −1 ] γ
     This yields the following algebraic equations
     γ = -1 equating powers of temperature
     α + γ = 0 equating powers of mass or time. From the previous equation we get α = 1
     Next −α + β + γ = 0 equating powers of length. This yields β = 2.
                                   Qa2
     Thus we obtain Tc − Ts =            where we insert the dimensionless factor 1/4 as sug-
                                    4λ
     gested in the problem. No penalty if the factor 1/4 is not written.

     Note: Same credit for alternate ways of obtaining α, β, γ.

A5 The desired temperature of the coolant is 5.770 ×102 K. Estimate the upper limit au on
   the radius a of the pin.                                                                        [1.0]

     Solution: au = 8.267 × 10−3 m.

     Detailed solution: The melting point of UO2 is 3138 K and the maximum tempera-
     ture of the coolant is 577 K. This sets a limit on the maximum permissible temperature
     (Tc − Ts ) to be less than (3138 - 577 = 2561 K) to avoid “meltdown”. Thus one may
     take a maximum of (Tc − Ts ) = 2561 K.
     Noting that λ = 3.28 W/m - K, we have
                                               2561 × 4 × 3.28
                                       a2u =
                                                 4.917 × 108
     Where we have used the value of Q from A2. This yields au w 8.267 × 10−3 m. So
     au = 8.267 × 10−3 m constitutes an upper limit on the radius of the fuel pin.

     Note: The Tarapur 3 & 4 NR in Western India has a fuel pin radius of 6.090 × 10−3
     m.

                        Theoretical Task 3 (T-3) : Solutions                                            4 of 9

B. The Moderator
Consider the two dimensional elastic collision between a neutron of mass 1 u and a moderator
atom of mass A u. Before collision all the moderator atoms are considered at rest in the
laboratory frame (LF). Let −→
                            vb and −
                                   →
                                   va be the velocities of the neutron before and after collision
                            −
                            →
respectively in the LF. Let vm be the velocity of the center of mass (CM) frame relative to LF
and θ be the neutron scattering angle in the CM frame. All the particles involved in collisions
are moving at non-relativistic speeds

B1 The collision in LF is shown schematically with θL as the scattering angle (Fig-IV). Sketch
   the collision schematically in CM frame. Label the particle velocities for 1, 2 and 3 in
   terms of −→
             vb , −
                  →
                  va and −
                         v→
                          m . Indicate the scattering angle θ.                                                   [1.0]
                                                             
                                Fig-IV                       va
                                                             2      Collision in the Laboratory Frame
                                                                   1-Neutron before collision
                       
                       vb                          L              2-Neutron after collision
                                                                   3-Moderator Atom before collision
                        1                 3                        4-Moderator Atom after collision
                                                              4

      Solution:

                            Laboratory Frame                          Center of Mass Frame
                                                                                        
                                                        va                              va  vm
                                                        2
                                                                  
                  vb                          L                  vb  vm           

                   1                  3                                                      
                                                                                 vm
                                                         4

B2 Obtain v and V , the speeds of the neutron and the moderator atom in the CM frame after
   the collision, in terms of A and vb .                                                                         [1.0]

      Solution: Detailed solution: Before the collision in the CM frame (vb − vm ) and
      vm will be magnitude of the velocities of the neutron and moderator atom respectively.
                                                                                vb
      From momentum conservation in the CM frame, vb − vm = Avm gives vm = A+1        .

      After the collision, let v and V be magnitude of the velocities of neutron and moderator
      atom respectively in the CM frame. From conservation laws,
                                               1              1 2   1     1
          v = AV                and              (vb − vm )2 + Avm = v 2 + AV 2 .(→ [0.2 + 0.2])
                                               2              2     2     2

                    Theoretical Task 3 (T-3) : Solutions                                         5 of 9

                         Avb              vb
     Solving gives v = A+1    and V = A+1    . (OR) From definition of center of mass frame
             vb                                                      Avb
     vm = A+1 . Before the collision in the CM frame vb − vm = A+1       and vm will be mag-
     nitude of the velocities of the neutron and moderator atom respectively. In elastic
     collision the particles are scattered in the opposite direction in the CM frame and so
                                    Avb            vb
     the speeds remain same v = A+1      and V = A+1  (→ [0.2 + 0.1]).

     Note: Alternative solutions are worked out in the end and will get appropriate weigh-
     tage.

B3 Derive an expression for G(α, θ) = Ea /Eb , where Eb and Ea are the kinetic energies of the
   neutron, in the LF, before and after the collision respectively, and α ≡ [(A − 1)/(A + 1)]2 ,          [1.0]

     Solution:
                             Ea   A2 + 2A cos θ + 1  1
                 G(α, θ) =      =             2
                                                    = [(1 + α) + (1 − α) cos θ] .
                             Eb       (A + 1)        2

     Detailed solution: Since −  →
                                 va = − →v +−v→      2     2    2
                                                m , va = v + vm + 2vvm cos θ (→ [0.3]). Substi-
                                              2
                                            A v  2       2
                                                        vb        2Avb2
     tuting the values of v and vm , va2 = (A+1)b 2 + (A+1)  2 + (A+1)2 cos θ (→ [0.2]), so

                                   va2   Ea   A2 + 2A cos θ + 1
                                       =    =                   .
                                   vb2   Eb       (A + 1)2

                                   A2 + 1       2A             1
                    G(α, θ) =             2
                                            +         2
                                                        cos θ = [(1 + α) + (1 − α) cos θ] .
                                  (A + 1)     (A + 1)          2
         Alternate form
                                      (1 − α)(1 − cos θ)
                              = 1−                       .
                                              2

     Note: Alternative solutions are worked out in the end and will get appropriate weigh-
     tage.

