Gr0877 Solutions 3w71

GR0877 SOLUTIONS Detailed solutions to the GRE Physics Test

by physicsworks (unless otherwise credited)

Version 1.1.

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September 17, 2012

1. (B) Once the ball has been released, the only acting force on it is a gravitational force (the problem ignores friction). Since gravity has a zero horizontal component, the ball, as viewed from above, moves in a straight line. Thus, one should eliminate all but choices (B) and (D). The car is moving to the right, while the ball is thrown perpendicularly to this direction, so the initial velocity of the ball is directed to the south-east (and preserve this direction in the future). This is choice (B). 2. (D) Horizontal and vertical motions of the object are independent (if we ignore friction). Therefore, no one really needs the initial horizontal component of the ball’s velocity to determine how long it will take to cover some distance in the vertical direction. The initial vertical component of the velocity is zero, so from H = gt2 /2 one gets H = (9.8 · 22 )/2 = 9.8 · 2 = 19.6 (m). 3. (E) The dissipating power is P = U 2 /R, where U is the voltage across the resistor R. So, if you double U , the power will quadruple. Comment: you cannot use P = U I, because if you change U the current I will also change. 4. (E) The loop exactly lies on a magnetic field line of the long wire. Thus, at every point of the loopR the magnetic field due to the wire is parallel to the direction of the current I2 . Fmag = I2 (dl × B). 5. (A) What else could be? 6. (E) The nth shell can accommodate 2n2 electrons. Thus, the total number of electrons is 2(12 + 22 ) = 2 · 5 = 10. p 7. (C) From mv 2 /2 = 3kT /2 it’s easy to obtain 3kT /m which is choice (C). 8. (D) This is a straightforward application of the Stefan-Boltzmann law. If the temperature is increased by the factor of two, the radiated power and the mass of the ice that can be melt will increase by the factor of 24 = 16. 9. (E) Statements II, I and III represent, respectively, the first, the second and the third Kepler’s law. Thus, choice (E). p 10. (B) From the conservation of energy one has ks2 /2 = mv 2 /2 and s = v m/k. 11. (C) The energy levels of a quantum harmonic oscillator are En = ~ω n + 12 , for n = 0, 1, 2, . . . The ground state is the lowest-energy state with n = 0 and E = 12 ~ω. 12. (C) One of the Bohr’s postulates is that the angular momentum of the electron is an integer multiple of ~: mvrn = n~. From this, one has mv = n~/rn which is (C). 13. (A) On a log-log plot a function of the form y = axm will appear as a straight line, where m is the slope of the line and a is the y-value corresponding to x = 1. Indeed, log10 y = log10 y2 − log10 y1 log10 (axm ) = log10 a + m log10 x. Thus, if x = 1, then y = a and m = = log10 x2 − log10 x1 log10 (y2 /y1 ) log10 (100/10) 1 ). In our case, for x = 1, y ≈ 6 and m = = . Therefore, log10 (x log10 (300/3) 2 √2 /x1 y = 6 x. 14. (B) (See a wonderful book by John R. Taylor An introduction to error analysis, Chapter 7) If we have N separate measurements of a quantity x: 1 ± σ1 , x2 ± σ2 , . . . , xN ± σN , then Px N w i xi , where wi = 1/σi2 , and the the best estimate is the weighted average xwav = Pi=1 N i=1 wi P −1/2 p √ N 1 1 −1/2 uncertainty in xwav is σwav = w . Thus, σ = + = 4/5 = 2/ 5. i wav 2 2 i=1 1 2 1

