[User]
Supposing that it takes five minutes to boil an egg. Estimate how long it takes to
    boil an ostrich egg, which has three times the diameter.
[2] Problem 27 (Cahn). A deep swimming pool is initially filled with water of depth D = 0.5 m with
    uniform temperature 0 ◦ C, during a cold winter where the air temperature is a constant T0 ≈ −10 ◦ C.
    The ice has thermal conductivity κ = 2.2 W/m K, and latent heat of fusion L = 3.4 × 105 J/kg. The
    density of ice is ρi = 0.9 × 103 kg/m3 , and the density of water is ρw = 1.0 × 103 kg/m3 . Assume
    for simplicity that the ice always forms a uniform layer over the water. Estimate how many days it
    takes the whole pool to freeze.

       Example 13

       Model a brick pizza oven as a sphere with inner radius R1 , outer radius R2 , and thermal
       conductivity κ. The inside of the oven is held at temperature T1 and the outside of the oven
       is held at temperature T2 . In equilibrium, find T (r), and the power needed to keep the inside
       of the oven at temperature T1 .

       Solution
       Consider a shell of radius r and thickness dr. The outward heat flow through it is
                                                               dT
                                           P (r) = −κ(4πr2 )      .
                                                               dr
       In addition, because the shells have constant temperature, there can be no net heat accumu-
       lating in any shell. Thus, P (r) must actually be independent of r. Renaming it to P , we can
       integrate the resulting differential equation to find
                                         Z T (r)        Z r
                                                               P
                                                 dT = −           2
                                                                    dr
                                           T1             R1 4πκr

       which gives the solution
                                                            r − R1 R2
                                  T (r) = T1 + (T2 − T1 )                .
                                                               r R2 − R1
       Now that we know T (r), we can evaluate dT /dr at any radius to find the total power, using
       the first equation. Another slicker way, which doesn’t even require knowing T (r), is to
       integrate the first equation all the way from the inside of the oven to the outside,
                                         Z T2        Z R2
                                                             P
                                              dT = −            2
                                                                  dr.
                                          T1          R1 4πκr

       This gives an expression for the power in terms of the other parameters, yielding
                                                             R1 R2
                                        P = 4πκ(T1 − T2 )           .
                                                            R2 − R1

Solution:
The thermal gradient dT /dr is of the order ∆T /R ∝ 1/R, where R is the radius of
    the object. The heat flux into the egg is proportional to A dT /dr ∝ R2 (1/R) = R. But the heat
    capacity of the egg scales as R3 . Thus, the time scales as R3 /R = R2 , and hence it takes about 45
    minutes to boil an ostrich egg. (This can also be shown by dimensional analysis.)

[2] Problem 27 (Cahn). A deep swimming pool is initially filled with water of depth D = 0.5 m with
    uniform temperature 0 ◦ C, during a cold winter where the air temperature is a constant T0 ≈ −10 ◦ C.
    The ice has thermal conductivity κ = 2.2 W/m K, and latent heat of fusion L = 3.4 × 105 J/kg. The
    density of ice is ρi = 0.9 × 103 kg/m3 , and the density of water is ρw = 1.0 × 103 kg/m3 . Assume
    for simplicity that the ice always forms a uniform layer over the water. Estimate how many days it
    takes the whole pool to freeze.
    Solution. Let the ice layer on the top of the lake have thickness x. The temperature difference
    across the ice layer is a constant ∆T = 10◦ C, so the rate of heat flow through the ice is
                                                κA∆T        dx
                                          P =        = Lρi A .
                                                  x         dt
    Separating and integrating gives
                                        Z Dρw /ρi            Z t
                                                                κ∆T
                                                    x dx =          dt
                                         0                    0 Lρi

    where the final height is Dρw /ρi ≈ 1.1 D because water expands
                                                              √       as it freezes. Notice that this
    implies the growth of the ice slows down over time, x ∝ t. This makes sense, since as the ice
    grows thicker, the temperature gradient gets smaller, slowing the heat transfer. By carrying out
    the integral, we find
                                            D2 Lρi ρ2w
                                        t=             = 25 days.
                                            2κ∆T ρ2i

       Example 13

       Model a brick pizza oven as a sphere with inner radius R1 , outer radius R2 , and thermal
       conductivity κ. The inside of the oven is held at temperature T1 and the outside of the oven
       is held at temperature T2 . In equilibrium, find T (r), and the power needed to keep the inside
       of the oven at temperature T1 .

       Solution
       Consider a shell of radius r and thickness dr. The outward heat flow through it is
                                                                   dT
                                             P (r) = −κ(4πr2 )        .
                                                                   dr
       In addition, because the shells have constant temperature, there can be no net heat accumu-
       lating in any shell. Thus, P (r) must actually be independent of r. Renaming it to P , we can
       integrate the resulting differential equation to find
                                         Z T (r)        Z r
                                                               P
                                                 dT = −           2
                                                                    dr
                                           T1             R1 4πκr

       which gives the solution
                                                             r − R1 R2
                                  T (r) = T1 + (T2 − T1 )                 .
                                                                r R2 − R1
       Now that we know T (r), we can evaluate dT /dr at any radius to find the total power, using
       the first equation. Another slicker way, which doesn’t even require knowing T (r), is to

       integrate the first equation all the way from the inside of the oven to the outside,
                                        Z T2            Z R2
                                                                 P
                                               dT = −                dr.
                                         T1              R1    4πκr2

       This gives an expression for the power in terms of the other parameters, yielding
                                                                R1 R2
                                        P = 4πκ(T1 − T2 )              .
                                                               R2 − R1


[User]
One wants to pump water out from a well with a depth $h=\SI{5.0}{m}$. The pump is on the surface and its water intake pipe (filled with water) has an inner diameter $d=\SI{16}{mm}$ and a length that is equal to the depth of the well.\\
a) How big has to be the power $P$ of the pump to pump water with a flow rate $q=\SI{30}{l/min}$? The efficiency of the pump is $\eta=\SI{25}{\percent}$.\\
b) What is the maximal depth $h_{m}$ of the well for which it is possible to pump water from the well with this type of pump?\\
Assume that besides other forces frictional force is also applied to the water column inside the water, which causes a decrease in the pressure per length $l$ of the pipe $\Delta p=c_{h}q^{2}l/d^{5}$ where $c_{h}=\SI{40}{Pa\cdot s^{2}/m^{2}}$. The density of water $\rho=\SI{1000}{kg/m^{3}}$, gravitational acceleration $g=\SI{9.8}{m/s^{2}}$ and the air pressure $p_{0}=\SI{100}{kPa}$.

Solution:
\osa Torus olevale veesambale mõjub raskusjõud $F_{r}$, takistusjõud $F_{h}$
ja pumba poolt avaldatav jõud $F_{p}$. Ühtlase pumpamise korral
kehtib jõudude tasakaal $F_{r}+F_{h}=F_{p}$. Raskusjõud on
arvutatav leides torus oleva vee massi
\[
F_{r}=mg=\rho Shg\approx\SI{9.85}{N},
\]
kus toru ristlõikepindala $S=\pi d^{2}/4.$ Hõõrdejõu, mis
on tingitud vee liikumisest torus, leiame ülesande tekstis antud rõhulangu
valemiga, kui korrutame selle toru ristlõikepindalaga.
\[
F_{h}=\Delta pS=c_{h}Q^{2}hS/d^{5}\approx\SI{9.59}{N}.
\]
Pumba võimsus on antud valemiga $P=F_{p}v/\eta,$ kus $v=Q/S$
on vee liikumise kiirus torus. Jõudude tasakaalust saame
\[
P=\left(F_{r}+F_{h}\right)\frac{Q}{S\eta}=\frac{Qh}{d^{5}\eta}\left(\rho gd^{5}+c_{h}Q^{2}\right)\approx\SI{193}{W}.
\]
\osa Kuna pump asub maapinnal, siis peab pump tekitama vee liigutamiseks
alarõhu. Maksimaalne alarõhk on juhul, kui pump tekitab vaakumi.
Sellisel piirjuhul surub õhurõhk veesammast ülespoole jõuga $p_{0}S$
ja pumba töötamiseks peab see jõud olema vähemalt sama suur kui veesambale
mõjuv raskusjõud $p_{0}S=\rho Sh_{m}g.$ Maksimaalne kaevu sügavus
on seega $h_{m}=p_{0}/\left(\rho g\right)\approx\SI{10.2}{m}$.

Pump

Kaevust sügavusega $h=\SI{5.0}{m}$ tahetakse pumbata vett. Pump asub
maapinnal ning selle veevõtutoru (täidetud veega) on siseläbimõõduga
$d=\SI{16}{mm}$ ja pikkusega, mis on võrdne kaevu sügavusega.\\
\osa Kui suur peab olema pumba võimsus $P$, et pumbata vett vooluhulgaga
$q=\SI{30}{l/min}$? Pumba kasutegur on $\eta=\SI{25}{\percent}$.\\
\osa Missugune on maksimaalne kaevu sügavus $h_{m}$, mille korral on
võimalik sellist tüüpi pumbaga kaevust vett pumbata?

Arvestada, et torus olevale veesambale mõjub lisaks teistele jõududele
ka hõõrdejõud, mis põhjustab rõhu vähenemist toru pikkuse $l$ kohta $\Delta p=c_{h}q^{2}l/d^{5},$
kus $c_{h}=\SI{40}{Pa\cdot s^{2}/m^{2}}.$ Vee tihedus $\rho=\SI{1000}{kg/m^{3}}$,
raskuskiirendus $g=\SI{9.8}{m/s^{2}}$ ja õhurõhk $p_{0}=\SI{100}{kPa}.$

\osa Torus olevale veesambale mõjub raskusjõud, takistusjõud ning pumba poolt avaldatud jõud. Ühtlase pumpamise korral kehtib jõudude tasakaal.\\
\osa Pump paneb vee alarõhku tekitades liikuma. Kuna pump asub maapinnal, tekib minimaalne alarõhk juhul, kui pump tekitab vaakumi.