B4 Assume that the above expression holds for D2 O molecule. Calculate the maximum pos-
   sible fractional energy loss fl ≡ EbE−E
                                         b
                                           a
                                             of the neutron for the D2 O (20 u) moderator.                [0.5]

     Solution: fl = 0.181

     Detailed solution: The maximum energy loss will be when the collision is head
     on ie., Ea will be minimum for the scattering angle θ = π.

     So Ea = Emin = αEb .

                                                                 Eb −Emin
     For D2 O, α = 0.819 and maximum fractional loss                 Eb
                                                                                = 1 − α = 0.181. [Ac-
     ceptable Range (0.170 to 0.190)]

                      Theoretical Task 3 (T-3) : Solutions                                    6 of 9

C. The Nuclear Reactor
To operate the NR at any constant neutron flux Ψ (steady state), the leakage of neutrons has to
be compensated by an excess production
                               h          of neutrons
                                                   i    in the reactor. For a reactor in cylindrical
                                       2         2
geometry the leakage rate is k1 2.405    + Hπ

                                  R
                                                     Ψ and the excess production rate is k2 Ψ. The
constants k1 and k2 depend on the material properties of the NR.

C1 Consider a NR with k1 = 1.021×10−2 m and k2 = 8.787×10−3 m−1 . Noting that for a
   fixed volume the leakage rate is to be minimized for efficient fuel utilisation obtain the
   dimensions of the NR in the steady state.                                                           [1.5]

       Solution: R = 3.175 m, H = 5.866 m.

       Detailed solution: For constant volume V = πR2 H,
                                    "       2   #
                                 d     2.405      π 2
                                               +       = 0,
                                dH       R        H

                              d 2.4052 πH  π2   2.4052 π    π2

                                          + 2 =          − 2 3 = 0,
                             dH     V      H       V        H

               2.405 2      π 2

       gives     R
                       =2   H
                                .

       For steady state,
                                     "           2                #
                                          2.405             π 2
                      1.021 × 10−2                     +                Ψ = 8.787 × 10−3 Ψ.
                                            R               H

       Hence H = 5.866 m [Acceptable Range (5.870 to 5.890)]
       R = 3.175 m [Acceptable Range (3.170 to 3.180)].

       Alternative Non-Calculus Method       to Optimize
                                                    2
                                             2.405          π 2
       Minimisation of the expression                  +        , for a fixed volume V =
                                               R           H
       πR2 H:
                                                    2.4052 πH     π2
       Substituting for R2 in terms of V, H we get             + 2,
                                                        V         H
                                      2           2          2
                                 2.405 πH 2.405 πH         π
       which can be written as,            +            + 2.
                                    2V          2V         H
       Since all the terms are positive applying AMGM inequality for three positive terms we
       get
             2.4052 πH   2.4052 πH   π2
                                           r                              r
                 2V
                       +    2V
                                   + H 2   3 2.4052 πH   2.4052 πH  π 2           4 4
                                                                           3 2.405 π
                                         ≥             ×           × 2 =              .
                          3                     2V          2V      H          4V 2

                    Theoretical Task 3 (T-3) : Solutions                                     7 of 9

     The RHS is a constant. The LHS is always greater or equal to this constant im-
     plies that this is the minimum value the LHS can achieve. The minimum is achieved
                                                                            2.4052 πH
     when all the three positive terms are equal, which gives the condition           =
                                                                               2V
                       2
     π2

               2.405          π 2
          ⇒              = 2        .
     H2          R            H
     For steady state,
                                    "           2                #
                                         2.405             π 2
                     1.021 × 10−2                     +                Ψ = 8.787 × 10−3 Ψ.
                                           R               H

     Hence H = 5.866 m [Acceptable Range (5.870 to 5.890)]
     R = 3.175 m [Acceptable Range (3.170 to 3.180)].

                                                                π2
     Note: Putting the condition in the RHS gives the minimum as 2 . From the condi-
                                         r                      H
                π3    2.4052 π 2   π2    3 2.4054 4
                                                  π
     tion we get 3 =             ⇒ 2 =              .
                H        2V        H         4V 2
     Note: The radius and height of the Tarapur 3 & 4 NR in Western India is 3.192 m and
     5.940 m respectively.

C2 The fuel channels are in a square arrangement (Fig-III) with nearest neighbour distance
   0.286 m. The effective radius of a fuel channel (if it were solid) is 3.617×10−2 m. Estimate
   the number of fuel channels Fn in the reactor and the mass M of UO2 required to operate
   the NR in steady state.                                                                            [1.0]

     Solution: Fn = 387 and M = 9.892 × 104 kg.

     Detailed solution: Since the fuel channels are in square pitch of 0.286 m, the ef-
     fective area per channel is 0.2862 m2 = 8.180 × 10−2 m2 .

     The cross-sectional area of the core is πR2 = 3.142 × (3.175)2 = 31.67 m2 , so the
     maximum number of fuel channels that can be accommodated in the cylinder is the
                     31.67
     integer part of 0.0818 = 387.

     Mass of the fuel=387×Volume of the rod×density
                     = 387 × (π × 0.036172 × 5.866) × 10600 = 9.892 × 104 kg.
     Fn = 387 [Acceptable Range (380 to 394)]
     M = 9.892 × 104 kg [Acceptable Range (9.000 to 10.00)]

     Note 1: (Not part of grading) The total volume of the fuel is 387 × (π × 0.036172 ×
     5.866) = 9.332 m3 . If the reactor works at 12.5 % efficieny then using the result of
     a-(iii) we have that the power output of the reactor is 9.332 × 4.917 × 108 × 0.125 =

                     Theoretical Task 3 (T-3) : Solutions                                  8 of 9

       573 MW.
       Note 2: The Tarapur 3 & 4 NR in Western India has 392 channels and the mass of the
       fuel in it is 10.15 ×104 kg. It produces 540 MW of power.

