39th International Physics Olympiad - Hanoi - Vietnam Theoretical Problem No. 1 /Solution. Solution

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1 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin Slutin. The structure f the mrtar.. Calculating the distance TG The vlume f water in the bucket is V = = cm m. The length f the bttm f the bucket is d = L htan 6 = (. 74. tan 6 ) m =. 53m. (as the initial data are given with tw significant digits, we shall keep nly tw significant digits in the final answer, but we keep mre digits in the intermediate steps). The height c f the water layer in the bucket is calculated frm the frmula: ( 3 / b) V = bcd + b ctan 6 c = 3 c d V / + d Inserting numerical values fr V, b and d, we find c =.8m. When the lever lies hrizntally, the distance, n the hrizntal axis, between the rtatin axis and the center f mass f water N, is TG = ( m/ M)TH =.57m (see the figure belw). d c tan =., and TH a 6 474m R S P H N K T Answer: TG =.6m... Calculating the values f α and α. When the lever tilts with angle, water level is at the edge f the bucket. At that α pint the water vlume is 3 3 m. Assume PQ < d. Frm gemetry V = hb PQ /, frm which P Q =.m. The assumptin PQ < d is bviusly satisfied ( d =.53m ). T cmpute the angle, we nte that tan α = h/ QS= h/( PQ+ 3h). Frm this we find α =.6. α

2 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin When the tilt angle is 3, the bucket is empty: α = 3. h T G R N S P I Q.3. Determining the tilt angle β f the lever and the amunt f water in the bucket m when the ttal trque μ n the lever is equal t zer Dente PQ = x(m). The amunt f water in the bucket is xhb m= ρwater = 9 x (kg). μ = when the trque cming frm the water in the bucket cancels ut the trque cming frm the weight f the lever. The crss sectin f the water in the bucket is the triangle PQR in the figure. The center f mass N f water is lcated at /3 f the meridian RI, therefre NTG lies n a straight line. Then: mg TN = Mg TG r m TN = M TG = 3.57 =.474 () Calculating TN frm x then substitute () : x x x TN = L+ a ( h 3 + ) = = which implies m TN = 9 x(.84 x/ 3) = 3x + 7.3x () S we find an equatin fr x : 3x + 7.3x =.474 (3) The slutins t (3) are x =.337 and x =.673. Since x has t be smaller than.53, we have t take x = x =.673 and m= 9x =.65kg. h tan β = =. 436, r β = x+ h 3 Answer: m =.6kg and β =3.6.. Parameters f the wrking mde

3 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin..Graphs f μ( α ), α( t), and μ(t) during ne peratin cycle. Initially when there is n water in the bucket, α =, μ has the largest magnitude equal t gm TG = = 4.64 N m. Our cnventin will be that the sign f this trque is negative as it tends t decrease As water flws int the bucket, the trque cming frm the water (which carries psitive sign) makes μ increase until μ is slightly psitive, when the lever starts t lift up. Frm that mment, by assumptin, the amunt f water in the bucket is cnstant. The lever tilts s the center f mass f water mves away frm the rtatin axis, leading t an increase f μ, which reaches maximum when water is just abut t verflw the edge f the bucket. At this mment α = α =.6. A simple calculatin shws that SI = SP + PQ / = / =.634m. TN = SI =.7644 m. 3 μ max = (. TN 3 TG) g cs.6 α. Therefre = ( ) 9. 8 cs. 6 =. 69 N m. μ =.7 N m. max As the bucket tilts further, the amunt f water in the bucket decreases, and when α = β, μ =. Due t inertia, α keeps increasing and μ keeps decreasing. The bucket is empty when α = 3, when μ equals 3 g TG cs 3 = 4. N m. After that α keeps increasing due t inertia t α ( μ = gm TGcsα = 4.6csα N m ), then quickly decreases t ( μ = 4.6N m). belw On this basis we can sketch the graphs f α( t), μ() t, and μ( α ) as in the figure 3