[User]
A rocket with a full fuel tank has a mass M and is initially stationary. The fuel is
    ejected at a rate σ, where σ has units of kg/s, at a relative velocity of u.

      (a) If the rocket begins in space, show that the velocity of the rocket when its total mass is M ′ is

                                                                M
                                                    v = u log      .
                                                                M′
           This is the Tsiolkovsky rocket equation.

      (b) Repeat part (a) for a rocket in a uniform gravitational field g. Do you get the best final
          velocity if σ is high or low? (Ignore gravity for the rest of this problem.)

        (c) In a multi-stage rocket, an empty fuel tank detaches from the rocket once it is used up, after
            which a second engine starts up. Explain why this can achieve a much higher final velocity

         than just firing both engines at once. (If you want a quantitative treatment of this, you can
         see INPhO 2016, problem 3.)

     (d) It is desired for a rocket to begin at zero speed and accelerate to speed v, to deliver a given
         payload. If the exhaust comes out with a relative velocity of u, how should u be chosen to
         minimize the fuel energy that must be spent to perform this maneuver? (Hint: let the final
         mass of the rocket be fixed, since that’s the mass of the payload we want to transport. You
         will have to solve an equation numerically.)

     (e) If u has this value, what fraction η of the spent fuel’s energy ends up in the rocket’s final
         kinetic energy?

      (f) Now suppose u can be freely varied over time. Qualitatively, how should it be chosen to

                   01m
          maximize η, and what is the maximum possible value of η?

Solution:
(a) Let p = mv be the momentum of the rocket and all the fuel instantaneously inside
         it. As some fuel of mass dm is ejected from the rocket, the total momentum is conserved, so

                                                 dp = (v − u) dm.

         On the other hand, we also have

                                                dp = m dv + v dm.

     Combining these equations gives
                                               m dv = −u dm
     so integrating gives
                                                      v
                                            log m = −   + C.
                                                      u
     Fixing C with the initial condition gives the desired result.

 (b) The reasoning is similar except that there is now an additional term representing the change
     in momentum due to the gravitational force. We have dp + (−dm)(v − u) = −mg dt, so
     m dv = −u dm − mg dt. Therefore,
                                           dm    dv g
                                              = − − dt,
                                           m     u  u
     so log m = −v/u − gt/u + log(M ). Solving for v gives
                                                  M     g
                                      v = u log      − (M − M ′ ).
                                                 M′ σ
     It’s better if σ is high, since you are constantly losing momentum to gravity.
                       M −M0      M
 (c) The idea is that M ′ −M

                               > M  ′ where M0 is the mass of the ejected tank, so the change in

     speed is higher. Basically, the empty fuel tank is now dead weight, so ejecting it means you
     don’t waste energy speeding it up.

 (d) Let the initial and final masses be M and M ′ . In order for the rocket to reach a velocity of v,
               M
     v = u log M               ′ v/u .
                 ′ , or M = M e

     Now, the energy released by burning a small mass dm of fuel is precisely (dm)u2 /2. One way
     to see this is to work in the frame instantaneously moving with the rocket; then the only final
     energy is in the kinetic energy (dm)u2 /2 of the ejected fuel itself, since the rocket picks up
     negligible speed. This energy must have come from the internal energy of the burning of the
     fuel, and this quantity is the same in all frames, as we’ve discussed in M3.
     Therefore, the total fuel energy burnt is, in any frame,
                                    1              1
                                 E = (M − M ′ )u2 = M ′ (ev/u − 1)u2
                                    2              2
     This is minimized when dE/du = 0 (treating M ′ as fixed), which gives
                                          2u(ev/u − 1) = vev/u .
     Letting x = v/u, we need to find the solution to
                                             x = 2(1 − e−x ).
     This can be done by plugging in 2(1 − e−Ans ) repeatedly in our calculator to get x = 1.5936.
     Thus the value of u should satisfy
                                                     v
                                            u≈
                                                  1.5936
 (e) At the end, the rocket will have a kinetic energy of 12 M ′ v 2 and the total fuel burnt will be
     1   ′ x      2   2
     2 M (e − 1)v /x , and divide the former by the latter to get

                                                  x2
                                          η=            = 0.6476
                                               (ex − 1)

      (f) We should always set u equal to the velocity of the rocket at that moment. Then when the
          fuel comes out, it’s at a dead stop, so all of the kinetic energy burned goes into the rocket.
          Thus the maximum value of η is 100%. This is called a “perfect rocket”, though it’s not the
          kind of thing one would want to use in practice. It’s not trivial to change u arbitrarily, from

                   01m
          an engineering point of view, and a perfect rocket at low speeds would have low power.

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


[User]
Graafikul on antud pulgakujulise keraamilisest pooljuhist (nn. PTC takisti) soojendi materjali erijuhtivuse $\sigma$ (\si{1/(\ohm. m)}) sõltuvus temperatuurist $t$(\si{\degreeCelsius}). Erijuhtivuseks nimetatakse eritakistuse pöördväärtust. Leida, millise temperatuurini kuumeneb avatud ruumis paiknev sellest materjalist soojendi, kui tema otstele rakendatakse pinge $U_1 = \SI{60}{V}$. Milliseks kujuneb soojendi temperatuur, kui otstele rakendatud pinge on $U_2 = \SI{36}{V}$? On teada, et kui soojendi otstele rakendatakse pinge $U_0 = \SI{30}{V}$, siis soojendi temperatuuriks kujuneb $t_0= \SI{70}{\degreeCelsius}$. Välisõhu temperatuur on $t_v = \SI{20}{\degreeCelsius}$.

\begin{center}
	\includegraphics[width=\linewidth]{2008-lahg-10-yl}
\end{center}

Solution:
Kütteelemendi takistus on
\[
R=\frac{\rho L}{S}=\frac{L}{\sigma S},
\]
kus $\rho$ on eritakistus, $L$ elemendi pikkus, $S$ elemendi ristlõikepindala ja $\sigma$ erijuhtivus. Elemendil eralduv võimsus on
\[
N=\frac{U^{2}}{R}=\frac{S}{L} U^{2} \sigma.
\]
Võimsus, millega keha annab soojust ümbritsevale keskkonnale, on võrdeline temperatuuride vahega ja tasakaalulise temperatuuri puhul peab võrduma kehal eralduva võimsusega. Seega
\[
N=k \Delta t=\frac{S}{L} U^{2} \sigma.
\]
Siit näeme, et
\begin{equation} \label{2008-lahg-10:eq1}
\frac{\sigma}{\Delta t} \cdot U^{2}=k \frac{L}{S}=\const,
\end{equation}
kus aga $\frac{\sigma}{\Delta t}$
on sirge tõus erijuhtivuse graafikul, kus algpunktiks on $t_v = \SI{20}{\degreeCelsius}$ ja
$\sigma = \num{0}$.

\begin{center}
	\includegraphics[width=\linewidth]{2008-lahg-10-lah}
\end{center}

Algtingimusi silmas pidades on pinge $U_0 = \SI{30}{V}$ juures see tõus $\left(\frac{\sigma}{\Delta t}\right)_{30}=\frac{7}{5}$ ning arvestades valemit (\ref{2008-lahg-10:eq1}) saame
\[
\left(\frac{\sigma}{\Delta t}\right)_{30} \cdot U_{0}^{2}=\left(\frac{\sigma}{\Delta t}\right)_{60} \cdot U_{1}^{2}=\left(\frac{\sigma}{\Delta t}\right)_{36} \cdot U_{2}^{2}.
\]
Siit
\[
\begin{aligned}\left(\frac{\sigma}{\Delta t}\right)_{60} &=\frac{U_{0}^{2}}{U_{1}^{2}}\left(\frac{\sigma}{\Delta t}\right)_{30}=\frac{7}{20}, \\\left(\frac{\sigma}{\Delta t}\right)_{36} &=\frac{U_{0}^{2}}{U_{2}^{2}}\left(\frac{\sigma}{\Delta t}\right)_{30}=\frac{35}{36}. \end{aligned}
\]
Kanname vastavad sirged (alguspunktiga $t = \SI{20}{\degreeCelsius}$ ja $\sigma = 0$) graafikule ja loeme vastavad tasakaalulised temperatuurid.

Saame $U_1 = \SI{60}{V}$ puhul $t_1 = \SI{277}{\degreeCelsius}$.

Tekib huvitav nähtus, et tänu mittelineaarsusele on $U_2 = \SI{36}{V}$ puhul pinge tõstmisel ja langetamisel tasakaalulised temperatuurid erinevad, vastavalt pinge tõusul $t_2 = \SI{112}{\degreeCelsius}$ ja langetamisel $t_2' = \SI{224}{\degreeCelsius}$.

Paneme tähele, \SI{36}{V} puhul on graafikult loetavaid lahendeid justkui kolm, aga keskmine lahend \SI{207}{\degreeCelsius} on ebastabiilne. Nimelt olukorras, kus küttekeha on sellel temperatuuril, viib väikenegi küttekeha temperatuuri tõus võimsuse suurenemiseni ja langus vastavalt võimsuse vähenemiseni, seega temperatuur kas kasvab temperatuurini \SI{224}{\degreeCelsius} või langeb temperatuurini \SI{112}{\degreeCelsius}.

