Solutions of Indian National Physics Olympiad 2018

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1 Solutions of Indian National Physics Olympiad 018 Date: 8 January 018 oll Number: Time : 09:00-1:00 (3 hours) Maximum Marks: 75 I permit/do not permit (strike out one) HBCSE to reveal my academic performance and personal details to a third party. Besides the International Physics Olympiad (IPhO) 018, do you also want to be considered for the Asian Physics Olympiad (APhO) 018? For APhO 018 and its pre-departure training, your presence will be required in Delhi and Vietnam from April 8 to May 15, 018. In principle, you can participate in both Olympiads. Yes/No. Full Name (BLOCK letters) Ms./Mr.: Extra sheets attached : 0 Date Centre (e.g. Kochi) Signature ================================================== (Do not write below this line) Instructions 1. This booklet consists of 16 pages (excluding this sheet) and total of 7 questions.. This booklet is divided in two parts: Questions with Summary Answer Sheet and Detailed Answer Sheet. Write roll number at the top wherever asked. 3. The final answer to each sub-question should be neatly written in the box provided below each sub-question in the Questions & Summary Answer Sheet. 4. You are also required to show your detailed work for each question in a reasonably neat and coherent way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of these pages. 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. 6. Adequate space has been provided in the answersheet for you to write/calculate your answers. In case you need extra space to write, you may request additional blank sheets from the invigilator. Write your roll number on the extra sheets and get them attached to your answersheet and indicate number of extra sheets attached at the top of this page. 7. Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators. 8. Use blue or black pen to write answers. Pencil may be used for diagrams/graphs/sketches. 9. This entire booklet must be returned at the end of the examination. Table of Constants Speed of light in vacuum c m s 1 Planck s constant h J s h/π Universal constant of Gravitation G N m kg Magnitude of electron charge e C est mass of electron m e kg est mass of proton m p kg Value of 1/4πɛ N m C Avogadro s number N A mol 1 Acceleration due to gravity g 9.81 m s Universal Gas Constant 8.31 J K 1 mol 1 Molar mass of water 18.0 g mol 1 Permeability constant µ 0 4π 10 7 H m 1 Question Marks Score Total 75 HOMI BHABHA CENTE FO SCIENCE EDUCATION Tata Institute of Fundamental esearch V. N. Purav Marg, Mankhurd, Mumbai,

2 INPhO 018 Page 1 Questions & Summary Answers ANY ALTENATIVE METHOD OF SOLUTION TO ANY QUESTION THAT IS SCIENTIFICALLY AND MATHEMATICALLY COECT, AND LEADS TO THE SAME ANSWE WILL BE ACCEPTED WITH FULL CEDIT. PATIALLY CO- ECT ANSWES WILL GAIN PATIAL CEDIT. 1. In a nucleus, the attractive central potential which binds the proton and the neutron is called the Yukawa potential. The associated potential energy U(r) is [4] U(r) = α e r/λ r Here λ = fm (fm = m), r is the distance between nucleons, and α = MeV fm is the nuclear force constant (1 MeV = J). Assume nuclear force constant α to be A c. Here = h/π and h is Planck s constant. In order to compare the nuclear force to other fundamental forces of nature within the nucleus of Deuterium ( H), let the constants associated with the electrostatic force and the gravitational force to be equal to B c and C c respectively. Here A, B and C are dimensionless. State the expression and numerical values of A, B and C. A = Value of A = B = Value of B = C = Value of C = A = α c = 0.44 e B = 4πɛ 0 c = or q p q n 4πɛ 0 c = 0 C = Gm p or Gm nm p = to c c Detailed answers can be found on page numbers:. An opaque sphere of radius lies on a horizontal plane. On the perpendicular through the point of contact, there is a point source of light at a distance above the top of the sphere (i.e. 3 from the plane). (a) Find the area of the shadow of the sphere on the plane. [] Area =