4 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin μ.7 N.m A 3.6 O E α.6 B 3 α -4.6 csα N.m D -4. N.m C -4.6 N.m F.. The infinitesimal wrk prduced by the trque μ( α ) is dw = μ( α) dα. The energy btained by the lever during ne cycle due t the actin f μ( α ) is W = μ( α) dα, which is the area limited by the line μ( α ). Therefre is equal W ttal t the area enclsed by the curve (OABCDFO) n the graph μ( α ). The wrk that the lever transfers t the mrtar is the energy the lever receives as it mves frm the psitin α = α t the hrizntal psitin α =. We have Wpunding equals t the area f (OEDFO) n the graph gm TG sin α = 4. 6sinα (J). μ( α ). It is equal t.3. The magnitudes f α can be estimated frm the fact that at pint D the energy f the lever is zer. We have area (OABO) = area (BEDCB) Apprximating OABO by a triangle, and BEDCB by a trapezid, we btain: (/ ) = 4. [( α 3.6) + ( α 3)] (/ ), which implies α = Frm this we find W = area (OEDFO) = Mg TG csα dα = 46. sin 347. = 63. punding

5 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin Thus we find Wpunding.6 J. μ 3. The rest mde The bucket is always verflwn with water. The tw branches f μ( α ) in the vicinity f α = β crrespnding t increasing and decreasing α cincide with each ther. The graph implies that α = β is a stable equilibrium f the mrtar Find the expressin fr the trque μ when the tilt angle is α = β +Δα ( Δ α is small ). The mass f water in bucket when the lever tilts with angle α is m= (/ ) ρbhpq, where PQ = h. A simple calculatin shws that tanα tan 3 when α increases frm β t β + Δα, the mass f water increases by Δ bh bh m = ρ ρ α α sin α Δ sin β Δ. The trque μ acting n the lever when the tilt β α is β +Δ α equals the trque due t Δm. We have μ =Δ m g TN cs( β +Δα). TN is fund frm the equilibrium cnditin f the lever at tilting angle β : TN = M TG / m= 3.57/.65 =.779m. We find at the end μ = 47. Δα N m 47 Δα N m Equatin f mtin f the lever d μ = where μ = 47 Δ α, α = β +Δ α, and I is the sum f mments α I dt f inertia f the lever and f the water in bucket relative t the axis T. Here I is nt cnstant the amunt f water in the bucket depends n α. When Δα is small, ne can cnsider the amunt and the shape f water in the bucket t be cnstant, s I is apprximatey a cnstant. Cnsider water in bucket as a material pint with mass.6 kg, a simple calculatin gives I = =.36.4 kg m. We have 47.4 d Δα Δ α =. That is the equatin fr a harmnic scillatr with perid dt 5

6 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. /Slutin.4 τ = π = 3.7. The answer is therefreτ = 3. s Harmnic scillatin f lever (arund α = β ) when bucket is always verflwn. Assume the lever scillate harmnically with amplitude Δ α arund α = β. At time t =, Δ α =, the bucket is verflwn. At time dt the tilt changes by dα. We are interested in the case dα <, i.e., the mtin f lever is in the directin f decreasing α, and ne needs t add mre water t verflw the bucket. The equatin f mtin is: Δ α = Δα sin( πt / τ), therefre d( Δ α ) = dα = Δα ( π / τ)cs( πt/ τ) dt. Fr the bucket t be verflwn, during this time the amunt f water falling t the bucket shuld be at least bh ρ Δαπ bh ρdt π t dm = dα = cs sin β τsin β τ ; dm is πbh ρδα maximum at t =, dm = dt. τsin β The amunt f water falling t the bucket is related t flw rate Φ ; dm =Φdt, πbh ρδα therefre Φ=. τ sin β An verflwn bucket is the necessary cnditin fr harmnic scillatins f the lever, therefre the cnditin fr the lever t have harmnic scillatins with ampltude π/36 rad is Φ Φ with r S Φ =.3kg/s. Φ = πbh ρπ. 39kg/s 36τsin β = 3.3 Determinatin f Φ If the bucket remains verflwn when the tilt decreases t.6, then the amunt f water in bucket shuld reach kg at this time, and the lever scillate harmnically with amplitude equal = 3 3Φ. The flw shuld exceed, therefre Φ = 3.3.7kg/s. This is the minimal flw rate fr the rice-punding mrtar nt t wrk. 6