Pooljuht

Graafikul on antud pulgakujulise keraamilisest pooljuhist (nn. PTC takisti) soojendi materjali erijuhtivuse $\sigma$ (\si{1/(\ohm. m)}) sõltuvus temperatuurist $t$(\si{\degreeCelsius}). Erijuhtivuseks nimetatakse eritakistuse pöördväärtust. Leida, millise temperatuurini kuumeneb avatud ruumis paiknev sellest materjalist soojendi, kui tema otstele rakendatakse pinge $U_1 = \SI{60}{V}$. Milliseks kujuneb soojendi temperatuur, kui otstele rakendatud pinge on $U_2 = \SI{36}{V}$? On teada, et kui soojendi otstele rakendatakse pinge $U_0 = \SI{30}{V}$, siis soojendi temperatuuriks kujuneb $t_0= \SI{70}{\degreeCelsius}$. Välisõhu temperatuur on $t_v = \SI{20}{\degreeCelsius}$.

\begin{center}
	\includegraphics[width=\linewidth]{2008-lahg-10-yl}
\end{center}

Kütteelement on soojuslikus tasakaalus ümbritseva keskkonnaga ning kaod keskkonda on võrdelised sise- ja välistemperatuuride vahega. Soojusliku tasakaalu temperatuur on määratav nn graafilise meetodiga, mis seisneb $\sigma (t)$ ja teatud funktsiooni pingest lõikepunkti leidmises.


[User]
A projectile of mass m is dropped from a height h above the ground. It falls and
    bounces elastically, experiencing the same quadratic drag as in problem 2. Find the maximum
    height to which it subsequently rises. (Hint: don’t try to use your results from problem 2.)

       Remark
       How does the top speed v of a rowboat depend on the number N of rowers? A light, fast-
       moving rowboat experiences quadratic friction, so that the drag force on it is proportional
       to v 2 A, where A is the submerged cross-sectional area of the boat. A boat designed for N
       rowers will have a submerged volume V ∝ N , and a streamlined shape so that A ∝ V 2/3 .
       Thus, the required power input is

                                               P = F v ∝ v 3 N 2/3 .

       The power output by the rowers scale as N , and combining these results gives the amazingly
       weak dependence v ∝ N 1/9 , which agrees decently with Olympic rowing times.

       Idea 1
       An ordinary differential equation is any equation involving a quantity x(t) and its derivatives.
       In introductory physics, we are usually concerned with a few very simple differential equations,
       with the following nice properties.

         • The differential equation is at most second order, meaning it can contain x, its first
           derivative ẋ = v, and its second derivative ẍ = a, but no higher derivatives. This implies
           the solution can be determined by an initial position and initial velocity. (We’ll focus on
           second order differential equations for the rest of this section; most first order differential
           equations can simply be solved by separation and integration, as we’ve seen above.)

         • The differential equation is linear, meaning that terms don’t contain products of x, ẋ,
           and ẍ. For example, a damped driven harmonic oscillator with time-dependent drag,

                                             mẍ = −b(t)ẋ − kx + f (t)

           is a second order linear differential equation. Solutions to such differential equations obey
           the superposition principle: if x1 (t) and x2 (t) are both solutions, so is c1 x1 (t) + c2 x2 (t).

         • The differential equation is homogeneous, meaning that each term is proportional to ex-
           actly one power of x or its derivatives. The above differential equation is not homogeneous,
           but it would be if we removed the driving f (t).

    • The differential equation is time-translation invariant, meaning that no functions of time
      appear except for x and its derivatives. The above equation isn’t, but it would be if we
      set f (t) and b(t) to constants.

  Idea 2
  Linear, homogeneous, time-translation invariant differential equations are very special, and
  they can all be solved by the exact same method. First, note that we can promote x(t) to
  a complex variable x̃(t) and solve the differential equation over the complex numbers. As
  long as we have a complex solution, we can recover a real solution by taking the real part.
  Now, the method of solution, which works for almost all equations of this form, is to guess a
  complex exponential solution
                                          x̃(t) = eiωt .
  Plugging this into the differential equation will yield the allowed values of ω, and the general
  solution can be found by superposing the complex exponentials.

  Example 2

  Solve the simple harmonic oscillator, mẍ + kx = 0, using the above principles.

  Solution
  First, we pass to a complex differential equation,

                                           ¨ + kx̃ = 0.
                                          mx̃

  We guess x̃(t) = eiωt . Plugging this in and using the chain rule gives

                                      m(iω)2 eiωt + keiωt = 0

  and canceling eiωt and solving gives two solutions,
                                                        p
                                    ω = ±ω0 ,    ω0 =    k/m.

  Since this a second-order linear differential equation, the general solution is given by the
  superposition of these two complex exponentials,

                                     x̃(t) = Aeiω0 t + Be−iω0 t

  where A and B are general complex numbers. The real part of x̃(t) satisfies the original real
  differential equation ma + kx = 0, and is

                                Re x(t) = C cos(ω0 t) + D sin(ω0 t)

  where C and D are real numbers.

       Remark
       What’s the meaning of promoting the real variable x(t) to a complex variable x̃(t)? You can
       regard it as just a mathematical trick, which makes the solutions look a bit simpler. But
       for the specific case of the simple harmonic oscillator, there’s a more satisfying explanation:
       x̃(t) encodes the position of the particle in its real part, and the momentum of the particle
       (up to a scaling constant) in its imaginary part. As the system evolves, position rotates into
       momentum and vice versa. Energy conservation corresponds to the fact that the complex
       magnitude |x̃|2 stays constant. We’ll see a bit more of this perspective in M4.

Solution:
The reason you shouldn’t try to use the results from problem 2 is that they are in terms
of time. Given how complicated the implicit expressions for v(t) are, the expressions for x(t) would
be extremely clunky. And they’re not necessary, because in this problem we don’t care about the
time-dependence at all; we just want to know the final height.
    Another way to say this is that we aren’t interested in v(t), we’re interested in v(x). While
the projectile is moving downward, we can integrate dv/dx to find the speed v0 at the moment it
hits the ground. Then, when it’s moving upward, we integrate dv/dx until it has zero speed again,
which is its final height. This will be a lot simpler than integrating dv/dt.
    For the upward and downward trajectories, Newton’s second law says
                                             dv
                                                = −g ± αv 2
                                             dt
and multiplying both sides by dt/dx gives

                                            dv    g
                                               = − ± αv.
                                            dx    v
Separating and integrating, on the way down we have
                                   Z −v0
                                                    1 −v0 v dv
                         Z 0                         Z
                                            dv
                             dx =                 =                .
                           h        0    αv − g/v   α 0  v 2 − g/α

Carrying out the integral and simplifying,

                                      h=−        log(1 − αv02 /g).
                                              2α
Now, on the way up, we have
                           Z h′       Z 0                     Z v0
                                             dv       1                v dv
                                  dx =              =
                            0          v0 −g/v − αv   α        0     v 2 + g/α

and carrying out the integral gives

                                      h′ =       log(1 + αv02 /g).
                                              2α
Combining the two equations gives

                                       h′ =      log(2 − e−2αh )
                                              2α
which you can check has the right limits. Also note that g drops out, as required by dimensional
analysis.

   Remark
   How does the top speed v of a rowboat depend on the number N of rowers? A light, fast-
   moving rowboat experiences quadratic friction, so that the drag force on it is proportional
   to v 2 A, where A is the submerged cross-sectional area of the boat. A boat designed for N
   rowers will have a submerged volume V ∝ N , and a streamlined shape so that A ∝ V 2/3 .

  Thus, the required power input is

                                          P = F v ∝ v 3 N 2/3 .

  The power output by the rowers scale as N , and combining these results gives the amazingly
  weak dependence v ∝ N 1/9 , which agrees decently with Olympic rowing times.

  Idea 1
  An ordinary differential equation is any equation involving a quantity x(t) and its derivatives.
  In introductory physics, we are usually concerned with a few very simple differential equations,
  with the following nice properties.

    • The differential equation is at most second order, meaning it can contain x, its first
      derivative ẋ = v, and its second derivative ẍ = a, but no higher derivatives. This implies
      the solution can be determined by an initial position and initial velocity. (We’ll focus on
      second order differential equations for the rest of this section; most first order differential
      equations can simply be solved by separation and integration, as we’ve seen above.)

    • The differential equation is linear, meaning that terms don’t contain products of x, ẋ,
      and ẍ. For example, a damped driven harmonic oscillator with time-dependent drag,

                                        mẍ = −b(t)ẋ − kx + f (t)

      is a second order linear differential equation. Solutions to such differential equations obey
      the superposition principle: if x1 (t) and x2 (t) are both solutions, so is c1 x1 (t) + c2 x2 (t).

    • The differential equation is homogeneous, meaning that each term is proportional to ex-
      actly one power of x or its derivatives. The above differential equation is not homogeneous,
      but it would be if we removed the driving f (t).

    • The differential equation is time-translation invariant, meaning that no functions of time
      appear except for x and its derivatives. The above equation isn’t, but it would be if we
      set f (t) and b(t) to constants.

  Idea 2
  Linear, homogeneous, time-translation invariant differential equations are very special, and
  they can all be solved by the exact same method. First, note that we can promote x(t) to
  a complex variable x̃(t) and solve the differential equation over the complex numbers. As
  long as we have a complex solution, we can recover a real solution by taking the real part.
  Now, the method of solution, which works for almost all equations of this form, is to guess a
  complex exponential solution
                                          x̃(t) = eiωt .
  Plugging this into the differential equation will yield the allowed values of ω, and the general
  solution can be found by superposing the complex exponentials.