1 F



1 R1

1 R2



− . Here R1 is the radius of 15. (E) a general formula for lenses = (n − 1) curvature of the lens surface closest to the light source and R2 is the radius of curvature of the lens surface farthest from the light source. Sign convention: the radius of curvature is positive if the center of spherical surface lies to the right from the lens and is negative if the center of the spherical surface lies to the left. Let the source of light be situated to the left from each of the lenses. Then, for (A) one has R1 = −R, R2 = R and FA ∼ −R/2. For (B): R1 = ∞, R2 = R, FB ∼ −R. (C): R1 = ∞, R2 = −R, FC ∼ R. For (D) and (E): R1 = R, R2 = −R, F ∼ R/2. But RE < RD , therefore FE < FD . 16. (D) When unpolarized light es through the first polarizer it loses one-half of its intensity. Why half? Here is an explanation. A beam of unpolarized light is nothing but a uniform mixture of linear polarizations at all possible angles. When a perfect polarizer is placed in a polarized beam of light, the intensity of the light that es through is I = I0 cos2 θ (Malus’ law). The average value of cos2 θ is 1/2, therefore the intensity is half of the initial: If irst = I0 /2. After the first polarizer the light is linearly polarized and the fraction of light that es through the second polarizer is (again, according to Malus’ law): I = If irst cos2 45◦ = I0 /4 or 25%. 17. (A) Let the Gaussian surface be a cylinder of radius r and length L. The axis of the 1 λ . cylinder coincides with the wire. Gauss’s law: 2πRLE = λL/0 , from which E = 2π 0 r The answer can also be found by elimination of choices, since (A) is the only choice with correct dimensions. 18. (C) This is an application of Lenz’s law. As the magnet enters the loop, the flux through the loop increases. The induced current generates a magnetic field that is opposing the bar magnet’s field. This current is counter-clockwise (that is, form b to a). As the magnet leaves the loop, the flux decreases and the current flows clockwise. Choice (E). 19. (A) According to Wien’s law λ1 T1 = λ2 T2 = b, from which λ2 = λ1 T1 /T2 = 500 · 6000/300 = 104 (nm) or 10 µm. 20. (A) The temperature dependence of the radiation is Tr ∝ a−1 , where a is a cosmic scale factor. Thus, when the temperature was higher by the factor of 4 (12/3 = 4), the distances were four times less than today. See Misner, Thorne and Wheeler, Gravitation, Chapter 28. 21. (C) For an adiabatic process P V γ = const. If one use P V /T = const for an ideal gas, one has P = const · T /V . Plugging this into P V γ = const yields T V γ−1 = const. p √ 22. (C) E = γme c2 ≡√ 4me c2 . From this, γ = 4 and v = c 1 − 1/γ 2 = 15c/4. The momentum: p = γme v = 15me c. 23. (B) Imagine the earth (S frame, x-axis is to the right and es through spaceships) and two spaceships approach to the earth from the left and right with equal speeds v. Let’s move to a reference frame of the left spaceship (system S 0 , two systems are oriented the same way). Then, according to the velocity addition rule, velocity of the right spaceship in this frame of ur − V reference is: u0r = , where ur = −v is the velocity of the right spaceship in the S 1 − ur V /c2 −2v frame and V denotes the speed of S 0 relative to S, V = v. Thus, u0r = . But this 1 + v 2 /c2 is exactly the speed with which two spaceships approach one another (as seeing from their reference frames). And we know from the length contraction formula that, if the relative speed between two frames is u, a stick at rest with respect to one reference pframe is observed from the other reference frame to be contracted by the factor of γ = 1/ 1 − u2 /c2 = l0 /l. Problem statement suggests γ = l0 /l = 1/0.6 = 5/3. Substituting u0r into the equation for γ and working with units in which c = 1: 2