3 INPhO 018 Page Questions & Summary Answers Last four digits of oll No.: S O Q (b) M P MP is the radius r of the shadow on ground. Let, MSP = θ. Triangles OSQ and SMP are similar. Area of the shadow = 3π. MP MS = OQ SQ r 3 = = OS OQ SQ = r = 3 3 A transparent liquid of refractive index 3 is filled above the plane such that the sphere is just covered with liquid. Find the area of the shadow of the sphere on the plane now. Area = S O i N Q [6] liquid M N P ay from S is refracted at N. New radius of shadow is MP = r. Also, N NP = θ. In NP N cos θ = NN NP = r + tan i r + tan i = = sec θ cos θ r = P N + N M = tan θ + tan i

4 INPhO 018 Page 3 Questions & Summary Answers which yields sec θ tan θ = tan i i = π θ sin θ = sin i 3 = 1 3 r = tan θ + tan i = = Area of the shadow = π. Detailed answers can be found on page numbers: 3. Consider an infinite ladder of resistors. The input current I 0 is indicated in the figure. (a) r r I 0 I n 1 I n Find the equivalent resistance of the ladder. Equivalent resistance = Let equivalent resistance be X Taking positive sign r I n 1 eq = X r + I n r r X + X = X X X = r ± r + 4r [] X = r + r + 4r (b) Find the recursion relation obeyed by the currents through the horizontal resistors r. You will get a relationship where I n will be related to (may be several) I i s, i < n, n > 0. elation : []

5 INPhO 018 Page 4 Questions & Summary Answers Last four digits of oll No.: From Kirchhoff s circuit law I n 1 = I n + I n (1) Using Kirchhoff s voltage law ri n + I n+1 I n = 0 () From Eq. (1) we see that I n+1 I n = (I n I n+1 ) (I n 1 I n ) = I n I n+1 I n 1 So that Eq. () becomes ( I n+1 + r ) I n + I n 1 = 0 (3) (c) which is the required recursion relation. Solve this for the special case = r to obtain I n and I n as explicit functions of n. You may have to make a reasonable assumption about the behaviour of I n as n becomes large. I n = I n = First method: One can see that I n 1, I n, I n+1 form geometric progression. Also I 1 I 0 I 0 I 1 X XI I 1 = I 0 (see the figure to the left) + X ( ) I = I 1 + X = I 0 (see the figure to the right) + X ( ) n Hence I n = I 0 + X ( ) n ( ) n For r =, I n = I 0 = I 0 k n = I 0 [4] Using Eq. () I n = I 0 ( 5 1 ) ( ) n 1 ( = I 0 ) n 1 Second method: For = r, Eq. (3) becomes I n+1 3I n + I n 1 = 0 (4)

6 INPhO 018 Page 5 Questions & Summary Answers We can solve this linear recursion relation by assuming the ansatz I n ρ n. This leads to ρ 3ρ + 1 = 0 so that we have ρ = 3 ± 5 This means that the general solution to (4) is ( ) n ( ) n I n = A + B (5) The second term grows without bound with increasing n - so that for an infinite ladder we must have B = 0. Also when n = 0, I 0 = A. Thus (d) ( ) n 3 5 I n = I 0 ( ) ( I n = I n 1 I n = I 0 ) n 1 If the ladder is chopped off after the N-th node (so that I N+1 = 0) what will the form of I n I N be for n N? I n I N = First method: Consider the end part of the ladder as shown below. I N 3 F I N I N I N 1 I N 1 E C A Voltage drop across AB = I N = I N I N = I N and I N 1 = 3I N. Voltage drop across CD = 5 3 3I N = I N 1 I N 1 = 5I N and I N = 8I N. Voltage drop across EF = I N = I N I N = 13I N and I N 3 = 1I N.. This is depicted in figure below. 1I N F 8I N 13I N D D 3I N 5I N I N B B I N I N I N [4] E C A