Alternative Solutions to sub-parts B2 and B3: Let σ be the scattering angle of the
Moderator atom in the LF, taken clockwise with respect to the initial direction of the neutron
before collision. Let U be the speed of the Moderator atom, in the LF, after collision. From
momentum and kinetic conservation in LF we have
                                   vb = va cos θL + AU cos σ,                                 (1)
                                    0 = va sin θL − AU sin σ,                                 (2)
                                 1 2    1         1
                                   vb =   AU 2 + va2 .                                        (3)
                                 2      2         2
Squaring and adding eq(1) and (2) to eliminate σ and from eq(3) we get
                               A2 U 2 = va2 + vb2 − 2va vb cos θL ,
                               A2 U 2 = Avb2 − Ava2 ,                                         (4)
which gives
                             2va vb cos θL = (A + 1)va2 − (A − 1)vb2 .                        (5)
(ii) Let v be the speed of the neutron after collision in the COMF. From definition of center
                      vb
of mass frame vm =         .
                     A+1
va sin θL and va cos θL are the perpendicular and parallel components of va , in the LF, resolved
along the initial direction of the neutron before collision. Transforming these to the COMF
gives va sin θL and va cos θL − vm as the perpendicular
                                                    p and parallel components of v. Substitut-
ing for vm and for 2va vb cos θL from eq(5) in v = va2 sin2 θL + va2 cos2 θL + vm2 − 2v v cos θ
                                                                                a m      L
                               Avb                                                            2
and simplifying gives v =           . Squaring the components of v to eliminate θL gives va =
                              A+1
v 2 + vm
       2
         + 2vvm cos θ. Substituting for v and vm and simplifying gives,
                                 va2   Ea   A2 + 2A cos θ + 1
                                     =    =                   .
                                 vb2   Eb       (A + 1)2
                     Ea    A2 + 1       2A             1
           G(α, θ) =    =         2
                                    +         2
                                                cos θ = [(1 + α) + (1 − α) cos θ] .
                     Eb   (A + 1)     (A + 1)          2

(OR)
                                                   vb
(iii) From definition of center of mass frame vm =     . After the collision, let v and V
                                                 A+1
be magnitude of the velocities of neutron and moderator atom respectively in the COMF.
From conservation laws in the COMF,
                                            1              1 2   1     1
               v = AV          and            (vb − vm )2 + Avm = v 2 + AV 2 .
                                            2              2     2     2
                    Avb          vb
Solving gives v = A+1   and V = A+1 . We also have v cos θ = va cos θL − vm , substituting for vm
and for va cos θL from eq(5) and simplifying gives
                                 va2   Ea   A2 + 2A cos θ + 1
                                     =    =                   .
                                 vb2   Eb       (A + 1)2

                        Theoretical Task 3 (T-3) : Solutions                                     9 of 9

                        Ea    A2 + 1       2A             1
            G(α, θ) =      =         2
                                       +         2
                                                   cos θ = [(1 + α) + (1 − α) cos θ] .
                        Eb   (A + 1)     (A + 1)          2

(OR)
                                                     vb
(iv) From definition of center of mass frame vm =        . After the collision, let v and V
                                                   A+1
be magnitude of the velocities of neutron and moderator atom respectively in the CM frame.
From conservation laws in the CM frame,
                                             1              1 2   1     1
                v = AV           and           (vb − vm )2 + Avm = v 2 + AV 2 .
                                             2              2     2     2
                   Avb             vb
Solving gives v = A+1  and V = A+1    . U sin σ and U cos σ are the perpendicular and parallel
components of U , in the LF, resolved along the initial direction of the neutron before collision.
Transforming these to the COMF gives U sin σ and −U cos σ + vm as the perpendicular and
parallel components of V . So we get U 2 = V 2 sin2 θ +V 2 cos2 θ +vm
                                                                    2
                                                                      −2V vm cos θ. Since V = vm
          2    2
we get U = 2vm (1 − cos θ). Substituting for U from eq(4) and simplifying gives

                                  va2   Ea   A2 + 2A cos θ + 1
                                      =    =                   .
                                  vb2   Eb       (A + 1)2

                        Ea    A2 + 1       2A             1
            G(α, θ) =      =         2
                                       +         2
                                                   cos θ = [(1 + α) + (1 − α) cos θ] .
                        Eb   (A + 1)     (A + 1)          2

                      √
                        A2 + 2A cos θ + 1
Note: We have va =                        vb . Substituting for va , v, vm in v cos θ = va cos θL −vm
                            A+1
gives the relation between θL and θ,
                                                 A cos θ + 1
                                  cos θL = √                       .
                                               A2 + 2A cos θ + 1
Treating the above equation as quadratic in cos θ gives,
                                                    p
                                  − sin2 θL ± cos θL A2 − sin2 θL
                          cos θ =                                 .
                                                  A
For θL = 0◦ the root with the negative sign gives θ = 180◦ which is not correct so,
                                          p
                                    cos θL A2 − sin2 θL − sin2 θL
                            cos θ =                               .
                                                  A
                                                                       va2
Substituting the above expression for cos θ in the expression for          gives an expression in terms
                                                                       vb2
of cos θL                                       p
                        va2   Ea   A2 + 2 cos θL A2 − sin2 θL + cos 2θL
                            =    =                                      .
                        vb2   Eb                 (A + 1)2

The Design of a Nuclear Reactor1                          (Total Marks: 10)


[User]
USAPhO 2015, problem A2.
    Now we give a few problems on current flow through continuous objects. Fundamentally, all one
    needs for these problems is the definition J = σE, and superposition.

       Example 3

       Consider two long, concentric cylindrical shells of radii a < b and length L. The volume
       between the two shells is filled with material with conductivity σ(r) = k/r. What is the
       resistance between the shells, and the charge density?