5 1 1 + v2 =q = , 4v 2 3 1 − v2 1 − (1+v 2 )2 5 − 5v 2 = 3 + 3v 2 , 2 = 8v 2 , v = 0.5 or v = 0.5c as in choice (B). 24. (B) To the observer, a meter stick l0 that es through with the speed v = 0.8c is observed p 5 (sorry for tautology) to be contracted by the factor of γ = 1/ 1 − (0.8)2 = . Thus, the 3 l0 3 time it takes the stick to the observer is ∆t = = = 2.5 (ns). γv 5 · 0.8 · 3 · 108 25. (E) The only choice which guarantees that the functions are both normalized and mutually orthogonal is (E). 26. (D) The probability that the electron would be found between r and r + dr is P = |ψ|2 dV = |ψ|2 4πr2 dr = p(r)dr. The most probable value is given by the maximum of the probability density p(r) (take the second derivative if you want to convince yourself that it is indeed 4πr2 2 the maximum): dP/dr = 4πr2 d|ψ|2 /dr + 8πr|ψ|2 = 0. Plugging in: − 3 e−2r/r0 + πa0 a0 8πr −2r/r0 = 0. From this, one has r = a0 which is nothing else but the Bohr radius. e πa30 27. (C) This is an application of the energy-time uncertainty principle: ∆E∆t & h. Rewriting 1 1 1 ∆E as h∆ν one has ∆ν ∼ = = ≈ 600 (MHz), which is closest to (C). ∆t τ 1.6 · 10−9 28. (D) 2(kx2 /2) = K(x/2)2 /2 gives K = 8k. 29. (C) In elastic√ collisions energy is conserved. So M v 2 /2 = M (v/2)2 /2 + M u2 /2 from which one has u = 23 v. 30. (D) I bet one half of the test takers who did this wrong (according to the official practice book only 51 per cent of all 14,395 examinees who took the Physics Test between July 1, 2007 and June 30, 2010 answered this question correctly) simply mixed up choices (C) and (D). 31. (C) Applying Archimedes’ principle: ρgV = ρwater · g(3V )/4 + ρoil · gV /4 =⇒ ρ = 34 ρwater + 1 ρ = 750 + 200 = 950 (kg/m3 ). 4 oil 32. (A) Who says Bernoulli’s principle is unlikely on the PGRE? According to this principle, 2 ρv 2 one has P0 + 20 = P + ρv2 (there is no ρgz term here because of horizontality of the pipe). Conservation of mass gives: ρv0 S = ρvS =⇒ v = 4v0 , since S = πr2 = (πr02 )/4 = S0 /4. ρv 2 ρv 2 Plugging v = 4v0 into the first equation one finally obtain P = P0 + 20 −16 20 = P0 − 15 ρv02 . 2 + νR dV , where 33. (E) According to the first law of thermodynamics dS = T1 dU + PT dV = mc dT T V c is the specific heat (per one kilogram). Assuming water is incompressible fluid one has dS = mc dT . Integrating this from T1 to T2 one obtain mc ln TT12 . T 34. (C) The first law of thermodynamics: Q = 32 νR∆T , Q0 = 32 νR∆T + p∆V = 32 νR∆T + νR∆T = 52 νR∆T . Here we have used P V = νRT to rewrite the second term. Thus, Q0 = 5Q/3. 35. (B) Assuming a heat pump is an ideal Carnot engine for its efficiency one has η = W/QH = 1 − TC /TH , where W is the work done by the system, QH is the heat put into the system, TC , TH are the absolute temperatures of the cold and hot reservoirs. W = QH (1 − TC /TH ) = 15000 · (1 − 280/300) = 15000/15 = 1000 (J). 3

36. (A) The magnetic energy is LI 2 /2. In LC-contour with initial conditions q(t = 0) = q0 and I0 ≡ (dq/dt)0 = 0 the charge on the capacitor is a cosine function of time and the current is, therefore, a sine function. Thus, magnetic energy is a square of the sine function of time with some proportionality factor. That is, it es through the origin and has a form similar to (A). 37. (E) Clearly, the electric field is in the −x direction, therefore, only (C) and (E) survive. The magnitude of the electric field at P is E = 2Eq cos θ = 2E−q cos θ where Eq = E−q are the magnitudes of the electric field at point P due to the charges q and −q, respectively; θ is the angle between x-axis and the line ing through the point P and charge +q. 1 1 ql 2q l/2 l/2 p =⇒ E = . For r l we have E ≈ . cos θ = p 2 2 2 2 2 2 4π0 r + l /4 r + l /4 4π0 r3 r + l /4 µ0 I . From the 2πa picture (and right-hand rule) it is obvious that the magnetic field at point P due to the horizontal wire has the same magnitude, but opposite direction, than of the magnetic field at this point due to the vertical wire. Thus, the net magnetic field is zero. p 39. (C) The lifetime of a muon in the laboratory is τl = γτ , where γ = 1/ 1 − 16/25 = 5/3. · 3 · 108 · 2.2 · 10−6 = 880 (m). The mean distance traveled is l = vτl = vγτ = 4·5 5·3