7 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. / Slutin Slutin. D θ E A C B D Figure Let us cnsider a plane cntaining the particle trajectry. At t =, the particle psitin is at pint A. It reaches pint B at t = t. Accrding t the Huygens principle, at mment < t < t, the radiatin emitted at A reaches the circle with a radius equal t AD and the ne emitted at C reaches the circle f radius CE. The radii f the spheres are prprtinal t the distance f their centre t B: ( ) / v ( ) CE ct t n = = = cnst CB t t β n The spheres are therefre transfrmed int each ther by hmthety f vertex B and their envelpe is the cne f summit B and half aperture π ϕ = Arcsin = θ, β n where θ is the angle made by the light ray CE with the particle trajectry... The intersectin f the wave frnt with the plane is tw straight lines, BD and BD'... They make an angle ϕ = Arcsin with the particle trajectry. β n

8 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. / Slutin. The cnstructin fr finding the ring image f the particles beam is taken in the plane cntaining the trajectry f the particle and the ptical axis f the mirrr. We adpt the ntatins: S the pint where the beam crsses the spherical mirrr F the fcus f the spherical mirrr C the center f the spherical mirrr IS the straight-line trajectry f the charged particle making a small angle α with the ptical axis f the mirrr. P θ A I M F C θ α S N O Q CF = FS = f CO//IS CM//AP CN//AQ Figure FCO = α FO = f α MCO = OCN = θ MO = f θ We draw a straight line parallel t IS passing thrugh the center C. The line intersects the fcal plane at O. We have FO f α. Starting frm C, we draw tw lines in bth sides f the line CO making with it an angle θ. These tw lines intersect the fcal plane at M and N, respectively. All the rays f Cherenkv radiatin in the plane f the sketch, striking the mirrr and being reflected,

9 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. / Slutin intersect at M r N. In three-dimensin case, the Cherenkv radiatin gives a ring in the fcal plane with the center at O (FO f α) and with the radius MO f θ. In the cnstructin, all the lines are in the plane f the sketch. Exceptinally, the ring is illustrated spatially by a dash line Fr the Cherenkv effect t ccur it is necessary that c n > v, that is c n =. v min Putting 4 ζ= n = 7. P, we get Because ζ min 4 c = 7. Pmin = = v β () Mc Mc Mc β = = = = K pc p Mv β β () then K =.94 ;.5 ;.4 fr prtn, kan and pin, respectively. Frm () we can express β thrugh K as β = + K (3) Since K << fr all three kinds f particles we can neglect the terms f rder higher than in K. We get β = K + K = Mc p (3a) K K β = + = Mc p (3b) Putting (3b) int (), we btain 3

10 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. / Slutin P = K 7. min 4 (4) We get the fllwing numerical values f the minimal pressure: P min = 6 atm fr prtns, P min = 4.6 atm fr kans, P min =.36 atm fr pins. 3.. Fr θπ = θ we have κ π κ κ csθ = cs θ = cs θ (5) We dente ε = β = K + K (6) Frm (5) we btain = (7) β n β π κ n Substituting β = ε and n = + ζ int (7), we get apprximately: ζ P ( K K ) 4ε ε, κ π = = 4 κ π = 4.(. 5) (. 4) = ζ = 6atm The crrespnding value f refractin index is n =.6. We get: θ =.6 ; θ = θ = 3.. κ We d nt bserve the ring image f prtns since P = 6atm < 6atm = P in fr prtns. m π κ 4

11 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. / Slutin Taking lgarithmic differentiatin f bth sides f the equatin csθ =, we btain βn sinθ Δθ csθ Δ = β β (8) Lgarithmically differentiating equatin (3a) gives Δβ Δp = β p (9) Cmbining (8) and (9), taking int accunt (3b) and putting apprximately tanθ = θ, we derive Δθ β K = = Δp θ pβ θ p () We btain -fr kans K κ = 5., π θ κ = 6. = 6. rad, and s, 8 Δθ Δp κ = 5., GeV / c -fr pins K π =. 4, θ π = 3., and Δθ Δp π =. GeV / c. 4.. Δ θκ +Δθπ Δp Δ θ = ( ) = 53.. Δp GeV / c GeV / c The cnditin fr tw ring images t be distinguishable is Δ θ <.( θ θ ) = 6. π κ. It fllws 6. Δ p < = 3GeV. / c