       Example 2

       Solve the simple harmonic oscillator, mẍ + kx = 0, using the above principles.

       Solution
       First, we pass to a complex differential equation,

                                                 ¨ + kx̃ = 0.
                                                mx̃

       We guess x̃(t) = eiωt . Plugging this in and using the chain rule gives

                                            m(iω)2 eiωt + keiωt = 0

       and canceling eiωt and solving gives two solutions,
                                                              p
                                          ω = ±ω0 ,    ω0 =    k/m.

       Since this a second-order linear differential equation, the general solution is given by the
       superposition of these two complex exponentials,

                                           x̃(t) = Aeiω0 t + Be−iω0 t

       where A and B are general complex numbers. The real part of x̃(t) satisfies the original real
       differential equation ma + kx = 0, and is

                                      Re x(t) = C cos(ω0 t) + D sin(ω0 t)

       where C and D are real numbers.

       Remark
      What’s the meaning of promoting the real variable x(t) to a complex variable x̃(t)? You can
      regard it as just a mathematical trick, which makes the solutions look a bit simpler. But
      for the specific case of the simple harmonic oscillator, there’s a more satisfying explanation:
      x̃(t) encodes the position of the particle in its real part, and the momentum of the particle
      (up to a scaling constant) in its imaginary part. As the system evolves, position rotates into
      momentum and vice versa. Energy conservation corresponds to the fact that the complex
      magnitude |x̃|2 stays constant. We’ll see a bit more of this perspective in M4.


[User]
Question A3
This problem inspired by the 2008 Guangdong Province Physics Olympiad
     Two infinitely long concentric hollow cylinders have radii a and 4a. Both cylinders are insulators;
the inner cylinder has a uniformly distributed charge per length of +λ; the outer cylinder has a
uniformly distributed charge per length of −λ.
     An infinitely long dielectric cylinder with permittivity  = κ0 , where κ is the dielectric constant,
has a inner radius 2a and outer radius 3a is also concentric with the insulating cylinders. The
dielectric cylinder is rotating about its axis with an angular velocity ω  c/a, where c is the speed
of light. Assume that the permeability of the dielectric cylinder and the space between the cylinders
is that of free space, µ0 .

   a. Determine the electric field for all regions.
  b. Determine the magnetic field for all regions.

Solution:
Question A3
This problem inspired by the 2008 Guangdong Province Physics Olympiad
     Two infinitely long concentric hollow cylinders have radii a and 4a. Both cylinders are insulators;
the inner cylinder has a uniformly distributed charge per length of +λ; the outer cylinder has a
uniformly distributed charge per length of −λ.
     An infinitely long dielectric cylinder with permittivity  = κ0 , where κ is the dielectric constant,
has a inner radius 2a and outer radius 3a is also concentric with the insulating cylinders. The
dielectric cylinder is rotating about its axis with an angular velocity ω  c/a, where c is the speed
of light. Assume that the permeability of the dielectric cylinder and the space between the cylinders
is that of free space, µ0 .

   a. Determine the electric field for all regions.
  b. Determine the magnetic field for all regions.

                                              Solution
   a. Consider a Gaussian cylinder of radius r and length l centered on the cylinder axis. The
      electric field is radial, so Gauss’s Law states that
                                     I
                                                 qin                  λin l
                                       E · dA =          ⇒ 2πrEl =
                                                  0                   0
      where λin is the linear charge density enclosed in the cylinder, so
                                                         λin
                                                 E=          r̂.
                                                       2πr0
      The field due to the hollow cylinders alone is therefore
                                                           
                                                  λ        0 r < a
                                                           
                                     Eapplied =       r̂ × 1 a < r < 4a
                                                2πr0      
                                                             0 r > 4a.
                                                           

      However, the field within the dielectric is reduced by a factor κ, so that in total
                                                    
                                                    
                                                    0     r<a
                                                    
                                                    1     a < r < 2a
                                                    
                                                    
                                                    
                                           λ
                                   E=          r̂ × 1/κ 2a < r < 3a
                                        2πr0       
                                                     1     3 < r < 4a
                                                    
                                                    
                                                    
                                                    
                                                    
                                                     0     r > 4a.
                                                    

                    Copyright c 2012 American Association of Physics Teachers

2012 Semifinal Exam                           Part A                                               8

  b. We can apply the results of the previous section to obtain the enclosed charge density λin as
     a function of radius,                  
                                            
                                             0     r<a
                                            
                                            λ      a < r < 2a
                                            
                                            
                                            
                                      λin = λ/κ 2a < r < 3a
                                            
                                              λ     3 < r < 4a
                                            
                                            
                                            
                                            
                                            
                                              0     r > 4a.
                                            

     Defining
                                                          1
                                            λi =       1−         λ
                                                          κ
     we conclude that a charge density −λi exists on the inner surface of the dielectric, a charge
     density λi exists on the outer surface, and there is no charge on the interior.
     As with the case of a very long solenoid, we expect the magnetic field to be entirely parallel
     to the cylinder axis ẑ, and to go to zero for large r. Consider an Amperian loop of length l
     extending along a radius, the inner side of which is at radius r and the outer side of which is
     at a very large radius. We have on this loop
                                             I
                                               B · dl = µ0 Iin .

     Letting B be the magnitude of the magnetic field at radius r, we find
                                                       µ0 Iin
                                               B=             .
                                                         l
     For r > 3a, Iin = 0, since the charge on the hollow cylinders is not moving. For 2a < r < 3a,
     the loop now encloses the outer surface of the dielectric. In time 2π  ω a charge λi l passes
     through the loop, so the current due to the outer surface is
                                                         λi lω
                                               Iout =
                                                          2π
     and thus this is Iin for 2a < r < 3a. For r < 2a, the loop now encloses both surfaces of the
     dielectric; the inner surface contributes a current that exactly cancels the outer one, so again
     Iin = 0. Putting this together,
                                                     
                                          µ0 ωλi     0 r < 2a
                                                     
                                     B=          ẑ × 1 2a < r < 3a
                                            2π       
                                                      0 r > 3a.
                                                     

                  Copyright c 2012 American Association of Physics Teachers

2012 Semifinal Exam                           Part A                                               9

USAPhO 2012 Question A3


[User] Water of density $\rho_w$ has been poured inside a U-tube with an even cross-sectional area S so that over half of the U-tube is filled with the water and the length of each of the unfilled part is $h$. One end of the U-tube is closed hermetically and into the other end oil is slowly poured up to the edge. How big was the density $\rho_{o}$ of the oil if it is known that the height of the added oil column was $l$? The atmosphere pressure is $p_0$.

Solution:
Pärast õli kallamist torusse langeb vee tase $\Delta h$ võrra selles torus, kuhu kallati õli, ning teises torus veetase tõuseb $\Delta h$ võrra. Õli ja vee piirpinna kõrgusel on rõhk mõlemas U-toru harus sama. Õliga täidetud harus avaldab õli vee ja õli piirpinnale rõhku
\[ p_1 = \rho_{\text{õ}}gl + p_0.\]

Teises (suletud) U-toru harus tekitab kokkusurutud õhk rõhu $p_{\text{õhk}}$. Õhu kokkusurumist võime vaadelda isotermilise protsessina, kus $pV = \const$. Seega
\[ p_0\cdot Sh = p_{\text{õhk}}\cdot S(h-\Delta h) \quad\Rightarrow\quad p_{\text{õhk}} = \frac{p_0h}{h-\Delta h}.\]

Vee ja õli nivoo kõrgusel avaldavad teises harus vesi ja õhk rõhku
\[ p_2 = \rho_vg(2\Delta h) + p_{\text{õhk}}.\]

Rõhud $p_1$ ja $p_2$ on võrdsed ning $\Delta h = l - h$, seega saame kirja panna seose
\[ \rho_{\text{õ}}gl + p_0 = \rho_vg2(l-h) + \frac{p_0h}{h-(l-h)}.\]

Avaldades viimasest seosest õli tiheduse $\rho_{\text{õ}}$, saame
\[ \rho_{\text{õ}} = \frac{l-h}{l}\left(2\rho_v+\frac{p_0}{g(2h-l)}\right).\]

U-toru

U-torusse ühtlase ristlõikepindalaga S on valatud vesi tihedusega $\rho_v$, nii et üle poole U-torust on veega täidetud ja kummagi täitmata osa pikkus on $h$. U-toru üks ots suletakse hermeetiliselt ning teise torusse valatakse aeglaselt õli kuni U-toru ülemise servani. Kui suur oli õli tihedus $\rho_{\text{õ}}$, kui on teada, et lisatud õlisamba kõrgus oli $l$? Atmosfäärirõhk on $p_0$.

Õli valamise tulemusena langeb vee tase $l - h$ võrra selles torus, kuhu õli kallati, ning tõuseb sama taseme võrra teises torus. Kuna süsteem on tasakaalus, peavad mõlemad vedelikusambad U-toru alumises punktis sama rõhku avaldama.