7 INPhO 018 Page 6 Questions & Summary Answers Last four digits of oll No.: This looks like the terms of Fibonacci sequence. Fibonacci sequence is 1,1,,3,5,8,13...etc in which nth term of Fibonacci sequence is the sum of previous two terms i.e. F n = F n 1 + F n, F 1 = F = 1. In other words: In general I N 1 = F 4, I N = F 6, I N 3 = F 8, I N 4 = F Second method: See Eq. 5. If I N+1 = 0, we must have I N r = F r+ I N (6) I n = F I (N n)+ N (7) where ρ,1 = 3± 5. Thus B = A and thus I n = ρn 1 ρn I N ρ N 1 ρn I n = A Aρ N Bρ N+1 = 0 ( ρ1 ρ ( ( ) ρ1 N+1 ρ ( ρ1 ) N+1 and thus ρ n 1 ρ n ( ) ) N+1 ρ1 ρ ρ 1 ρ [ ] ) N+1 = ρ1 n N 1 ρ n N 1 ρ ρ 1 ρ Substituting the values, we get I n = 1 ( ) N+1 n ( ) N+1 n (8) I N 5 Equations (7) and (8) are equivalent. Generalization if r : Solution to Eq. (3) is Thus ρ,1 = b ± b 4 where b = + r I n = ρ [ ] 1ρ ρ n N 1 1 ρ n N 1 I N ρ ρ 1 1 [ ] = ρ N+1 n ρ N+1 n 1 ρ ρ 1 (9) (10) Alternate form of Eqs. (7), (8) and (10) are also accepted. Detailed answers can be found on page numbers:

8 INPhO 018 Page 7 Questions & Summary Answers 4. An hour glass is placed on a weighing scale. Initially all the sand of mass m 0 kg in the glass is held in the upper reservoir (ABC) and the mass of the glass alone is M kg. At t = 0, the sand is released. It exits the upper reservoir at constant rate dm dt = λ kg/s where m is the mass of the sand in the upper reservoir at time t sec. Assume that the speed of the falling sand is zero at the neck of the glass and after it falls through a constant height h it instantaneously comes to rest on the floor (DE) of the hour glass. Obtain the reading on the scale for all times t > 0. Make a detailed plot of the reading vs time. eading on the scale : A D C E B h [9] eading (kg) time Velocity of grain when reaches to bottom v = gh. At t = 0, sand is released. Time taken t 1 = h/g. 0 < t < t 1 : For t < 0, the scale was (m + M)g. As t = 0 t 1, more sand will enter the mid air

9 INPhO 018 Page 8 Questions & Summary Answers Last four digits of oll No.: portion and hence the reading will drop, λ = dm m = λt dt (11) Weight (W 1 ) = (M + m)g λt 1 g (1) t 1 < t < t Here t = m/λ is the time when all the sand has left the upper reservoir. Just after t = t 1, force on the scale will partly due to weight of sand as given in above equation and partly due to impulse provided by falling sand. Impulse force = v dm dt = λ gh (13) Weight (W ) = [(M = m)g λt 1 g] + λ gh = (M + m)g (14) t < t < t 3 Here t 3 = m/λ+ h/g. This is the time span between the sand has left upper reservoir to when all sand has reached the floor of the hour glass. The weight is For t > t 3, weight is eading (kg) (m + M) + λ h/g (m + M) (m + M) λ h/g (0, 0) W 3 = W + λ(t t )g (15) W 4 = (M + m)g t 1 t t 3 time Detailed answers can be found on page numbers: 5. (a) Consider two short identical magnets each of mass M and each of which maybe considered as point dipoles of magnetic moment µ. One of them is fixed to the floor with its magnetic moment pointing upwards and the other one is free and found to float in equilibrium at a height z above the fixed dipole. The magnetic field due to a point dipole at a distance r from it is B( r) = µ 0 (3( µ ˆr)ˆr µ) 4πr3 Obtain an expression for the magnitude of the dipole moment of the magnet in terms of z and related quantities. [3] µ =

10 INPhO 018 Page 9 Questions & Summary Answers Given B( r) = µ 0 [3( µ ˆr)ˆr µ] 4πr3 ˆr = ˆk,r = z B(z) = µ 0 4πz 3 [3µˆk µˆk] = µ 0 4πz 3 [µˆk] B(z) = µ 0µ πz 3 The magnets are oppositely facing along ẑ. The force on the upper magnet due to the lower fixed magnet is F = 3µ 0µ πz 4 The repulsive magnetic force is balanced by the weight acting down. µ 0 Mg = 3 πz 4 µ (16) µ = πz 4 Mg 3 µ 0 ( ) Mg 4π 1/ µ = z (17) 6 (b) i. Consider a ring of mass M r rotating with uniform angular speed about its axis. A charge q is smeared uniformly over it. elate its angular momentum S r to its magnetic moment ( µ r ). µ r = µ 0 Let radius of the ring to be and it is rotating with the angular speed ωẑ. Dipole moment can be expressed as µ r = current area = qω π π ẑ S r = M r ω ẑ µ r = q Sr M r [1 1 / ] ii. Assume that the electron is a sphere of uniform charge density rotating about its diameter with constant angular speed. Also assume that the same relation as in the previous part holds between its angular momentum S and its magnetic dipole moment ( µ B ). Further assume that S = h/(π) where h is Planck s constant. Calculate µ B. [1] µ B = µ B = e m e = to A m