12 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. / Slutin The lwer limit f β giving rise t Cherenkv effect is β = n = 33.. () The kinetic energy f a particle having rest mass M and energy E is given by the expressin Mc T = E Mc = Mc = Mc. () β β Substituting the limiting value () f β int (), we get the minimal kinetic energy f the particle fr Cherenkv effect t ccur: Tmin = Mc 57. M = c (3) Fr α particles, T min = GeV =. 96 GeV. Fr electrns, T min = 57. 5MeV. = 64MeV.. Since the kinetic energy f the particles emitted by radiactive surce des nt exceed a few MeV, these are electrns which give rise t Cherenkv radiatin in the cnsidered experiment. 6. Fr a beam f particles having a definite mmentum the dependence f the angle θ n the refractin index n f the medium is given by the expressin csθ = (4) nβ 6.. Let δθ be the difference f θ between tw rings crrespnding t tw wavelengths limiting the visible range, i.e. t wavelengths f.4 µm (vilet) and.8 µm (red), respectively. The difference in the refractin indexes at these wavelengths is n n δ n. ( n ) v r = =. Lgarithmically differentiating bth sides f equatin (4) gives 6

13 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. / Slutin sinθ δθ δ n = (5) csθ n Crrespnding t the pressure f the radiatr P = 6 atm we have frm 4.. the values θ π = 3., n =.6. Putting apprximately tanθ = θ and n =, we get δ n δθ = = 33.. θ The bradening due t dispersin in terms f half width at half height is, accrding t (6.), δθ = The bradening due t achrmaticity is, frm 4..,. 3GeV/c. 6 GeV/c =., that is three times smaller than abve The clr f the ring changes frm red t white then blue frm the inner edge t the uter ne. 7

14 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. 3 / Slutin Slutin. Fr an altitude change dz, the atmspheric pressure change is : dp = ρgdz () where g is the acceleratin f gravity, cnsidered cnstant, ρ is the specific mass f air, which is cnsidered as an ideal gas: Put this expressin in () : m pμ ρ = = V RT dp μg = dz p RT.. If the air temperature is unifrm and equals T, then dp p μg = dz RT After integratin, we have :.. If then μg z RT ( ) p( ) e p z = () ( ) ( ) T z = T Λ z (3) dp μg = dz p R T( ) Λ z (4)... Knwing that : ( ) ( ) ( ) dz d T Λz = = ln Λ T Λz Λ T Λz Λ by integrating bth members f (4), we btain : ( ) ( ) ( ) Λ g Λz ( ) ( ) p z μg T z μ ln = ln = ln p RΛ T RΛ T ( T( ) z) Λz = T ( ) p( ) p z ( ) μg RΛ (5)

15 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. 3 / Slutin... The free cnvectin ccurs if: ρ ρ ( z) ( ) > The rati f specific masses can be expressed as fllws: ( ) ( ) ( ) ( ) ( ) Λ ( ) ( ) ρ z p z T z = = ρ p T z T μg RΛ The last term is larger than unity if its expnent is negative: μg < RΛ Then : μ g K Λ> = = 34. R 83. m. In vertical mtin, the pressure f the parcel always equals that f the surrunding air, the latter depends n the altitude. The parcel temperature T parcel depends n the pressure... We can write: dtparcel dtparcel dp = dz dp dz p is simultaneusly the pressure f air in the parcel and that f the surrunding air. Expressin fr dt parcel dp By using the equatin fr adiabatic prcesses pv γ = cnst and equatin f state, we can deduce the equatin giving the change f pressure and temperature in a quasi-equilibrium adiabatic prcess f an air parcel: T parcel p γ γ = cnst (6)