[User]
7. ZENER (7 points)

— Jaan Kalda. A Zener
diode is connected to a source of alternating cur-
rent as shown in the figure. The current is sinus-
oidal I = I0 cosωt with a constant amplitude.
The inductance L of the inductor is such that
LωI0 ≫ V1,V2, where V1 and V2 are the break-
down voltages (V1 > V2). The current-voltage
characteristic oftheZener diodeis shownin the
figure. In the following assume that a very long
timehaspassedsincethecurrentsourcewasfirst
turned on.
i) (5points)Find the average current 〈I〉 through
the inductor.
ii) (2points)Find the peak-to-peak amplitude of
the current changes ∆I in the inductor.
L
I0cos
I
V
V2
V1
t - ω

Solution:
7. ZENER

i) The diode limits the maximum amount
of voltage that can be applied to the inductor.
Whenever the current from the current source
is higher than the current through the inductor,
the voltage V1 is applied to the inductor, and
when the current current source is lower than
the current through the inductor a voltage −V2
is applied to the inductor. If the largest voltage
V1 would be applied to the inductor for one
period T = 2π
ω
, then the change in the current
through the inductor is
δI =
V1
L
T =
2πV1
Lω
.
Because we had the relation LωI0 ≫V1, we get
that δI ≪ I0. The current through the inductor
changes negligibly little compared to I0 dur-
ing one period and can be considered constant.
The equilibrium current can be found by finding
when the average voltage is zero. During one
period let T1 be the time when 〈I〉 > I0 cosωt,
and let T2 be the time when 〈I〉< I0 cosωt. We
get
T1 =
2
ω
arccos
(
〈I〉
I0
)
.
The average voltage is zero when V1T1 = V2T2
and by using T1 +T2 = T we have
V1T1 =V2(T −T1) =⇒ T1 =
V2
V1 +V2
T =⇒
2
ω
arccos
(
〈I〉
I0
)
=
V2
(V1 +V2)
2π
ω
=⇒
〈I〉= I0 cos
(
πV2
V1 +V2
)
.
ii) The peak-to-peak amplitude of the current
fluctuations can be calculated from dI
dt = U
L
∆I =
V2T2
L
=
V1T1
L
=
2πV1V2
ωL(V1 +V2)
.

ZENER


[User] A Ski jumping hill’s track is situated on a slope with an angle of $\alpha$. The track’s take-off platform is horizontal. The ski jumper began the in-run at a height of $h$ relative to the take-off platform, but right at the start, his wallet fell out of his pocket. How far from the take-off will the wallet land (along the horizontal axis) if the wallet is moving without friction?

Solution:
Hoovõturaja alumise otsani jõudes on rahakott omandanud kineetilise energia $E=\frac{mv^2}{2}=mgh$, kusjuures algkiirus $v=\sqrt{2gh}$ on horisontaalne. Kui valime koordinaatide alguspunktiks hoovõturaja alumise otsa, määravad rahakoti lennutrajektoori võrrandid $x=vt$ ja $y=-gt^2 / 2$. Rahakott maandub siis, kui selle trajektoor ja nõlva kirjeldav joon $y=-x\tan\alpha$ lõikuvad. Seega
$$-gt^2 / 2=-vt\tan\alpha,$$
$$t=\frac{2v \tan \alpha}{g},$$
ehk
$$x=4h \tan\alpha.$$

Kadunud rahakott

Suusahüppemäe hoovõturada asub nõlval, mille tõusunurk on $\alpha$. Hoovõturaja alumine
ots on horisontaalne. Suusahüppaja alustas hoovõttu kõrguselt $h$,
kuid kohe hoovõtu alguses kukkus tal rahakott taskust välja. Kui kaugele
äratõukepunktist (mööda
horisontaali) rahakott lendab, kui see liigub ilma takistuseta?

Hoovõturaja alumises otsas on rahakott omandanud teatud horisontaalse kiiruse. Edasi hakkab rahakott liikuma mööda paraboolset trajektoori, kuni see taas vastu mäe nõlva kukub.


[User] A point is marked on the outer surface of a glass cylinder. If that cylinder is observed at a great distance (much bigger than the radius of the cylinder), so that through the cylinder the point seems to be on the symmetry axis, then two additional images of the point can be seen. One point can be seen on one side of the symmetry axis, the second on the other side. If the cylinder were turned around its symmetry axis then at a particular moment those two points would merge together and disappear. The third image stays. If continuing to turn the cylinder, then at the moment when the angle of rotation with respect to the initial position is $\ang{15}$, the third image would disappear so that the marked point would not be seen anymore. How big is the refractive index of the glass?

Solution:
\begin{wrapfigure}{r}{0.35\textwidth}%
\includegraphics[trim = 0mm 0mm 12mm 0mm, clip, width=1\linewidth]{2014-v3g-10-silinder}
\end{wrapfigure}

Tähistame märgitud punkti $A$-ga ning murdugu sealt lähtunud kiir punktis $B$ nii, et suundub eemale läbi punkti $C$ (vt joonist).
Sihi $BC$ suunast kaugelt vaadatuna näeme tumeda punkti asukohana punkti $B$. Uurime, kuidas sõltub kiire $BC$ levikusuund, mida kirjeldame
$AO$ ja $BC$ vahelise nurga $2\alpha-\beta$ abil, kiire algsest levikusuunast $\alpha$:
$$2\alpha-\beta= 2\alpha-\arcsin (n\sin\alpha).$$

Silindri algasendi korral $2\alpha-\arcsin (n\sin\alpha) =0$, mille üks lahend $\alpha=0$ annab keskmise näiva punkti
ning kaks külgmist tumedat punkti vastavad võrrandi $\sin(2\alpha)=n\sin\alpha$ ülejäänud lahenditele vahemikus $-45^\circ <\alpha<45^\circ$.
Kui pöörata nüüd silindrit nurga $\delta$ võrra, siis vastavad näivad punktid võrrandi
$$2\alpha-\arcsin (n\sin\alpha) =\delta$$
lahenditele. Võrrandi vasakul pool on funktsioon, mis väikeste nurkade puhul käitub kui $(2-n)\alpha$; suuremate nurkade puhul teise liidetava suhteline mõju kasvab.
Seega juhul $2>n$, on tegemist väikeste nurkade puhul kasvava funktsiooniga, mis läheb suuremate nurkade puhul üle kahanevaks;
juhul $2<n$ on see aga monotoonselt kahanev funktsioon. Kuivõrd $\delta=0$ puhul on kolm lahendit, siis peab olema tegemist esimese juhtumiga, $2>n$.
Nende pöördenurkade $\delta$ puhul, mis on suuremad selle funktsiooni lokaalsest maksimumist, on võrrandil vaid üks lahend.
Funktsioon saavutab globaalse maksimumi täieliku sisepeegelduse piirjuhul
$$n\sin\alpha=-1,$$
mis annab pöördenurga
$$90^\circ-2\arcsin \frac 1n=\SI{15}{}^\circ\Rightarrow n=1/\sin \ang{37,5} = \SI{1,64}{}.$$

Klaassilinder

Klaassilindri välispinnal märgitakse markeriga punkt. Kui seda silindrit vaadata suurelt kauguselt (hulga suuremalt kui silindri raadius) nii, et punkt paistab läbi silindri selle sümmeetriateljel olevat, siis on lisaks näha veel kahte punkti kujutist. Üks kujutis on näha ühel ja teine teisel pool sümmeetriatelge. Kui silindrit keerata ümber oma sümmeetriatelje, siis teatud hetkel sulavad kaks punkti kujutist kokku ning kaovad ära. Kolmas kujutis jääb alles. Kui silindrit edasi keerata, siis hetkel, kui selle pöördenurk algasendi suhtes on \ang{15}, kaob ka kolmas kujutis, nõnda et markeriga tehtud punkti polegi enam näha. Kui suur on klaasi murdumisnäitaja?

Tähistame märgitud punkti $A$-ga ja silindri keskpunkti $O$-ga. Antud ülesande kontekstis tasub vaadelda funktsiooni $f(\alpha) = \delta$, mis kirjeldab punktist $A$ alguse saanud kiire silindrist väljumise nurka $\delta$ kiire $AO$ suhtes funktsioonina stardinurgast $\alpha$ kiire $AO$ suhtes. Silindri algasendit kirjeldab $f(\alpha) = 0$ kolme lahendiga. Lisaks on teada, et $f(\alpha) = \ang{15}$ on lahenditeta.


[User] Kui suure rõhuni $p_N$ võib pumbata jalgpalli palli kolbpumbaga, kui pumpa surutakse kokku $N = 40$ korda? Iga pumpamiskäigu jooksul võtab pump atmosfäärist õhu koguse ruumalaga $v = \SI{150}{cm^3}$. Atmosfääri rõhk $p_0 = \SI{0,1}{MPa}$, palli ruumala $V = \SI{3}{l}$. Lugeda, et õhu temperatuur pallis võrdub välistemperatuuriga.

Solution:
Iga pumpamiskäigu alguses täidab atmosfääri õhk rõhuga  $p_0$ pumba siseruumi ruumalaga $v$. Pumpamiskäigu lõpus on see õhk pallis, kus ta ruumala on $V$ ja osarõhk $p$. Viimase saame leida Boyle-Mariotte’i seadusest:
\[
p = \frac{p_0v}{V}.
\]
Pärast $N$ pumpamiskäiku on rõhk pallis võrdne osarõhkude summaga:
\[
p_{N}=N p=\frac{N v p_{0}}{V}=\frac{40 \cdot 150 \cdot \num{0,1}}{3000}=\SI{0,2}{MPa}.
\]

Jalgpall

Kui suure rõhuni $p_N$ võib pumbata jalgpalli palli kolbpumbaga, kui pumpa surutakse kokku $N = 40$ korda? Iga pumpamiskäigu jooksul võtab pump atmosfäärist õhu koguse ruumalaga $v = \SI{150}{cm^3}$. Atmosfääri rõhk $p_0 = \SI{0,1}{MPa}$, palli ruumala $V = \SI{3}{l}$. Lugeda, et õhu temperatuur pallis võrdub välistemperatuuriga.

Pumpamise käigus kehtib ideealse gaasi olekuvõrrand.