11 INPhO 018 Page 10 Questions & Summary Answers Last four digits of oll No.: iii. Assume that the sole contribution to the dipole moment of a ZnFe O 4 molecule comes from an unpaired electron. Also assume that the magnets in the part (5a) are 0.48 kg each of ZnFe O 4 and the unpaired electrons of the molecules are all aligned. Calculate the height z. (Note: The molecular weight of ZnFe O 4 = 11) [3] z = The system (dipole) is made of ZnFe O 4 and has one electron/molecule. The number of molecules is = Mass of ZnFe O 4 in gm Molecular weight of ZnFe O 4 N A (18) = M A N A (19) iv. Assuming that each molecule of ZnFe O 4 has one unpaired electron and each contributes to dipole moment. Further all of them are perfectly aligned. Hence dipole moment is Using Eq. (17) µ = MN Aµ B A 6.60 z (cm) 6.80 In an experiment µ B is aligned along a magnetic field of 1 T. It is flipped in a direction anti-parallel to the magnetic field by an incident photon. What should be the wavelength of this photon? Wavelength = (0) Wavelength = hc = 1.00 to 1.10 cm (1) µ B B Detailed answers can be found on page numbers: [1 1 / ] 6. The Van der Waals Gas: Consider n mole of a non-ideal (realistic) gas. Its equation of state maybe described by the Van der Waals equation ( ) (V ) P + an V n b = T where a and b are positive constants. We take one mole of the gas (n = 1). You must bear in mind that one is often required to make judicious approximations to understand realistic systems. (a) For this part only take a = 0. Obtain expressions in terms of V, T and constants for i. the coefficient of volume expansion (β); [1] β =

12 INPhO 018 Page 11 Questions & Summary Answers β = 1 V ( ) dv = V b dt p V T () ii. the isothermal compressibility (κ). κ = [1] κ = 1 V ( ) dv dp T = (V b) V T (b) Criticality: The Van der Waals gas exhibits phase transition. A typical isotherm at low temperature is shown in the figure. Here L (G) represents the liquid (gas) phase and at P LG there are three possible solutions for the volume (V L, V LG, V G ). As the temperature is raised, at a certain temperature T c, the three values of the volume merge to a single value, V c (corresponding pressure being P c ). This is called the point of criticality. As the temperature is raised further there exists only one real solution for the volume and the isotherm resembles that of an ideal gas. P P LG (L) V L V LG (LG) V Low temperature High temperature V G (G) Isotherms of a Van der Waals gas at low and high temperatures P V A family of isotherms of a Van der Waals gas (3) i. Obtain the critical constants P c, V c and T c in terms of a, b and. P c = V c = T c = [4] At criticality, the curve is a cubic like function. Hence ( ) dp dv T = ( d ) P dv T = 0

13 INPhO 018 Page 1 Questions & Summary Answers Last four digits of oll No.: Above condition along with Van der Waal s equation can be used to obtain V C = 3b (4) T C = 8a (5) 7b P C = a 7b (6) ii. Obtain the values of a and b for CO given T c = K and P c = N m. a = b = [1] a = 0.37 m 6 Pa/mol and b = m 3 /mol. iii. The constant b represents the volume of the gas molecules of the system. Estimate the size d of a CO molecule. d = b = N A d 3 or N A 4πr 3 /3 d = m to m (c) The gas phase: For the gaseous phase the volume V G b. Let the pressure P LG = P 0, the saturated vapour pressure. i. Obtain the expression for V G in terms of, T, P 0 and a. V G = [1] [1 1 / ] P 0 VG T V G + a = 0 V G = T ( ( 1 ± 1 4aP ) ) 1/ 0 P 0 T Taking positive sign since in the ideal gas limit a 0, V I = T P 0 for one mole. V G T P 0 a T ii. State the corresponding expression for V I for an ideal gas. V I = [ 1 / ]