16 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. 3 / Slutin where cp γ = is the rati f isbaric and ischric thermal capacities f air. By c V lgarithmic differentiatin f the tw members f (6), we have: dt T parcel parcel γ dp + = γ p Or dt T = (7) p γ parcel parcel γ dp Nte: we can use the first law f thermdynamic t calculate the heat received by the m parcel in an elementary prcess: dq = cv dtparcel + pdv, this heat equals zer in an μ adiabatic prcess. Furthermre, using the equatin f state fr air in the parcel m pv = RTparcel we can derive (6) μ dp Expressin fr dz Frm () we can deduce: where dp pgμ = ρg = dz RT T is the temperature f the surrunding air. On the basis f these tw expressins, we derive the expressin fr dt dz : parcel / dtparcel γ μg Tparcel = = G (8) dz γ R T In general, G is nt a cnstant If at any altitude, T = Tparcel, then instead f G in (8), we have : r γ μg Γ= = cnst (9) γ R 3

17 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. 3 / Slutin μg Γ= (9 ) c p... Numerical value: K K Γ= = m m..3. Thus, the expressin fr the temperature at the altitude z in this special atmsphere (called adiabatic atmsphere) is : ( ) ( ) T z = T Γ z ().3. Search fr the expressin f T ( z ) parcel Substitute T in (7) by its expressin given in (3), we have: dtparcel γ μg dz = T γ R T Λz parcel Integratin gives: parcel ( ) ( ) ( ) ( ) ( ) Tparcel z γ μg T Λz ln = ln T γ R Λ T Finally, we btain: Γ Λ ( ) T ( ) T Λz Tparcel ( z) = Tparcel ( ).4. Frm () we btain Λz Tparcel ( z) = Tparcel ( ) Λ T ( ) T ( ) Λ z << T, then by putting x =, we btain Λ z If ( ) Tparcel ( z) = Tparcel ( ) + x x Γ Γz T ( ) Γz T ( ) Γz parcel ( ) e parcel ( ) parcel T T T T ( ) ( ) Γz () 4

18 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. 3 / Slutin hence, ( ) ( ) T z T parcel parcel Γ z () 3. Atmspheric stability In rder t knw the stability f atmsphere, we can study the stability f the equilibrium f an air parcel in this atmsphere. At the altitude z T ( z ) T( z ), where parcel =, the air parcel is in equilibrium. Indeed, in this case the specific mass ρ f air in the parcel equals ρ '- that f the surrunding air in the atmsphere. Therefre, the buyant frce f the surrunding air n the parcel equals the weight f the parcel. The resultant f these tw frces is zer. Remember that the temperature f the air parcel T ( ) parcel z is given by (7), in which we can assume apprximately G = Γ at any altitude z near z = z. Nw, cnsider the stability f the air parcel equilibrium: Suppse that the air parcel is lifted int a higher psitin, at the altitude z (with d>), T ( z + d) = T ( z ) Γd and ( ) ( ) parcel parcel T z + d = T z Λ d. In the case the atmsphere has temperature lapse rate ( ) ( ) parcel Λ>Γ + d, we have T z + d > T z + d, then ρ < ρ '. The buyant frce is then larger than the air parcel weight, their resultant is riented upward and tends t push the parcel away frm the equilibrium psitin. Cnversely, if the air parcel is lwered t the altitude ( ) ( ) T z d < T z d and then ρ > ρ '. parcel z d (d>), The buyant frce is then smaller than the air parcel weight; their resultant is riented dwnward and tends t push the parcel away frm the equilibrium psitin (see Figure ) S the equilibrium f the parcel is unstable, and we fund that: An atmsphere with a temperature lapse rate Λ>Γis unstable. In an atmsphere with temperature lapse rate Λ <Γ, if the air parcel is lifted t a higher psitin, at altitude z + d (with d>), T ( z + d) < T( z + d), then parcel 5