[User]
\begin{wrapfigure}{r}{0.38\textwidth}
  \begin{center}
    \includegraphics[width=0.4\textwidth]{2014-v3g-04-periskoopprillid_yl_joonis_ing}
  \end{center}
\end{wrapfigure}
Reading a book for too long is known to cause neck strain. To avoid this, a special pair of glasses were invented that do not require tilting the head to look down. The main element of these glasses is a prism shown in the figure. The upper face of the prism is covered with a material that reflects light. The apex angle $\varphi$ of the prism is chosen so that if the light ray entering the prism is perpendicular to the surface, then the ray that exits is also perpendicular. The prism is made of material with a refractive index $n=\SI{1,5}{}$.\\
a) On the line $AD$ there are points $B$ and $C$ that divide it into three sections: $AB$, $BC$ and $CD$. Depending on what section the ray falls on there are in principal three different routes in the prism for the ray. Make a draft of the ray’s routes for all the cases. \\
b) Find the value of $\varphi$.\\
c) Let the length of the side $AD$ be $l$. How far are the points $B$ and $C$ distanced from the tip $A$?\\
d) Why is this quite complex system used in the glasses instead of using a simple plane mirror to tilt the beams?

Solution:
\osa Kuna sisenevad ja väljuvad kiired on prisma pinnaga risti, siis keskkondade lahutuspiiril nende suund ei muutu. Kui kiir langeb prismale lõigul $AB$, siis peegeldub see esmalt prisma ülemisel tahul, seejärel alumisel tahul ning väljub prismast läbi parempoolse tahu (vt joonist). Kui kiir langeb prismale lõigul $BC$, siis peegeldub see ülemisel tahul, siis parempoolsel tahul ning väljub läbi alumise tahu ja ei jõuagi silma. Kui kiir langeb prismale lõigul $CD$, siis peegeldub see esmalt parempoolsel tahul, seejärel ülemisel tahul ning väljub jällegi läbi alumise tahu.

\begin{center}
 \includegraphics[width=0.8\textwidth]{2014-v3g-04-periskoopprillid_lahendus_joonis1.pdf}
\end{center}

\osa Vaatleme kiirt, mis siseneb prismasse lõigul $AB$. Kuna sisenev kiir on pinnaga risti, siis tekib täisnurkne kolmnurk $AKL$ (vt joonist). Kolmnurga üheks teravnurgaks on $\varphi$ ja seega on teise teravnurga suurus $\ang{90}-\varphi$. Kuna viimane nurk on esimesel peegeldumisel langemisnurga täiendnurgaks, siis on ka langemisnurk $\varphi$. Peegeldumisseadusest järeldub, et esimene peegeldumisnurk on samuti $\varphi$. Kuna prismast väljuv kiir on paralleelne prisma ülemise tahuga, siis on teisel peegeldumisel peegeldumisdumisnurga täiendnurk ja seega ka langemisnurga täiendnurk $\varphi$. Näeme, et täisnurkse kolmurga $KLM$ teravnurgad on $\varphi$ ja $2\varphi$. Kuna kolmnurga sisenurkade summa on \ang{180}, siis $\varphi=(\ang{180}-\ang{90})/3=\ang{30}$.

\begin{center}
 \includegraphics[width=0.65\textwidth]{2014-v3g-04-periskoopprillid_lahendus_joonis2.pdf}
\end{center}

\emph{Märkus.} Nüüd, kui prisma tipunurk on leitud, saame veenduda, et kui pinnaga risti sisenenud kiir peegeldub prisma alumiselt või parempoolselt tahult, siis on need peegeldumised täielikud. Eelmisest alaülesandest on näha, et nendel juhtudel on langemisnurgaks $\ang{90}-\varphi=\ang{60}$. Täieliku sisepeegeldumise piirnurk on $\alpha=\arcsin(1/n)=\ang{42}$. Langemisnurk \ang{60} on sellest suurem ehk toimub täielik sisepeegeldumine.

\osa Kui kiir siseneb prismasse punktis $B$, siis läbib väljuv kiir punkti $D$ (vt joonist).

\emph{Märkus.} Väljuva kiire suund pole sel juhul üheselt määratud, kuid see ei oma ülesande lahendamisel tähtsust.\\
Kasutades eelmises alaülesandes saadud $\varphi$ väärtust, on näha, et kolmnurgad $ABL$ ja $BDL$ on teineteise peegeldused lõigu $BL$ suhtes (öeldakse, et need kolmnurgad on kongruentsed). Seega on lõikude $AB$ ja $BD$ pikkused võrdsed, millest järeldub, et $|AB|=l/2$. Kui kiir siseneb prismasse punktis $C$, siis peegeldub see punktis $O$ otse tagasi ning väljub prismast punktis $C$. Täisnurksetest kolmnurkadest saame, et $|AO|=\cos(\ang{30})|AD|$ ja $|AC|=\cos(\ang{30})|AO|$ ehk punkti $C$ kaugus punktist $A$ on
\[
|AC|=\cos^2(\ang{30})|AD|=\frac{3}{4}l.
\]

\begin{center}
 \includegraphics[width=0.65\textwidth]{2014-v3g-04-periskoopprillid_lahendus_joonis3.pdf}
\end{center}

\osa Ühe tasapeegli kasutamisel paistab tekst peegelpildis. Seetõttu tuleb teksti õigetpidi nägemiseks kasutada süsteemi, kus toimub paarisarv peegeldusi.

Periskoopprillid

\begin{wrapfigure}{r}{0.38\textwidth}
 \begin{center}
 \includegraphics[width=0.4\textwidth]{2014-v3g-04-periskoopprillid_yl_joonis.pdf}
 \end{center}
\end{wrapfigure}

Kui liiga kaua järjest raamatut lugeda, võib kael pikast allapoole vaatamisest ära väsida. Selle vältimiseks on välja mõeldud erilised prillid, mille abil saab pead kallutamata alla vaadata. Prillide põhiliseks elemendiks on joonisel kujutatud prisma, mille pealmine tahk on kaetud valgust peegeldava materjaliga. Prisma tipunurk $\varphi$ on valitud selliselt, et kui prismasse sisenev valguskiir on pinnaga risti, siis on seda ka väljuv kiir. Prisma on tehtud materjalist murdumisnäitajaga $n=\SI{1,5}{}$.

\osa Lõigul $AD$ on punktid $B$ ja $C$, mis jagavad selle kolmeks piirkonnaks: $AB$, $BC$ ning $CD$. Sõltuvalt sellest, millisele piirkonnale kiir langeb, on kiire käiguks prismas kolm põhimõtteliselt erinevat võimalust. Tehke joonis kiirte käigust kõigi juhtude jaoks.\\
\osa Leidke nurga $\varphi$ väärtus.\\
\osa Olgu külje $AD$ pikkus $l$. Kui kaugel asuvad punktid $B$ ja $C$ tipust $A$?\\
\osa Miks on prillides üldsegi vaja kasutada suhteliselt keerulist prismaga süsteemi, selle asemel et kiirte kallutamiseks kasutada ühte tasapeeglit?

Nurga $\varphi$ leidmiseks tuleb vaadelda kiirt, mis siseneb lõiku $AB$.


[User]
5. Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the
     Detailed Answer Sheet. Simple short answers and plots may be directly entered in the Summary
     Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer
     calculations. Strike out any rough work that you do not want to be considered for evaluation.

Solution:
5. A pair of long parallel metallic rails of negligible resistance and separation w are placed horizon-
   tally. A horizontal metal rod (dark thick line) of mass M and resistance R is placed perpendic-
   ularly onto the rails at one end (as shown). A uniform magnetic field B exists perpendicular to
   the plane of the paper (pointing into the page). One end of the rail track is connected to a key
   (K) and a capacitor of capacitance C charged to voltage V0 . At t = 0 the key is closed. Neglect
   friction and self inductance of the loop.

                                             C1                 9
   (a) What is the final speed (vfinal ) attained by the rod?                                             [6]

         vfinal =

                      HB
          Solution: At t = 0, Q0 = CV0 . At any instant, Q̇ = −I, where Q is the charge on
          capacitor and I is the current in RC circuit. From Newton’s second law and Lorentz
          force,
                                                   dv
                                               m      = IwB                                     (5.1)
                                                   dt
                                                      = −Q̇wB                                   (5.2)

          Induced emf in the circuit is Bwv. From KVL,
                                              Q
                                                = IR + Bwv                                      (5.3)
                                              C
                                                = −Q̇R + Bwv                                    (5.4)

INPhO 2019 Page 12                    Questions & Answers                       Last four digits of Roll No.:

         Differentiating Eq. (5.4) and using Eq. (5.2)

                                                 1       B 2 w2
                                         Q̈ = −     Q̇ −        Q̇                                    (5.5)
                                               RC "       mR #
                                                     1    B 2 w2
                                            = −Q̇       +                                             (5.6)
                                                    RC      mR
                                                 Q̇      1    1     B 2 w2
                                         Q̈ = −     where =       +                                   (5.7)
                                           !     τ       τ   RC      mR
                                        Q̇       t                         V0
                                log           = − where Q̇0 = I(t = 0) = −                            (5.8)
                                       Q̇0       τ                         R
                                                 V0
                                           Q̇ = − e−t/τ                                               (5.9)
                                                 R
         Combining Eqs. (5.2) and (5.9)

                                                    dv    Vo Bw −t/τ
                                               m        =       e                                   (5.10)
                                        Z vfinal
                                                    dt    Z ∞
                                                             R
                                                               V0 Bw −t/τ
                                                 m dv =              e    dt                        (5.11)
                                         0                 0     R
                                                                            ∞
                                                          BwV0
                                                                 (−τ )e−t/τ