       Solution
       To find the resistance, we compute the current I when a voltage V is applied between the
       shells. By symmetry, in the steady state the current density must be
                                                         I
                                               J(r) =         r̂.
                                                       2πrL
       On the other hand, we also know that
                                                  Z b
                                                                     I(b − a)
                                       Z
                                                         I
                                 V = E · dr =                   dr =
                                                    a 2πrLσ           2πkL
       from which we conclude
                                                      b−a
                                                 R=         .
                                                      2πkL
       Note that the radial electric field between the shells is constant, so
                                                         V
                                               E(r) =         r̂.
                                                       b−a

       This means that in the steady state, there must be a nonzero charge density between the
       shells. (If there weren’t, then we would have E(r) ∝ 1/r, rather than a constant.)

       To find the charge density explicitly, it’s easiest to use Gauss’s law in differential form in
       cylindrical coordinates. We use the form of the divergence derived in E1,

                                      1 ∂(rEr )                   1 V    ρ
                             ∇·E=               + (other terms) =      =
                                      r ∂r                        rb−a   ϵ0
       thus showing that the charge density is proportional to 1/r. Of course, we could also get this
       result by applying Gauss’s law in integral form, to concentric spheres.

[2] Problem 4 (Cahn). A washer is made of a material of resistivity ρ. It has a square cross section
    of length a on a side, and its outer radius is 2a. A small slit is made on one side and wires are
    connected to the faces exposed.

    Since the washer has an irregular shape, the current distribution inside it is complicated: it spreads
    out from the first wire, goes around the washer, and converges into the second wire. However, the
    situation is simpler if we glue a perfectly conducting square plate, of side length a, to each exposed
    face. Find the resistance in this case.

[3] Problem 5 (BAUPC 1995). An electrical signal can be transferred between two metallic objects
    buried in the ground, where the current passes through the Earth itself. Assume that these objects
    are spheres of radius r, separated by a horizontal distance L ≫ r, and suppose both objects are
    buried a depth much greater than L in the ground. If the Earth has uniform resistivity ρ, find the
    approximate resistance between the terminals. (Hint: consider the superposition principle.)

[3] Problem 6 (PPP 162). A plane divides space into two halves. One half is filled with a homogeneous
    conducting medium, and physicists work in the other. They mark the outline of a square of side
    a on the plane and let a current I0 in and out at two of its neighboring corners. Meanwhile, they
    measure the potential difference V between the two other corners.

    Find the resistivity ρ of the medium.
[3] Problem 7 (MPPP 174). We aim to measure the resistivity of the material of a large, thin,
    homogeneous square metal plate, of which only one corner is accessible. To do this, we chose points
    A, B, C and D on the side edges of the plate that form the corner.

    Points A and B are both 2d from the corner, whereas C and D are each a distance d from it. The
    length of the plate’s sides is much greater than d, which, in turn, is much greater than the thickness
    t of the plate. If a current I enters the plate at point A, and leaves it at B, then the reading on a
    voltmeter connected between C and D is V . Find the resistivity ρ of the plate material.

       Remark
       Setups like those in the previous two problems are commonly used to measure resistivities,
       but why do they use a complicated “four terminal” setup? Wouldn’t it have been easier to
       just attach two terminals, send a current I through them, and measure the voltage drop
       V ? The problem with this is that it also picks up the resistance R of the contacts between
       the terminals and the material, along with the resistances of the wires. By having a pair of
       terminals measure voltage alone, drawing negligible current, we avoid this problem.

Solution:
Since the plates are conducting, the potential doesn’t depend on the radius r at the plates
    themselves, so by rotational symmetry, it doesn’t depend on r anywhere in the washer. Therefore,
    there is no radial current; all the current flows tangentially, so we can think of the washer as a set
    of radial rings in parallel.
       Split the washer into a bunch of radial rings with width dr. We see that r ranges from a to 2a.
    Each little ring has resistance ρ(2πr)/(adr), and they are all effectively connected in parallel. Thus,
                                           Z 2a
                                     1                   a            a
                                       =          ρ−1       dr = ρ−1    log 2,
                                     R      a           2πr          2π

    giving the answer,
                                                           2π ρ
                                                   R=             .
                                                          log 2 a
[3] Problem 5 (BAUPC 1995). An electrical signal can be transferred between two metallic objects
    buried in the ground, where the current passes through the Earth itself. Assume that these objects

    are spheres of radius r, separated by a horizontal distance L ≫ r, and suppose both objects are
    buried a depth much greater than L in the ground. If the Earth has uniform resistivity ρ, find the
    approximate resistance between the terminals. (Hint: consider the superposition principle.)
    Solution. We can consider one object at a time, and then use superposition to find the combined
    effect of both. Suppose that current I comes out from one of the objects. Placing this object at the
    origin, we have
                                               I             ρI
                                         J=      2
                                                   r̂, E =       r̂.
                                             4πr            4πr2
    Therefore, the potential difference between this object and where the other object would be is
                                               ρI L dr
                                                    Z
                                                              ρI
                                           V =            ≈
                                               4π r r2       4πr
    where we used r ≪ L. Finally, the other object takes in current I, with its J, E, and V superposing
    with the first object. Thus, the total potential difference is ρI/2πr, so
                                                   V    ρ
                                              R=     =     .
                                                   I   2πr
[3] Problem 6 (PPP 162). A plane divides space into two halves. One half is filled with a homogeneous
    conducting medium, and physicists work in the other. They mark the outline of a square of side
    a on the plane and let a current I0 in and out at two of its neighboring corners. Meanwhile, they
    measure the potential difference V between the two other corners.