38. (E) Magnetic field at distance a from a wire carrying a current I is B =

40. (B) Conservation of momentum immediately suggests that after the decay particle m and massless particle have the same momentum p. From the relativistic energy-momentum equation (we use units in which c = p 1) E 2 − p2 = m2 one gets E = p for massless particle. The conservation of energy: M = p+ m2 + p2 . Moving p to the left-hand side and squaring results in (M − p)2 = m2 + p2 or M 2 − 2M p = m2 , from which one has p = (M 2 − m2 )/(2M ). Clearly, the decay is possible if M > m. 41. (B) According to Einstein’s photoelectric equation: hν = ϕ + Kmax , where Kmax is the maximum kinetic energy of the ejected electron. Kmax = eV , where V is the stopping potential. From these two equations: V = he ν − ϕ/e and the slope is he . 42. (E) From the picture we see that the period of oscillations (this is 2π phase difference) is approximately 6 cm, while the difference between two points of equal voltage-level (say, zero-level) for two waveforms is 2 cm. Thus, the phase difference between waveforms is 2 · 2π/6 = 2π/3. 43. (D) If you took solid state physics course, even at introductory level, you should this. 44. (D) The attraction in cooper pair is due to the electron-phonon interaction. c + vo f , where fo is observed frequency, c + vs c is the speed of sound waves in the medium, vs and vo are the speed of the source and the observer relative to the medium, respectively. Sign convention: the speed vs is taken to be positive if the source is moving away from the observer at speed vs , while vo is taken to be positive if the observer is moving toward the source at speed vo . In our case, the source and observer are moving in such a way that vo = vs (and equal to the speed of the wind w). Thus, we have no Doppler effect at all : fo = f .

45. (C) According to the general Doppler formula fo =

c 46. (D) The first minimum is determined by d sin θ = λ or d sin θ = c/ν. Thus, ν = ≈ d sin θ 350 500 = ≈ 7 · 500 = 3500 (Hz). 0.14 · 0.7 0.14

4

47. (D) For a pipe of length L, closed at one end and open at the other the resonant frequencies v nv = , where n = 1, 3, 5, . . . and v is the speed of the sound. Thus, are given by f = λ 4L v a fundamental frequency is f (n = 1) = and the next harmonic has a frequency f (n = 4L 3v . For this next harmonic f (n = 3) = 3f (n = 1) = 3 · 131 = 393 (Hz). 3) = 4L 48. (C) We have one NOR-gate with two inverted inputs, one AND-gate and one NAND-gate (you should this from your electrical engineering course to solve this problem). For NOR-gate with inverted inputs the output is A + B, for NAND-gate C · D. Thus, E = A + B · C · D. 49. (D) The only choice in which free atoms involved is (D). 50. (C) Dimension analysis works fine here. However, you can also derive needed expression using Newton’s second law and Bohr’s quantization rule. Indeed, the former gives mv 2 /r = Ze2 /(4π0 r2 ), while the later mvr = n~. From these two equations (excluding v): rn ∝ n2 . Potential energy: Epot = −Ze2 /(4π0 r). The total energy: E = Ekin + Epot = me2 Z mZ 2 e4 −Ze2 /(8π0 r). Substituting rn into the expression for the total energy yields E ∝ . n2 To for the motion of the nuclei we can treat m in the last equation as the reduced mass. Thus, choice (C). 51. (D) I: true; II: false, an atom can emit photons of light only with energy equal to the energy difference between two quantum states; III: true. 52. (C) According to Bragg’s law 2d sin θ = nλ, where θ is the angle between incident ray and λ ≈ 2λ = 0.500 (nm). scattering planes. In our case n = 1 and d = 2 sin θ 53. (D) According to the problem statement, the angular momenta are the same: m1 v1 R = m2 v2 R, therefore m1 /m2 = v2 /v1 . For the orbital periods one has T1 = 2πR/v1 and T2 = 2πR/v2 (orbits are circles!). Thus, m1 /m2 = v2 /v1 = T1 /T2 = 3. Think about why cannot we apply Kepler’s third law in its usual form. 54. (E) A solar-mass black hole would exert no more gravitational pull than our sun. Thus, the orbits would remain unchanged. s 1+β 4 λ0 = ≈ . Thus, 16 − 16β = 55. (A) Applying relativistic Doppler effect formula: λlab 1−β 3 9 + 9β =⇒ β = 7/25 = 0.28. 56. (D) To fly due north a pilot should point the plane in such a direction that v + u k SN , where v is the velocity of the plain in still air, u is the velocity of the wind, and SN is the south-north line. According to the velocity addition formula, the velocity of the plane in the 2 north√direction is V = v + u, √ V ⊥ u. From this (v)2 , V 2 + u2 √ = v2, √ we have (V − u) = √ V = v 2 − u2 , t = L/V = L/ v 2 − u2 = 500/ 2002 − 302 = = 50/ 400 − 9 = 50/ 391 (h). 57. (B) For both figures accelerations of masses 2m and m are the same and can be calculated via Newton’s second law for the whole system of two bodies: 3ma = F , a = F/(3m). For the first figure ma = F12 and F12 = F/3. For the second figure 2ma = F12 and F12 = 2F/3. Thus, choice (B).