                                        C1
                                              mvfinal =

                                                        =

                                                 vfinal =
                                                           9 R
                                                          BwV0 τ
                                                             R
                                                             BwCV0
                                                          m + B 2 w2 c
                                                                            0
                                                                                (5.12)

                                                                                (5.13)

                                                                                (5.14)

   (b) Consider the ratio r of the maximum kinetic energy attained by the rod to the energy                     [2]
       initially stored in the capacitor. What is rmax , the maximum possible value of r, by appro-

               HB
       priately choosing B?

         rmax =

         Solution:
                                               2 /2
                                            mvmax
                                        r=                                                          (5.15)
                                             CV02 /2
                                               x               B 2 w2 C
                                          =          where x =                                      (5.16)
                                            (1 + x)2              m
                                        1
         r has a maximum value of         at x = 1
                                        4

                                                     rmax = 1/4                                     (5.17)

    (c) Let M = 10.0 kg, w = 0.10 m, V0 = 1.00 × 104 V and a bank of capacitors ensures that C                  [1]
        = 1.00 F. If r = rmax , calculate the value of vfinal .
         vfinal (r = rmax ) =

INPhO 2019 Page 13               Questions & Answers

         Solution: For rmax ,

                                         B 2 w2 C
                                                  = 1 ⇒ B = 31.6 T                           (5.18)
                                            m
                                             vmax = 1.58 × 103 m/s                           (5.19)

                                Detailed answers can be found on page numbers:

INPhO 2019 Problem 5


[User]
Juku built a circuit diagram that is shown in the figure using six identical resistors of resistance $R=\SI{10}{\ohm}$, four identical lamps of resistance $r=\SI{20}{\ohm}$ and a voltage source with an electromotive force $\mathcal{E}=\SI{5}{\volt}$. Calculate the power dissipated by each lamp (in the figure marked as 1, 2, 3, 4). Do not account for internal resistance of the voltage source.
\begin{center}
\includegraphics[width=0.6\linewidth]{2013-lahg-09-lambidJoonis-crop}
\end{center}

Solution:
Paneme tähele, et skeemi sümmeetria tõttu ei läbi vool lampi 1, sest selle otstel on võrdsed potentsiaalid. Sama kehtib lambi 4 kohta. Joonistame skeemi ümber, ühendades lambi 1 otspunktid omavahel kokku, sama teeme ka lambi 4 otspunktidega. Saame võrdlemisi lihtsa skeemi, mis koosneb kolmest jadamisi ühendatud paralleelühendusest, nagu joonisel näidatud.

\begin{center}
\includegraphics[width=0.8\textwidth]{2013-lahg-09-ahelLah.pdf}
\end{center}

Arvutame ahela kogutakistuse
\[R_t = \frac{R}{2} + \frac{1}{\frac{1}{R}+\frac{1}{r}+\frac{1}{r}+\frac{1}{R}}+\frac{R}{2} =\frac{4}{3}R,\]
kus arvestasime, et $r=2R$, kusjuures keskmise paralleelühenduse takistuseks tuli $R_k=R/3$, mis moodustab $1/4$ osa kogutakistusest, seega langeb keskmisele osale ka $1/4$ osa patarei pingest ehk $V_k = \mathcal{E}/4 = 5/4\si{V}$. Nüüd on juba lihtne avaldada lampides 2 ja 3 eralduv võimsus
\[P_2=P_3=\frac{V_k^2}{r}=\frac{25}{16\cdot 20}\SI{}{W}\approx\SI{0,08}{W}.\]
Lampides 1 ja 4 mõistagi võimsust ei eraldu.

Lambid

Juku ehitas kodus niisuguse elektriskeemi nagu joonisel näidatud, kasutades
selleks kuut ühesugust takistit takistusega $R=\SI{10}{\ohm}$, nelja
ühesugust lampi takistusega $r=\SI{20}{\ohm}$ ning pingeallikat
elektromotoorjõuga
$\mathcal{E}=\SI{5}{\volt}$. Arvutage igas lambis (joonisel tähised 1, 2, 3,
4) eralduv võimsus. Pingeallika sisetakistusega mitte arvestada.
\begin{center}
\includegraphics[width=0.6\linewidth]{2013-lahg-09-lambidJoonis-crop}
\end{center}

Paneme tähele, et sümmeetria tõttu on lampide 1 ja 4 otspunktide potentsiaalid võrdsed.


[User]
A uniform string of length L and linear density ρ is stretched between two fixed
    supports. The tension in the string is T .

     (a) Find the standing wave solutions and angular frequencies for the given boundary conditions.

      (b) A very small mass m is now placed a distance ℓ from one end of the string. Find the change
          in the angular frequencies to first order in m, by using the fact that the average potential
          energy and average kinetic energy for each standing wave should remain equal.

      (c) How would you go about finding the exact standing wave frequencies for this setup?

Solution:
A uniform string of length L and linear density ρ is stretched between two fixed
    supports. The tension in the string is T .
      (a) Find the standing wave solutions and angular frequencies for the given boundary conditions.

      (b) A very small mass m is now placed a distance ℓ from one end of the string. Find the change
          in the angular frequencies to first order in m, by using the fact that the average potential
          energy and average kinetic energy for each standing wave should remain equal.

     (c) How would you go about finding the exact standing wave frequencies for this setup?
                                       p
    Answer. (a) ωn = nπv/L where v = T /ρ, (b) ∆ωn ≈ (−m/ρL) ωn sin2 (nπℓ/L)


[User]
9. SOLENOIDS (8 points)

(by S. Ainsaar).
A tightly-wound rigid solenoid coil has been
partly inserted into another. They are con-
nected to a constant current source so that
the same current I flows through them; both
generate a magnetic field in the same direc-
tion. The solenoids have N turns both, their
length is l, their cross-sectional areas are A1
and A2. It may be assumed that A1,A2 ¿ l2
.
You may find it useful to know that the mag-
netic field at the centre of one solenoid taken
separately is B =μ0IN/l, where μ0 is the per-
meability of vacuum.
i) (2 points) The centres of the coils are
at a distance x < l along their common axis
[A1,A2 ¿ (x−l)2
,x2
]. Find the total energy
Em of the magnetic field in the system.
ii) (4 points) Find the electromotive forces
E1 and E2 generated on the coils when one is
pulled out with a speed v.
iii) (2 points) Find the force F needed to pull
one coil outwards.

Solution:
9. SOLENOIDS

i) The overlapping region has magnetic field
2B, rest of the coils’ insides has B. Let
the inner coil have an area of A2. Energy
density, where the magnetic field is B, is
B2
2μ0
. Therefore Em = B2
2μ0
(A1l − A2(l − x)+
A2x)+(2B)2
2μ0
A2(l−x)= B2
2μ0
[A1l+A2(3l−2x)]=
μ0I2N2
2l2 [A1l+ A2(3l−2x)].
ii) The outer coil is intersected by all the
internal coil’s magnetic flux BA2. After time
dt this flux is enclosed by fewer turns of
the outer coil — by the Nvdt/l turns that
are on the length vdt. Thus the flux en-
closed by the outer coil changes with a rate
E1 = Φ̇1 = BA2Nv/l = μ0A2IN2
v/l2
. The in-
ner coil is also intersected by the flux BA2,
thus E2 = E1; those electromotive forces are
in the same direction.
iii) The work done when pulling a coil out
is balanced in two places: the magnetic field
energy decreases and our constant current
source dissipates some energy (because our
electomotive forces work against it). In total,
the applied mechanical power Fv =
dEm
dx v+
I(E1 +E2), from where F =−B2A2
μ0
+ 2IBA2N
l =
μ0I2N2A2
l2 .

SOLENOIDS


[User]
THEORETICAL PROBLEM No. 3

                           WHY ARE STARS SO LARGE?

The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen
into helium in their central parts. In this problem we use concepts of both classical and
quantum mechanics, as well as of electrostatics and thermodynamics, to understand
why stars have to be big enough to achieve this fusion process and also derive what
would be the mass and radius of the smallest star that can fuse hydrogen.

                                                            Figure 1. Our Sun, as most stars, shines
                                                            as a result of thermonuclear fusion of
                                                            hydrogen into helium in its central
                                                            parts.

 USEFUL CONSTANTS
 Gravitational constant = G = 6.7 ×10 −11 m3 kg-1 s2
 Boltzmann ́s constant = k = 1.4 × 10−23 J K-1
 Planck’s constant = h = 6.6 × 10−34 m2 kg s-1
 Mass of the proton = m p = 1.7 × 10 −27 kg
 Mass of the electron = me = 9.1 × 10 −31 kg
 Unit of electric charge = q = 1.6 × 10 −19 C
 Electric constant (vacuum permittivity) = ε 0 = 8.9 × 10 −12 C2 N-1 m-2
 Radius of the Sun = RS = 7.0 × 108 m
 Mass of the Sun = M S = 2.0 × 1030 kg

     1. A classical estimate of the temperature at the center of the stars.

Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons
in equal amounts), and that it behaves like an ideal gas. From the classical point of view,
to fuse two protons, they need to get as close as 10 −15 m for the short range strong
nuclear force, which is attractive, to become dominant. However, to bring them together
they have to overcome first the repulsive action of Coulomb’s force. Assume classically
that the two protons (taken to be point sources) are moving in an antiparallel way, each
with velocity vrms , the root-mean-square (rms) velocity of the protons, in a one-
dimensional frontal collision.

1a   What has to be the temperature of the gas, Tc , so that the distance of 1.5
     closest approach of the protons, d c , equals 10 −15 m? Give this and all
     numerical values in this problem up to two significant figures.