    Find the resistivity ρ of the medium.
    Solution. The surface of the plane enforces the boundary condition J · n̂ = 0, since current can’t
    come out of it, which is equivalent to setting E · n̂ = 0. Then the uniqueness theorems in E1 apply.
       Now, for simplicity, suppose there is only current coming in at A. Then one possible solution is
    that J points radially outward from A and has (hemi)spherical symmetry, with magnitude
                                                                I0
                                       J · 2πr2 = I0 =⇒ J =         .
                                                               2πr2
    This obeys the boundary condition, so it must be the unique solution. In this case,
                                                      I0
                                               E=         r̂
                                                    2πσr2
    where σ is the conductivity of the material, which implies
                                         Z √2a                          √
                                                  I0         I0
                             VD − VC =               2
                                                       dr =      (1 − 1/ 2).
                                          a     2πσr        2πσa

    Similarly, for the case where the current is coming out of B, we have
                                                     I0         √
                                        VD − VC =        (1 − 1/ 2).
                                                    2πσa
    The actual voltage drop is the superposition of the two,
                                                    I0 ρ     √
                                             ∆V =        (2 − 2)
                                                    2πa
    where ρ = 1/σ is the resistivity. Then ρ can be calculated as
                                                              √
                                           2πa∆V       πa(2 + 2)∆V
                                     ρ=         √ =                .
                                         I0 (2 − 2)          I0

[3] Problem 7 (MPPP 174). We aim to measure the resistivity of the material of a large, thin,
    homogeneous square metal plate, of which only one corner is accessible. To do this, we chose points
    A, B, C and D on the side edges of the plate that form the corner.

    Points A and B are both 2d from the corner, whereas C and D are each a distance d from it. The
    length of the plate’s sides is much greater than d, which, in turn, is much greater than the thickness
    t of the plate. If a current I enters the plate at point A, and leaves it at B, then the reading on a
    voltmeter connected between C and D is V . Find the resistivity ρ of the plate material.

    Solution. This problem is a harder than the previous one because it’s harder to guess a current
    configuration that satisfies the boundary conditions, i.e. that the current density at the edges of the
    plate is parallel to the plate. The key is that we can use the following “image current” configuration
    to automatically satisfy the original problem’s boundary conditions, but on an infinite plate.

    We have a new current source and sink respectively at the reflections of A and B in O. The current
    sources and sinks all have magnitude 2I, rather than I, because only half of the currents at A and
    B actually enter and exit the physical plate, shaded in gray.
                                                                              2I
       Now, if a current 2I enters the plate, the current at a distance r is 2πrt , so the electric field at a
                  ρI                                                   ρI
    distance r is πrt (pointing radially outward), so the potential is πt log(r0 /r) for some arbitrary r0 ,
    which we’ll take to be the same for all current sources and sinks. Then
                             ρI                                       √       ρI
                      VC =      (log(r0 /d) + log(r0 /3d) − 2 log(r0 / 5d)) =    log(5/3).
                             πt                                               πt
    By symmetry, VC = −VD . Thus,

                                           2ρI                          πt     V
                              V = 2VC =        log(5/3),       ρ=                .
                                            πt                      2 log(5/3) I

       Remark
       Setups like those in the previous two problems are commonly used to measure resistivities,
       but why do they use a complicated “four terminal” setup? Wouldn’t it have been easier to
       just attach two terminals, send a current I through them, and measure the voltage drop
       V ? The problem with this is that it also picks up the resistance R of the contacts between
       the terminals and the material, along with the resistances of the wires. By having a pair of
       terminals measure voltage alone, drawing negligible current, we avoid this problem.


[User]
1. DC-DC CONVERTER (8 points)

In order
to obtain high voltage supply using a battery,
the following circuit is used.
An electromagnetic switch K1 connects a bat-
tery of electromotive force E to an inductor
of inductance L: it is closed if there is no cur-
rent in the inductor (a spring keeps it closed),
but if the inductor current reaches a critical
value I0, magnetic field created by the in-
ductor pulls it open. Due to inertia, once the
key is open, it takes a certain time τK to close
again even if the current falls to zero.
For the diode D you may assume that
its current is zero for any reverse voltage
(VD < 0), and also for any forward voltage
smaller than the opening voltage V0 (i.e. for
0 < VD < V0). For any non-zero forward cur-
rent, the diode voltage VD remains equal to
V0.
You may express your answers in terms
of L, E, I0, V0, and the capacitance C (see
figure).
i) (1 point) At first, let the key K2 be open. If
the initial inductor current is zero, how long
time τL will it take to open the key K1?
ii) (1 point) Assuming (here and in what
follows) that L/R ¿ τK ¿ τL, plot the in-
ductor current as a function of time t (for
0≤ t<3τL).
iii) (1 point) What is the maximal voltage
Vmax on the resistor R?
iv) (2 points) Assuming that Vmax À V0,
what is the average power dissipation on the
diode?
v) (2 points) Now, let the key K2 be closed,
and let us assume simplifyingly that V0 =0;
also, RC ÀτL and τK >π
p
LC. Suppose that
the circuit has been operated for a very long
time. Find the average voltage on the res-
istor.
vi) (1 point) Find the amplitude of voltage
variations on the resistor.