5

58. (A) The only force that provides acceleration a for the block B is the friction force due to the block A. Therefore, by Newton’s second law, this force is equal to mB a = 10 kg · 2 m/s2 = 20 N. 59. (C) Elimination of choices. (A): wrong, the period must depends on a; (B): for a particular case a = g the period blows up — wrong; (C): reasonable; (D): for a = 0, T = 0 — wrong; (E): for a = 0, T blows up — wrong. Thus, choice (C). 60. (C) Take a point (x, 0, 0) on the x-axis. At this point the magnetic field due to the wire µ0 I along zˆ is µ0 I/(2πx); due to the wire crossing the first and third quadrants is ; 2πx sin 45◦ µ0 I due to the wire crossing the second and fourth quadrants is . Thus, the net 2πx sin 45◦ √ µ0 I magnetic field at point (x, 0, 0) is (1 + 2 2). 2πx q 61. (E) Newton’s second law: mv 2 /R = qvB or · d = const (v and B are constants). If we m q double , the value of d should decrease by the factor of 2. m 1 · 10−9 +100 ≈ 100+100 = 8.85 · 10−12 200 (N · m2 /C). Check the table of information in your test book for the numeric value of 0 (if you don’t ).

62. (E) Gauss’s law: Φtotal = ΦA +Φ = q/0 ; Φ = q/0 −ΦA = =

63. (D) Given example is a β + decay. Beta decay is mediated by the weak force. √ 64. (D) Eigenvalues of L2 are ~2 l(l+1), eigenvalues of Lz are m~. Thus, ( 2~)2 = ~2 l(l+1), from which l = 1. For a given l, there are 2l + 1 different values of m: m = −l, −l + 1, . . . , l − 1, l. In our case, m = −1, 0, 1 and, therefore, the possible values of Lz are −~, 0, ~. 65. (C) Step-by-step elimination. I: true: En = ~ω(n + 1/2), this is indeed evenly spaced spectrum; II: false, the potential function is a quadratic function, not linear; III: false, the expectation value of both potential and kinetic energy is half the total energy hV i = hT i = 1 ~ω(1 + 1/2). Thus, for n = 0, hT i = 6 0. IV: true. 2 66. (D) To for the fact that the nucleus has non-infinite mass and moves around atom’s CM one must replace the electron mass (muon mass for the muonic atom) with the reduced mass µ. Noting that Rydberg R is proportional to µ we conclude that if we replace electron mp mµ mp + me µmuon = . me with the muon mµ the R is changed by the factor of µelectron mp + mµ mp me 67. (D) At any instant of time an electric field inside the parallel-plate capacitor is E =

σ = 0

Q , where σ is a surface density of a charge Q on a positive plate. Differentiating this with 0 S dE dQ/dt I 9 1012 V 12 respect to t gives ≈ = = = = 4 · 10 . dt 0 S 0 S 8.85 · 10−12 · (0.5)2 0.25 m·s 68. (D) The total resistance of the circuit is equal to the resistance of three R +R = 2R resistors 1 1 1 1 3 2R connected in parallel: = + + = , Rtotal = . The current flowing Rtotal 2R 2R 2R 2R 3 V 3V through the battery is I = = . Rtotal 2R 69. (D) The impedance of an ideal resistor is ZR = R, and the impedance of an ideal capacitor is 1 ZC = , where i is imaginary unit. The total impedance of the circuit is Z = ZR + ZC = iωC 6

1 Vi . The total current is I = Vi /Z = 1 . The amplitude Vo of the output voltage iωC R + iωC Vi Vo 1 Vi = . From this one has G = = . is Vo = IZC = 1 iRCω + 1 Vi iRCω + 1 R + iωC iωC Therefore, when ω → ∞, G → 0 and when ω → 0, G → 1. The only graph with such a behavior of G is (D). R+

∆Φ ∆q = −E. Substituting E = IR, I = and ∆t ∆t B∆S BS 0.5 · 10 · 10−4 ∆Φ = B∆S one has ∆q = − = = = 10−4 (C). R R 5 v1 r1 1 71. (B) According to Newton’s second law: mv 2 /R = qvB or v ∝ R. Thus, = = . v2 r2 3