19 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. 3 / Slutin ρ > ρ '. The buyant frce is then smaller than the air parcel weight, their resultant is riented dwnward and tends t push the parcel back t the equilibrium psitin. Cnversely, if the air parcel is lwered t altitude z d (d > ), ( ) ( ) T z d > T z d and then ρ < ρ '. The buyant frce is then larger than the parcel air parcel weight, their resultant is riented upward and tends t push the parcel als back t the equilibrium psitin (see Figure ). S the equilibrium f the parcel is stable, and we fund that: An atmsphere with a temperature lapse rate Λ<Γis stable. z T T parcel Λ > Γ z +d z z -d Tparcel Tparcel > T ρ parcel < T ρ parcel < ρ up > ρ dwn Γ Λ unstable T( z ) T Figure z T parcel T Λ < Γ z +d z z -d Tparcel Tparcel < T ρ parcel > T ρ parcel > ρ dwn < ρ up Λ Γ stable T( z ) T Figure 6

20 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. 3 / Slutin In an atmsphere with lapse rate Λ =Γ, if the parcel is brught frm equilibrium psitin and put in any ther psitin, it will stay there, the equilibrium is indifferent. An atmsphere with a temperature lapse rate Λ =Γis neutral 3.. In a stable atmsphere, with Λ <Γ, a parcel, which n grund has temperature T parcel ( ) > T ( ) and pressure p( ) equal t that f the atmsphere, can rise and reach a maximal altitude h, where ( ) T h = ( ) parcel T h. In vertical mtin frm the grund t the altitude h, the air parcel realizes an adiabatic quasi-static prcess, in which its temperature changes frm parcel ( ) ( ) T h = T h. Using (), we can write: Γ ( ) ( ) ( ) Λh Λ Tparcel Tparcel = = T( ) T h Λh T ( ) T ( ) Γ Λ Λh = Tparcel T ( ) ( ) T ( ) T parcel ( ) t ( ) Λ Λ ΛΓ - - parcel Λh = T T ( ) T ΛΓ( ) Λ Λ h= T( - - ) T ΛΓ parcel ( ) T ΛΓ ( ) Λ Λ Γ T( ) T Λ Γ parcel ( ) T Γ Λ = ( ) Λ S that the maximal altitude h has the fllwing expressin: h= T( ) Λ Γ ( T ( ) ) ( ) ( Tparcel ) Λ Γ Λ (3) 7

21 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. 3 / Slutin 4. Using data frm the Table, we btain the plt f z versus T shwn in Figure 3. 3 Altitude [m] D(.6 C;4 m) C(.8 C; 9 m) B(. C; 96 m) A( C; m) Temperature [ C] Figure We can divide the atmsphere under m int three layers, crrespnding t the fllwing altitudes: ) < z < 96 m, K Λ = = m ) 96 m < z < 9 m, Λ =, isthermal layer.. K 3) 9 m < z < 5 m, Λ 3 = = m In the layer ), the parcel temperature can be calculated by using () ( ) T parcel 96 m = K 94. K that is. C In the layer ), the parcel temperature can be calculated by using its expressin in Γz = exp T isthermal atmsphere T ( z) T ( ) parcel parcel ( ). 8

22 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. 3 / Slutin The altitude 96 m is used as rigin, crrespnding t m. The altitude 9 m crrespnds t 3 m. We btain the fllwing value fr parcel temperature: h And ( ) T parcel 9 m = K that is.8 C 4.. In the layer 3), starting frm 9 m, by using (3) we find the maximal elevatin = 3 m, and the crrespnding temperature 93.6 K (r.6 C). Finally, the mixing height is H = = 4 m. ( ) T parcel 4 m = K that is.6 C Frm this relatin, we can find ( ) Tparcel 9 m K and h = 3 m. Nte: By using apprximate expressin () we can easily find T ( ) parcel z = 94 K and 93.8 K at elevatins 96 m and 9 m, respectively. At 9 m elevatin, the difference between parcel and surrunding air temperatures is.7 K (= ), s that the maximal distance the parcel will travel in the third layer is.7/( Γ Λ 3 ) =.7/.3 = 3 m. 5. Cnsider a vlume f atmsphere f Hani metrplitan area being a parallelepiped with height H, base sides L and W. The emissin rate f CO gas by mtrbikes frm 7: am t 8: am M = 8 5 /36 = 3 3 g/s The CO cncentratin in air is unifrm at all pints in the parallelepiped and dented by C( t ). 5.. After an elementary interval f time dt, due t the emissin f the mtrbikes, the mass f CO gas in the bx increases by Mdt. The wind blws parallel t the shrt sides W, bringing away an amunt f CO gas with mass LHC ( t) udt. The remaining part raises the CO cncentratin by a quantity r Mdt LHC ( t) udt = LWHdC dc in all ver the bx. Therefre: 9