Solution:
1. DC-DC converter

1) (1 pt) From the Kircho's voltage law for
the loop consisting of L and E, E = LdI
dt
, hence
I = Et/L. From I0 = EτL/L we obtain
τL = LI0/E.
2) (1 pts) Once the current I0 is reached, the key
is opened; the current trough L cannot change
instantaneously and therefore is forced to flow
through the resistor R. Since the characteristic
time of this current loop (consisting of L and R
is very short (L/R fi τK), the current decays
very fast and becomes essentially equal to zero
while the key is still open. Now there is no cur-
rent through the inductor, so that the key will
close again and the process will start repeating
from the beginning. As a result we'll have a
periodic graph as shown in figure.
I
I0
t
3) (1 pt) The voltage through the resistor is
maximal when the current is maximal, which
happens immediately after the switch is opened;
the maximal current is I0 so that Vmax = RI0.
4) (2 pts) Due to Vmax fl V0, we can neglect
the eect of the diode; so we have the Kircho's
voltage law LdI
dt
= RI = Rdq
dt
(here we ex-
pressed the current via the charge q which flows
through the resistor). Integration over a single
cycle (when the inductor current drops from I0
down to 0) yields LI0 = R∆q, hence the charge
flown through the resistor (and through the di-
ode) ∆q = I0L/R. During that cycle, the diode
had a constant voltage V0, so the electric field
performed work A = V0∆q which was released
as heat in the diode. So, the average power dis-
sipation
P =
A
τL
=
V0I0L
RτL
=
V0E
R
.
5) (2 pts) Now, since the characteristic time of
the RC-loop is very large, the capacitor main-
tains its charge (and voltage) during that period
of time when the diode is closed. Once the key
K1 opens, the diode will open, and the capacitor
is connected to the inductor so that a LC-loop is
formed. That loop admits oscillations of period
TLC = 2π
√
LC and as the current to the resistor
can be neglected (since RC fl TLC), we can neg-
lect the presence of the resistor. During the time
interval when the diode is open, we can also neg-
lect the presence of the diode since V0 fi Vmax .
Hence, the current i in the LC-loop changes in
time sinusoidally, starting from i = I0 and end-
ing when i = 0 (then the diode will close discon-
necting the LC-loop). During that process, the
magnetic energy of the inductor
1
2
LI2
0 is conver-
ted into the electric field energy of the capacitor,
which is later released as heat on the resistor.
When the stationary regime is achieved, the en-
ergy lost by the capacitor during one period (of
duration τL) as the heat dissipation on the res-
istor Q = V 2
avτL/R must be equal to the energy
received from the inductor; so,
V 2
avτL
R
=
1
2
LI2
0 ⇒ Vav = I0
r
LR
2τL
=
r
EI0R
2
.
6) (1 pt) The charge which flows away from
the capacitor when the diode is closed can be
found as qC = τLVav/R (owing to RC fl τLC,
the relative change of the capacitor's voltage
is small). Hence, the voltage drop is found as
∆V = qC/C = τLVav/(RC). The amplitude is
half of the dierence between the minimal and
maximal values, so that the amplitude
U0 =
∆V
2
=
τLVav
2RC
=
I0L
2C
r
I0
2RE
.

DC-DC CONVERTER


[User]
APhO 2013, problem 2. This is a challenging question that ties together
    everything you’ve learned about kinematics.

Solution:
01h APhO 2013, problem 2. This is a challenging question that ties together
    everything you’ve learned about kinematics.


[User]
Question B2
Consider a parallel plate capacitor with the plates vertical. The plates of the capacitor are rigidly supported
in place. The distance between the plates is d. The plates have height h and area A ≫ d2 . Assume
throughout this problem that the force of air resistance may be neglected; however, the force of gravity
cannot be neglected. Neglect any edge effects as well as any magnetic effects.
                                                                                d/2
                                      d

                                                                Rigid Support

                                                                String
                                                  L
                                                                                h
                     h

                                                 h/2

  a. A small metal ball with a mass M and a charge q is suspended from a string of length L that is tied
     to a rigid support. When the capacitor is not charged, the metal ball is located at the center of the
     capacitor— at a distance d/2 from both plates and at a height h/2 above the bottom edge of the plates.
     If instead a constant potential difference V0 is applied across the plates, the string will make an angle
     θ0 to the vertical when the metal ball is in equilibrium.
        i. Determine θ0 in terms of the given quantities and fundamental constants.
       ii. The metal ball is then lifted until it makes an angle θ to the vertical where θ is only slightly
           greater than θ0 . The metal ball is then released from rest. Show that the resulting motion is
           simple harmonic motion and find the period of the oscillations in terms of the given quantities
           and fundamental constants.
       iii. When the ball is at rest in the equilibrium position θ0 , the string is cut. What is the maximum
            value for V0 so that the ball will not hit one of the plates before exiting? Express your answer in
            terms of the given quantities and fundamental constants.
  b. Suppose instead that the ball of mass M and charge q is released from rest at a point halfway between
     the plates at a time t = 0. Now, an AC potential difference V (t) = V0 sin ωt is also placed across the
     capacitor. The ball may hit one of the plates before it falls (under the influence of gravity) out of the
     region between the plates. If V0 is sufficiently large, this will
                                                                    p only occur for some
                                                                                p range of angular
     frequencies ωmin < ω < ωmax . You may assume that ωmin ≪ g/h and ωmax ≫ g/h. Making these
     assumptions, find expressions for ωmin and ωmax in terms of the given quantities and/or fundamental
     constants.
   c. Assume that the region between the plates is not quite a vacuum, but instead humid air with a uniform
      resistivity ρ. Ignore any effects because of the motion of the ball, and assume that the humid air doesn’t
      change the capacitance of the original system.

        i. Determine the resistance between the plates.
       ii. If the plates are originally charged to a constant potential source V0 , and then the potential is
           removed, how much time is required for the potential difference between the plates to decrease to
           a value of V0 /e, where ln e = 1?
       iii. If the plates are instead connected to an AC potential source so that the potential difference
            across the plates is V0 sin ωt, determine the amplitude I0 of the alternating current through the
            potential source.

                         Copyright c 2008 American Association of Physics Teachers

2008 Semifinal Exam                            Part B                             10

   STOP: Do Not Continue to Part C
If there is still time remaining for Part B, you should review your work for
    Part B, but do not continue to Part C until instructed by your exam
                    supervisor. You may not return to Part A

                      Copyright c 2008 American Association of Physics Teachers

2008 Semifinal Exam                                Part C                                                   11

Optical Society of America Bonus Question
Researchers have developed a lens made of liquid. The spherical lens consists of a droplet of transparent
liquid resting on an electrically controllable surface. When the voltage of the surface is changed, the droplet
itself changes shape; it either tries to “ball-up” more strongly or it becomes flatter. The figure below is a
sketch of the liquid lens and several parameters that describe it, including the thickness of the lens (t), the
radius of curvature of the top surface (R) and the contact angle (θ), which represents the angle between the
flat surface beneath the droplet and the tangent to the curved surface at the point of contact.