70. (A) According to Faraday’s law of induction

72. (D) A: bosons do not obey the Pauli exclusion principle; B: bosons do not have antisymmetric wave functions; C: fermions do not have symmetric wave functions; D: true; E: bosons do not obey the Pauli exclusion principle. 73. (D) See David Griffiths’ Introduction to elementary particles, Chapter I (for both editions). It is strongly recommended to read and understand first two chapters from this book before your test to be in position to answer questions like this one. 1 2 1 + = − (note the minus sign). From this R l R equation l = − R3 . The only choice that corresponds to this distance is (E), but if you don’t want to invoke lensmakers equation just drop two rays onto the mirror, say, one from the tip of the object parallel to the lens’ axis and the second from the tip of the object to the point where the lens’ axis intersects the mirror, and you will see where the image is formed (and why it is upright).

74. (E) Using lensmakers equation one has

75. (B) The shift for the wave reflecting off the top surface of the film is ∆top = λ/2 (since the light is reflecting from a higher-n medium). The shift for the wave reflecting off the bottom surface of the film is 2 t, where t is the thickness of the film. Thus, ∆ = ∆bottom − ∆top = 2t − λ/2. For constructive interference one has 2t − λ/2 = mλ. But λ here is the wavelength λ λ = m or 4t = nλ (2m + 1). The film in the medium, so we must replace it with λ/n: 2t − 2n n λ first appears bright for λ and m = 0 =⇒ 4t = . Then it appears bright for λ0 and m = 1 n 3λ0 and λ0 = λ/3 = 540/3 = 180 (nm). =⇒ 4t = n 76. (B) According to Snell’s law sin θ = n sin β, where β is the angle of refraction. The critical angle αc for the total internal reflection is determined by sin αc = 1/n, where αc + β = π/2. √ From these three equations it is easy to obtain the critical angle of incidence: θ = −1 sin n2 − 1. Thus, we must choose between (A) and (B). Now, if we increase θ, β will increase and α will decrease, which prevents the total internal reflection. Therefore, choice (B) is correct. 77. (C) The average time between intermolecular collisions is τ = l/v = 1/(σnv), where l is r 3kT the mean free path, v = is the root mean square speed, σ is the molecule’s effective m collision area. Thus, the time varies as the square root of m. X 1 Ei Ei 78. (B) According to the Gibb’s distribution Pi = exp − , where Z = exp − . Z kT kT i E2 exp − kT . Thus, P2 = E1 E2 exp − kT + exp − kT 7

Z V2 Z V2 dV a dV 1 1 RT V2 − b − −a +a − . 79. (D) P = , A = RT0 = RT0 ln 2 V −b V2 V1 − b V2 V1 V1 V V1 V − b r r 1 k 1 2k 80. (C) Clearly, ν0 = ,ν = , where ν0 = 1 Hz, m0 = 1 kg, m = 8 kg. Thus, 2π 2π m r r m0 2m0 2·1 ν = ν0 =1· = 1/2 (Hz). m 8 v 2 1 1 1 1 1 81. (B) Kinetic energy of the moving disk is T = mv 2 + Ic.m. ω 2 = mv 2 + · mR2 = 2 2 2 2 2 R 1 2 1 2 3 mv + mv = mv 2 . The conservation of energy gives T = mgh, from which one has 2 4 4 3v 2 h= . 4g 82. (C) Kinetic energy of the mass is T = 12 mr2 ϕ˙ 2 + 12 mr˙ 2 . Potential energy of the spring is U = 12 k(r − s)2 . Lagrangian is L = T − U . Choice (D). 83. (E) From the Hamilton’s equation p˙φ = − ∂H = 0 one has pφ = const. ∂φ 1 84. (E) The x-coordinate of the rod’s center of mass is xc = M Z L 3 2 2 L 2L x2 dx = 2 · = . 2 L 0 L 3 3