23 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Theretical Prblem N. 3 / Slutin dc u M + C() t = (4) dt W LWH 5.. The general slutin f (4) is : ut M C() t = Kexp + W LHu (5) Frm the initial cnditin C ( ) =, we can deduce : M ut C() t = exp LHu W (6) 5.3. Taking as rigin f time the mment 7: am, then 8: am crrespnds t t =36 s. Putting the given data in (5), we btain : 3 ( 36 s) (. 64). 3 mg/m C = =

24 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin Slutin Task... T = 5± C Vsamp ( T ) = mv With different experiment sets, V samp may differ frm the abve value within ±4 mv. Nte fr errr estimatin: δv and δv are calculated using the specs f the multimeter: ±.5% reading digit + n the last digit. Example: if V = 5mV, the errr δv = 5.5% +. =.7 mv 3 mv. ( ) Thus, Vsamp T = 574± 3 mv. All values f Vsamp ( T ) within mv are acceptable... Frmula fr temperature calculatin: Frm Eq (): V = V ( T ) α( T T ) samp samp V samp samp ( 5 C ) ( 7 C ) = 53.9 mv V = mv samp ( 8 C ) V = mv Errr calculatin: ( ) ( ) δv = δv T + T T δα samp samp Example: V = 495. mv, then δ Vsamp = ( 5 5) = 345mV. 35mV. samp Thus: samp ( 5 C ) V = 54±4 mv samp ( 7 C ) V = 484±4 mv

25 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin V samp ( 8 C ) = 464±5 mv The same rule fr acceptable range f V samp as in. is applied.... Data f cling-dwn prcess withut sample: t (s) Vsamp (mv) (±3mV) ΔV (mv) (±.mv)

26 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin The acceptable range f ΔV is ±4 mv. There is n fixed rule fr the change in ΔV with T (this depends n the psitins f the dishes n the plate, etc.).. Graph 53 5 V samp [mv] t [s] 3

27 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin The crrect graph shuld nt have any abrupt changes f the slpe..3. Graph - ΔV[mV] V samp [mv] The crrect graph shuld nt have any abrupt changes f the slpe Dish with substance t (s) Vsamp (mv) (±3mV) ΔV (mv) (±.mv)

28 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin

29 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin 3.. Graph [mv] 5 V samp t [s] The crrect Graph 3 shuld cntain a shrt plateau as marked by the arrw in the abve figure Graph ΔV[mV] V samp [mv] 6

30 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin The crrect Graph 4 shuld have an abrupt change in ΔV, as shwn by the arrw in the abve figure. Nte: when the dish cntains the substance, values f ΔV may change cmpared t thse withut the substance V is shwn in Graph 3. Value V = (53±3) mv. Frm that, T = 6.5 C can s be deduced. s 4.. V is shwn in Graph 4. Value V = (53±3) mv. Frm that, T = 6.5 C can s be deduced. s s s 4.3. Errr calculatins, using rt mean square methd: Errr f T s : variable. Therefre errr f the errr. δ A VT ( ) VT ( ) = + = + α s Ts T T A T s Errr fr A is calculated separately: s, in which A is an intermediate ( ) ( ) can be written as δt = δt + δa [ ] VT ( ) VT ( ) δα VT ( ) VT ( ) δ α VT ( ) VT ( s ) α s s = + in which we have: δ [ VT ( ) VT ( )] = [ δvt ( )] + [ δ VT ( )] s s Errrs f ther variables in this experiment: dt = C δ V( T ) = 3 mv, read n the multimeter. da =.3 mv/ C dv(t s )ª 3 mv Frm the abve cnstituent errrs we have: [ ] δ VT ( ) VT ( s ) 44. mv, in which d is 7

31 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin δ A. C Finally, the errr f T is: δ 5. C s T s Hence, the final result is: Ts =6±.5 C Nte: if the student uses any ther reasnable errr calculatin methd that leads t apprximately the same result, it is als accepted. 8