14 INPhO 018 Page 13 Questions & Summary Answers V I = T P 0 iii. Obtain (V G V I )/V I for water given T = C, P 0 = Pa, b = m 3 mol 1 and a = 0.56 m 6 Pa mol. Comment on your result. V G V I V I = Comment: [] Comment 1: For ideal gas, a T vanishes as T. V G V I = ap 0 V I (T ) to Comment : Answer is negative, indicating attractive intermolecular forces. (d) The liquid phase: For the liquid phase P a/vl. i. Obtain the expression for V L. V L = In this phase P a V L V L = Hence a VL (V L b) = T a 1 ± 1 4bT T a [1 1 / ] Taking negative sign since as T 0, V L b. V L = a 1 1 4bT T a ( b 1 + bt ) a ii. Obtain the density of water (ρ w ). You may take the molar mass to be kg mole 1. ρ w = [1 1 / ]

15 INPhO 018 Page 14 Questions & Summary Answers Last four digits of oll No.: ρ w b = to Kg/m 3 (7) iii. The heat of vaporization is the energy required to overcome the attractive intermolecular force as the system is taken from the liquid phase (V L ) to the gaseous phase (V G ). The term a/v represents this. Obtain the expression for the specific heat of vaporization per unit mass (L) and obtain its value for water. [] L = Value of L = VG a L molar mass = V L V dv a V L L to J/Kg Detailed answers can be found on page numbers: 7. A small circular hole of diameter d is punched on the side D and the near the bottom of a transparent cylinder of diameter d D. The hole is initially sealed and the cylinder is filled with water of density ρ w. It is then inverted onto a bucket filled to the brim with water. The seal is removed, air rushes in and height h(t) of the water level (as measured from the surface h level of the water in the bucket) is recorded at different times (t). The figure below and the table in part (c) illustrates this process. Assume that air is an incompressible fluid with density ρ a and its motion into the cylinder is a streamline flow. Thus its speed v is related to the pressure difference P by the Bernoulli relation. Take the outside pressure P 0 to be atmospheric pressure = Pa. (a) Obtain the dependence of the instantaneous speed v w of the water level in the cylinder on h. v w = [3] at time t from pressure balance equation P 0 = p + hρ w g P = P 0 p = hρ w g From Bernoulli s principle 1 v ρ a = hρ w g = P ρw g v = h ρ a

16 INPhO 018 Page 15 Questions & Summary Answers The speed of air gushing in through the small hole is related to speed v w of fall of water level by πd 4 v = πd 4 v w (8) v w = d ρw g h (9) D ρ a (b) Obtain the dependence of h on time. h = [3] (c) (d) v w = dh dt = d ρw g h D ρ a h = h0 d ρw g D t ρ a The table gives the height h as function of time t. Draw a suitable linear graph (t on x axis) from this data on the graph paper provided. Two graph papers are provided with this booklet in case you make a mistake. t(sec) h(cm) From the graph and the following data: D = 6.66 cm, ρ a = 1.14 kg/m 3, ρ w = kg/m 3 obtain i. The height h 0 at t = 0. h 0 = [4] [5].0 h 0 (cm) 4.0 ii. The value of d. d = 0.7 slope( cm/s) d (cm) 0.14 iii. The initial speed (v w ) of the water level. v w (t = 0) =

17 INPhO 018 Page 16 Questions & Summary Answers Last four digits of oll No.:.35 v w (cm/s).65 Detailed answers can be found on page numbers:

Indian National Physics Olympiad 2017

Indian National Physics Olympiad 2017 Indian National Phsics Olmpiad 2017 Date: 29 th Januar 2017 Solutions Roll Number: 1 7 0 0-0 0 0 0-0 0 0 0 Time : 09:00-12:00 (3 hours) Maimum Marks: 75 I permit/do not permit (strike out one) HBCSE to