                            t

                                                                              θ

                                           R                      R

  a. When a certain voltage is applied, both the contact angle and lens thickness increase (and the lens
     becomes more curved). In this case, is the liquid attracted or repelled by the surface?
  b. Express the contact angle as a function of R and t.
   c. The total volume of the liquid lens is an important parameter because as the liquid lens changes shape,
      its volume is conserved. Calculate the volume of the lens as a function of R and t.
  d. Use your result to part (b) to eliminate the variable t from your expression for the volume and find
     V (R, θ).
   e. By changing the voltage on the control surface, the contact angle, θ, can be changed, which in turn
      changes the focal length of the lens, f . The lensmaker’s formula can be used to calculate the focal
      length and is given by
                                        1                        1    1
                                           = (nliquid − nair )      −      ,
                                        f                        R1   R2
     where nliquid and nair are the refractive indices of the liquid in the lens and air around it, and R1 and
     R2 are the radii of curvature of the two surfaces of the lens. In figure 1, R1 is the curved face and R2
     is the flat face. Use the lensmaker’s formula to calculate the focal length of the lens in terms of the
     total volume of the liquid, the contact angle, and the relevant refractive indices.
     Sidenote: liquid lenses are interesting because they are electrically controllable, variable focus lenses
     that can be very compact. People are working on putting them into cell phone cameras for ultra-
     compact zoom lenses. For more information on this type of liquid lens, see T. Krupenkin, S. Yang,
     and P. Mach, “Tunable liquid microlens,” Appl. Phys. Lett. 82, 316-318 (2003).

                        Copyright c 2008 American Association of Physics Teachers

Solution:
Question B2
Consider a parallel plate capacitor with the plates vertical. The plates of the capacitor are rigidly supported
in place. The distance between the plates is d. The plates have height h and area A  d2 . Assume
throughout this problem that the force of air resistance may be neglected; however, the force of gravity
cannot be neglected. Neglect any edge effects as well as any magnetic effects.
                                                                                d/2
                                      d

                                                                Rigid Support

                                                                String
                                                  L
                                                                                h
                     h

                                                 h/2

  a. A small metal ball with a mass M and a charge q is suspended from a string of length L that is tied
     to a rigid support. When the capacitor is not charged, the metal ball is located at the center of the
     capacitor— at a distance d/2 from both plates and at a height h/2 above the bottom edge of the plates.
     If instead a constant potential difference V0 is applied across the plates, the string will make an angle
     θ0 to the vertical when the metal ball is in equilibrium.

        i. Determine θ0 in terms of the given quantities and fundamental constants.
       ii. The metal ball is then lifted until it makes an angle θ to the vertical where θ is only slightly
           greater than θ0 . The metal ball is then released from rest. Show that the resulting motion is
           simple harmonic motion and find the period of the oscillations in terms of the given quantities
           and fundamental constants.

                         Copyright ©2008 American Association of Physics Teachers

2008 Semifinal Exam                                Part B                                                   14

       iii. When the ball is at rest in the equilibrium position θ0 , the string is cut. What is the maximum
            value for V0 so that the ball will not hit one of the plates before exiting? Express your answer in
            terms of the given quantities and fundamental constants.
  b. Suppose instead that the ball of mass M and charge q is released from rest at a point halfway between
     the plates at a time t = 0. Now, an AC potential difference V (t) = V0 sin ωt is also placed across the
     capacitor. The ball may hit one of the plates before it falls (under the influence of gravity) out of the
     region between the plates. If V0 is sufficiently large, this will
                                                                    p only occur for some
                                                                                p range of angular
     frequencies ωmin < ω < ωmax . You may assume that ωmin  g/h and ωmax  g/h. Making these
     assumptions, find expressions for ωmin and ωmax in terms of the given quantities and/or fundamental
     constants.
  c. Assume that the region between the plates is not quite a vacuum, but instead humid air with a uniform
     resistivity ρ. Ignore any effects because of the motion of the ball, and assume that the humid air doesn’t
     change the capacitance of the original system.
         i. Determine the resistance between the plates.
        ii. If the plates are originally charged to a constant potential source V0 , and then the potential is
            removed, how much time is required for the potential difference between the plates to decrease to
            a value of V0 /e, where ln e = 1?
       iii. If the plates are instead connected to an AC potential source so that the potential difference
            across the plates is V0 sin ωt, determine the amplitude I0 of the alternating current through the
            potential source.

                                                Solution
  a.    i. The electric field between the plates is E = V0 /d. In equilibrium, the horizontal and vertical
           components of the tension balance the electrostatic and gravitational forces. Consequently,
                                                                qV0
                                                     tan θ0 =        .
                                                                M gd

       ii. The simplest way to see this is to note that the electric field, for the purpose of this metal ball
           alone, acts like an effective extra horizontal contribution to the gravitational field. The system
           is thus a simple pendulum experiencing tilted gravity, which we know performs simple harmonic
           motion for small amplitudes. Then
                                        s           s                          s
                                            L                  L                  L cos θ0
                                T = 2π          = 2π p                    = 2π             .
                                           geff         g 2 + (qV0 /M d)2            g

       iii. Upon cutting the string the ball will move in a straight line, tangent to the angle the string
            originally made. So it will leave the region between the plates a distance

                                                 x1 = (L + h/2) tan θ0

          away from the center line. Setting this equal to d/2 and solving for V0 gives

                                                             M gd2
                                                    V0 =             .
                                                           (2L + h)q

  b. The ball will experience an oscillating force in the x direction of
                                                         V0
                                                  FE =      sin ωt.
                                                         d