32 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin Task.. t (s).. T = 6 ±C.. Measured data with the lamp ff ΔV(T ) (mv) (±.mv) Values f ΔV(T ) can be different frm ne experiment set t anther. The acceptable values lie in between mv... Measured data with the lamp n t (s) ΔV (mv) (±.mv)

33 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin When illuminated (by the lamp) values f ΔV may change mv cmpared t the initial situatin (lamp ff)..3. Measured data after turning the lamp ff t (s) ΔV (mv) (±.mv) Pltting graph 5 and calculating k ( ) ( ) 3.. x = t ; y = ln ΔV T ΔV t Nte: ther reasnable ways f writing expressins fr x and y that als leads t a linear relatinship using ln are als accepted. 3.. Graph 5

34 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin. Graph 5 y = -.9x +.95, max dev:.335, r =.998 ln[δv(t )-ΔV(t)] t (s) k 3.3. Calculating k: =.9 s - and C =.69 J/K, thus: k = W/K C Nte: Errr f k will be calcu lated in 5.5. Students are nt asked t give errr f k in this step. The acceptable value f k lies in between W/K depending n the experiment set. 4. Pltting Graph 6 and calculating E 4.. kt x = exp C ; y =ΔV ( T ) ( ) Δ V t 4.. y = +9.7x, r =.994 Graph 6 ΔV(T )-ΔV(t)] 8 4 Graph 6 shuld exp(-k*t/c)

35 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin be substantially linear, with the slpe in between 5 5 mv, depending n the experiment set Frm the slpe f Graph 6 and the area f the detectr rifice we btain E = 4 W/m. The area f the detectr rifice is 3 4 δ Rdet Sdet = πrdet = π ( 3 ) = 5. 3 m with errr: R det = 5% Errr f E will be calculated in 5.5. Students are nt asked t give errr f E in this step. The acceptable value f E lies in between 6 W/m, depending n the experiment set Circuit diagram: Slar cell mv ma 5.. Measurements f V and I V (mv) (±.3 3mV) I (ma) (±.5.mA) P (mw) 8.6 ± ± ± ±

36 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin Graph W] P [m I [ma] 5.4. P max = 3.7±. mw Th e acceptable value f Pmax lies in between mw, depending n the experiment set Expressin fr the efficiency = 9 4 mm 6 S = 45 m cell Then P η = max max E S = cell. 58 Errr calculatin: δη max slar cell. δp max δ E δscell = ηmax + + Pmax E Scell, in which S cell is the area f the δ P P max max is estimated frm Graph 7, typical value ª 6 % 3

37 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin δ S S cell cell : errr frm the millimeter measurement (with the ruler), typical value ª 5 % E is calculated frm averaging the rati (using Graph 6): ΔVT ( ) ΔVt ( ) Eπ R B = = k k exp t C det α in which B is an intermediate variable, R det is the radius f the detectr rifice. E = kb π R α det Calculatin f errr f E: δ E δk δb δ R det δα = E k B Rdet α k is calculated frm the regressin f: We set k Δ T =ΔT( )exp t C, hence ln Δ = ln Δ ( ) k T T C t k/ C = m then k = mc Frm the regressin, we can calculate the errr f m: δ m ( r). % m δ k δ m δ C = k + m C We derive the expressin fr the errr f η max : δη max δp max δs cell B δr det m C = ηmax Pmax Scell B Rdet m C δ δ δ δα α Typical values fr ηmax 58. η max and ther cnstituent errrs: δ P max P max = 5% δ B δ m ; 6. % ;. % ; B m δ S S cell cell δ Rdet 5% ; 5% ; R det 4

38 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin δc C Finally: δ k E 3% ; 3%. 5% k ; δ δα 5 E ; α. % and δη max =.7% ; δηmax. 74 η max ( ) η = ± max % Nte: if the student uses any ther reasnable errr methd that leads t apprximately the same result, it is als accepted. 5

39 39th Internatinal Physics Olympiad - Hani - Vietnam - 8 Experimental Prblem / Slutin 6