     2. Finding that the previous temperature estimate is wrong.
To check if the previous temperature estimate is reasonable, one needs an independent
way of estimating the central temperature of a star. The structure of the stars is very
complicated, but we can gain significant understanding making some assumptions. Stars
are in equilibrium, that is, they do not expand or contract because the inward force of
gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the
equation of hydrostatic equilibrium at a given distance r from the center of the star, is
given by
∆P    G M r ρr
   =−          ,
∆r      r2
where P is the pressure of the gas, G the gravitational constant, M r the mass of the star
within a sphere of radius r , and ρ r is the density of the gas in the slab.

                                                                    Figure 2. The stars
                                                                    are in hydrostatic
                                                                    equilibrium, with the
                                                                    pressure difference
                                                                    balancing gravity.

An order of magnitude estimate of the central temperature of the star can be obtained
with values of the parameters at the center and at the surface of the star, making the
following approximations:
∆ P ≈ Po − Pc ,

where Pc and Po are the pressures at the center and surface of the star, respectively.
Since Pc >> Po , we can assume that

∆ P ≈ − Pc .
Within the same approximation, we can write
∆r ≈ R,
where R is the total radius of the star, and
Mr ≈ MR = M ,
with M the total mass of the star.
The density may be approximated by its value at the center,
ρ r ≈ ρc .
You can assume that the pressure is that of an ideal gas.
2a   Find an equation for the temperature at the center of the star, Tc , in terms   0.5
     of the radius and mass of the star and of physical constants only.

We can use now the following prediction of this model as a criterion for its validity:

2b Using the equation found in (2a) write down the ratio M / R expected for            0.5
   a star in terms of physical constants and Tc only.

2c   Use the value of Tc derived in section (1a) and find the numerical value          0.5
     of the ratio M / R expected for a star.

2d Now, calculate the ratio M ( Sun) / R ( Sun) , and verify that this value is        0.5
   much smaller than the one found in (2c).

     3. A quantum mechanical estimate of the temperature at the center of the
        stars

The large discrepancy found in (2d) suggests that the classical estimate for Tc obtained
in (1a) is not correct. The solution to this discrepancy is found when we consider
quantum mechanical effects, that tell us that the protons behave as waves and that a
single proton is smeared on a size of the order of λ p , the de Broglie wavelength. This
implies that if d c , the distance of closest approach of the protons is of the order of λ p ,
the protons in a quantum mechanical sense overlap and can fuse.

3a                          λp                                                         1.0
     Assuming that d c =           is the condition that allows fusion, for a proton
                            21 / 2
     with velocity vrms , find an equation for Tc in terms of physical constants
     only.

3b   Evaluate numerically the value of Tc obtained in (3a).                            0.5

3c    Use the value of Tc derived in (3b) to find the numerical value of the 0.5
     ratio M / R expected for a star, using the formula derived in (2b). Verify
     that this value is quite similar to the ratio M ( Sun) / R ( Sun ) observed.

Indeed, stars in the so-called main sequence (fusing hydrogen) approximately do follow
this ratio for a large range of masses.

     4. The mass/radius ratio of the stars.

The previous agreement suggests that the quantum mechanical approach for estimating
the temperature at the center of the Sun is correct.

4a    Use the previous results to demonstrate that for any star fusing hydrogen,    0.5
     the ratio of mass M to radius R is the same and depends only on physical
     constants. Find the equation for the ratio M / R for stars fusing hydrogen.

     5. The mass and radius of the smallest star.
The result found in (4a) suggests that there could be stars of any mass as long as such a
relationship is fulfilled; however, this is not true.
The gas inside normal stars fusing hydrogen is known to behave approximately as an
ideal gas. This means that d e , the typical separation between electrons is on the average
larger that λe , their typical de Broglie wavelength. If closer, the electrons would be in a
so-called degenerate state and the stars would behave differently. Note the distinction in
the ways we treat protons and electrons inside the star. For protons, their de Broglie
waves should overlap closely as they collide in order to fuse, whereas for electrons their
de Broglie waves should not overlap in order to remain as an ideal gas.
The density in the stars increases with decreasing radius. Nevertheless, for this order-of-
magnitude estimate assume they are of uniform density. You may further use that
m p >> me .

5a    Find an equation for ne , the average electron number density inside the      0.5
     star.

5b    Find an equation for d e , the typical separation between electrons inside    0.5
     the star.

5c                   λe                                                             1.5
     Use the d e ≥   1/ 2
                            condition to write down an equation for the radius of
                     2
     the smallest normal star possible. Take the temperature at the center of the
     star as typical for all the stellar interior.

5d    Find the numerical value of the radius of the smallest normal star                0.5
     possible, both in meters and in units of solar radius.

5e    Find the numerical value of the mass of the smallest normal star possible,        0.5
     both in kg and in units of solar masses.

     6. Fusing helium nuclei in older stars.

As stars get older they will have fused most of the hydrogen in their cores into helium
(He), so they are forced to start fusing helium into heavier elements in order to continue
shining. A helium nucleus has two protons and two neutrons, so it has twice the charge
                                                                                λp
and approximately four times the mass of a proton. We saw before that d c =              is the
                                                                                21 / 2
condition for the protons to fuse.

6a    Set the equivalent condition for helium nuclei and find vrms ( He) , the rms 0.5
     velocity of the helium nuclei and T ( He) , the temperature needed for
     helium fusion.

Solution:
Answers
                                  Theoretical Problem No. 3

                                    Why are stars so large?
     1) A first, classic estimate of the temperature at the center of the stars.

1a   We equate the initial kinetic energy of the two protons to the electric       1.5
     potential energy at the distance of closest approach:
       1              q2
             2
     2( m p vrms )           ; and since
       2            4  0 dc
     3       1
       k Tc  m p vrms
                   2
                       , we obtain
     2       2
                 q2
     Tc                    5.5109 K
            12  0 dc k

      2) Finding that the previous temperature estimate is wrong.

2a   Since we have that                                                            0.5
     P    G M r r
                  , making the assumptions given above, we obtain that:
     r      r2
            G M c
     Pc           . Now, the pressure of an ideal gas is
              R
            2 c k Tc
     Pc              , where k is Boltzmann ́s constant, Tc is the central
               mp
     temperature of the star, and m p is the proton mass. The factor of 2 in the
     previous equation appears because we have two particles (one proton and
     one electron) per proton mass and that both contribute equally to the
     pressure. Equating the two previous equations, we finally obtain that:
            G M mp
     Tc 
              2k R

2b From section (2a) we have that:                                                 0.5
     M 2 k Tc
      
     R G mp

2c   From section (2b) we have that, for Tc  5.5109 K:                          0.5

      M 2 k Tc
               1.4 1024 kg m-1.
      R G mp

2d For the Sun we have that:                                                      0.5
      M ( Sun)
                2.9 1021 kg m-1 , that is, three orders of magnitude smaller.
      R( Sun)

     3) A quantum mechanical estimate of the temperature at the center of the stars

3a   We have that                                                                 1.0
                h
     p              , and since
             m p vrms

      3       1
        k Tc  m p vrms
                    2
                        , and
      2       2
                  q2
     Tc                      , we obtain:
            12   0 d c k

                q4 mp
     Tc                        .
            24  2  02 k h 2

3b              q4 mp                                                             0.5
     Tc                         9.7 106 K.
            24  2  02 k h   2

3c   From section (2b) we have that, for Tc  9.7 106 K:                         0.5

      M 2 k Tc
               2.41021 kg m-1 ; while for the Sun we have that:
      R G mp

      M ( Sun)
                2.9 1021 kg m-1.
      R( Sun)

     4) The mass/radius ratio of the stars.

4a   Taking into account that                                                     0.5

     M 2 k Tc
             , and that
     R G mp

                   q4 mp
     Tc                                , we obtain:
            24  2  02 k h 2

     M       q4
                        .
     R 12  2  02 G h 2

     5) The mass and radius of the smallest star.

5a                  M                                                                  0.5
     ne 
            (4 / 3)  R3 m p

5b                                                           1 / 3                    0.5
                                 M       
     de  n   1 / 3
                                        
              e          (4 / 3)  R3 m 
                                       p 

5c   We assume that                                                                    1.5
              e
     de             . Since
            21 / 2
                      h
     e                         ,
             me vrms (electron )
     3       1
       k Tc  me vrms
                  2
                      (electron ) ,
     2       2
                   q4 mp
     Tc                                ,
            24  2  02 k h 2

     M       q4
                        , and
     R 12  2  02 G h 2
                                                1 / 3
                   M      
     de                                               ,
           (4 / 3)  R m 
                       3
                        p 
     we get that
                         o1 / 2 h 2
     R
            41 / 4 q me3 / 4 m5p / 4 G1 / 2

5d                        o1 / 2 h 2                                                  0.5
      R      1/ 4        3/ 4       5/ 4   1/ 2
                                                          6.9  107 m = 0.10 R(Sun)
             4       qm   e      m   p      G

5e   The mass to radius ratio is:                                                  0.5
     M     q4
                     2.4  1021 kg m-1, from where we derive that
     R 12  0 G h
          2 2      2

     M  1.7  1029 kg = 0.09 M (Sun)

     6) Fusing helium nuclei in older stars.

6a   For helium we have that                                                       0.5
              4 q2                h
                            1/ 2            ; from where we get
     4   0 mHe vrms ( He) 2 mHe vrms ( He)
                   2

                    21 / 2 q 2
     vrms ( He)                2.0  106 m s-1.
                     0 h
     We now use:
                  2
                 vrms ( He) mHe
     T ( He)                    6.5 108 K.
                        3k
     This value is of the order of magnitude of the estimates of stellar models.

WHY ARE STARS SO LARGE?