                       Copyright ©2008 American Association of Physics Teachers

2008 Semifinal Exam                                      Part B                                                  15

       Hence, the x component of the acceleration will be
                                                              qV0
                                                       ax =       sin ωt.
                                                              Md
       The ball is released from rest, so this can be directly integrated to give the x component of the velocity,
                                                         qV0
                                                  vx =        (1 − cos ωt) .
                                                         M dω
       This can be integrated again to find the position relative to the center, which we will define as x = 0,

                                                  qV0         1
                                             x=          t − sin ωt .
                                                 M dω         ω

       The ball hits one of the platespif this value exceeds d/2 while it is still in the region between the plates.
       It exits this region when t = h/g.
       For ωmin , we may perform a small angle approximation,
                                                                 1
                                                    sin ωt ≈ ωt − ω 3 t3
                                                                 6
       yielding
                                    M d2 ω  1                               3M d2 p 3 3
                                           = ω 2 t3       ⇒       ωmin =           g /h .
                                    2qV0    6                                qV0
       As for ωmax , we may neglect the (sin ωt)/ω term entirely, for

                                           M d2 ω                          2qV0 p
                                                  =t     ⇒      ωmax =           h/g.
                                           2qV0                            M d2
       At higher frequencies the ball will miss, falling out of the plates before it hits either side.

  c.     i. The resistance between the plates is given by

                                                              R = ρd/h2 .

         ii. This is an RC circuit, with characteristic time
                                                              d
                                                   τ = RC = ρ 2 0 h2 d = ρ0 .
                                                             h
        iii. The AC circuit is effectively a capacitor in parallel with a resistor. The current through each is
             ninety degrees out of phase, so if the magnitudes of the current through the capacitor IC and the
             resistor IR are known, then                   q
                                                         I0 =      IC2 + IR
                                                                          2.

            Since the devices are in parallel, the potential drop across either are equal. We have

                                              IR = V0 /R,      IC = V0 /|ZC | = ωCV0 .

            Then                       r
                                           1              V0 p             V0 h2 p
                             I0 = V0         2
                                               + ω2 C 2 =     1 + (ωRC)2 =         1 + (ωρ0 )2 .
                                           R              R                 ρd

                         Copyright ©2008 American Association of Physics Teachers

2008 Semifinal Exam                           Part B                             16

   STOP: Do Not Continue to Part C
If there is still time remaining for Part B, you should review your work for
    Part B, but do not continue to Part C until instructed by your exam
                    supervisor. You may not return to Part A

                      Copyright ©2008 American Association of Physics Teachers

2008 Semifinal Exam                                   Part C                                                17

Optical Society of America Bonus Question
Researchers have a developed a lens made of liquid. The spherical lens consists of a droplet of transparent
liquid resting on a electrically controllable surface. When the voltage of the surface is changed, the droplet
itself changes shape; it either tries to “ball-up” more strongly or it becomes flatter. Figure 1 is a sketch of
the liquid lens and several parameters that describe it, including the thickness of the lens (t), the radius
of curvature of the top surface (R) and the contact angle (θ), which represents the angle between the flat
surface beneath the droplet and the tangent to the curved surface at the point of contact.

                            t
                                                                               θ

                                           R                       R

  a. When a certain voltage is applied, both the contact angle and lens thickness increase (and the lens
     becomes more curved). In this case, is the liquid attracted or repelled by the surface?
  b. Express the contact angle as a function of R and t.
   c. The total volume of the liquid lens is an important parameter because as the liquid lens changes shape,
      its volume is conserved. Calculate the volume of the lens as a function of R and t.
  d. Use your result to part (b) to eliminate the variable t from your expression for the volume and find
     V (R, θ).
   e. By changing the voltage on the control surface, the contact angle, θ, can be changed, which in turn
      changes the focal length of the lens, f . The lensmaker’s formula can be used to calculate the focal
      length and is given by
                                        1                        1    1
                                           = (nliquid − nair )      −      ,
                                        f                        R1   R2
     where nliquid and nair are the refractive indices of the liquid in the lens and air around it, and R1 and
     R2 are the radii of curvature of the two surfaces of the lens. In figure 1, R1 is the curved face and R2
     is the flat face. Use the lensmaker’s formula to calculate the focal length of the lens in terms of the
     total volume of the liquid, the contact angle, and the relevant refractive indices.
     Sidenote: liquid lenses are interesting because they are electrically controllable, variable focus lenses
     that can be very compact. People are working on putting them into cell phone cameras for ultra-
     compact zoom lenses. For more information on this type of liquid lens, see T. Krupenkin, S. Yang,
     and P. Mach, “Tunable liquid microlens,” Appl. Phys. Lett. 82, 316-318 (2003).

                                                 Solution
  a. Because the center of mass of the liquid rises, it is repelled by the surface.

  b. The angle θ is also the angle between the vertical and the radii drawn above, so cos θ = 1 − t/R.
   c. Let z be the vertical distance from the top of the surface. Then
                                    Z t          Z t
                               V =      πr2 dz =     π(2Rz − z 2 ) dz = π(Rt2 − t3 /3).
                                     0            0

                        Copyright ©2008 American Association of Physics Teachers

2008 Semifinal Exam                                Part C                                               18

  d. By rearranging, we have t/R = 1 − cos θ. Thus,

                   πR3                    πR3                            πR3
             V =       (t/R)2 (3 − t/R) =     (1 − cos θ)2 (2 + cos θ) =     (cos3 θ − 3 cos θ + 2).
                    3                      3                              3

  e. The radius of curvature of the flat face is R2 = ∞, and R1 = R, so
                                                                                1/3
                                   R                 1                       3V
                        f=                  =                                                       .
                             nliquid − nair   nliquid − nair       π(cos3 θ − 3 cos θ + 2)

     This matches equation (3) of the cited paper.

                      Copyright ©2008 American Association of Physics Teachers

USAPhO 2008 Question B2