Z 0

L

1 xdm = M

Z

L

x 0

2M xdx = L2

r

2 (this is from normal85. (B) As one can check for himself/herself the constant A is equal to Z L ZL 2L/3 2 3πx 2 2 ization condition ψ (x)dx = 1). The probability in question is: sin dx = L L/3 L 0 "Z # Z 2L/3 2L/3 6πx 1 1 1 L cos dx − −0 = . dx = L L/3 L L 3 3 L/3 86. (B) The eigenvalues of the matrix A are the solutions λto the equation det(A − λI) = 0, 2−λ i where I is identity matrix. Thus, A−λI = and det(A−λI) = (2−λ)2 +i2 ≡ −i 2 − λ 0. From this, one has λ1 = 1, λ2 = 3. 0 1 0 −i 0 −i 0 1 i 0 −i 0 87. (D) σx σy − σy σx = − = − = 0 −i 0 i 1 0 i 0 i 0 1 0 1 0 = 2i = 2iσz . 0 −1 2 88. (D) Let us first determine the normalization constant A: χ† χ = |A| ((1 − i)(1 + i) + 4) = √ 2 1 + i 1 0 +√ , where χ+ represents 6|A|2 = 1 =⇒ A = 1/ 6. Now, χ = aχ+ +χ− = √ 6 0 6 1 spin up and χ− is spin down. When you measure Sz on a particle in the state χ, you could get +~/2 with probability |a|2 or −~/2 with probability |b|2 . Thus, the probability of finding 2 2 the particle with spin projection Sz = −~/2 is | √ |2 = . For more on this, see section 3 6 4.4 of Griffiths’ Introduction to quantum mechanics (2nd ed.) and, in particular, identical Example 4.2.

89. (D) The reflection coefficient R is defined in of the incident and reflected probability |jref l | ~k ~k 2 |B|2 current density j: R = , where jref l = − |B|2 and jinc = |A| . Thus, R = . |jinc | m m |A|2 8

According to the standard boundary conditions, ψ is always continues and dψ/dx is continues (except at points where the potential becomes infinite). Therefore, for the point x = 0 one has A + B = C for continuity of ψ and Ak1 − Bk2 = Ck2 for continuity of dψ/dx. From 2 k1 − k2 |B|2 B k1 − k2 = and R = . these two equations = A k1 + k2 |A|2 k1 + k2 Z r Z r Q dr 1 1 Q Edr = − 90. (D) For a < r < b one has ϕ(r) = − − ; for r > b = 2 4π 4π r a 0 r 0 a a Z b Z r Z b Q dr 1 1 Q Edr = − Edr − ϕ(r) = − − . −0= 2 4π0 b a a 4π0 r b a 91. (C) When ∇ × E = 0, the line integral of E around any closed loop, according to Stokes’ theorem, is zero. Because of this, we can unambiguously talk about a scalar function ϕ(r) = Z r

−

E · dl such that E = −∇ϕ. O

H 92. (E) According to Ampere’s law B · dl = µ0 Ienc , where Ienc is the total current enclosed by R the integration path: Ienc = J · da, J is the current density. For r 2R the enclosed current is Ienc = I and B = or B ∼ 1r . The only graph that reflects 2πr such a behavior of B is (E). 93. (B) The capacitance of the parallel-plate capacitor filled with dielectric exceeds the vacuum value by a factor of the dielectric constant: C = kCvac . Thus, the electromagnetic energy 1 1 U = CV02 (if V0 is unchanged) is increased by the same factor k: U = kCvac V02 = kU0 . We 2 2 are left with (B) and (E), but do not rush to choose the second one. You might think that, when you insert a dielectric between the plates, the electric field between them (that is, inside V0 the dielectric) is reduced by the factor of k in comparison with the vacuum value . But you d forgot about the battery! In fact, when the dielectric is inserted, the surface R charge density on the lates of the capacitor is changed due to the battery. Let’s apply D · da = Qf enc to the gaussian pillbox one surface of which is inside the positive plate of the capacitor and the other one is inside the dielectric material (Qf enc here is a free charge enclosed by the gaussian surface; pillbox’s surfaces are parallel to the plate’s surface). Noting that D = 0 inside the metal plate, one has DA = σA =⇒ D = σ. The electric field inside the dielectric D σ D = =⇒ E = . We can find σ using the fact that V0 is unchanged during is E = 0 k 0 k Qvac 0 S Q 0 kS Q Qvac the insertion process. Cvac = = ,C = = , so σ = =k = kσvac . V0 d V0 d S S σ σvac V0 Finally, E = = ≡ . 0 k 0 d v 94. (C) Lorentz transformation t0 = γ(t − 2 x) suggests that for the observer O0 moving at c constant speed v parallel to x-axis the event (flash of light) that occurs at (t0 , x0 ) = (0, 10) for v the observer O is at time t0 = γ(0 − 2 x0 ) (we assume O0 is moving in the positive direction c of x and at a zero time according to both observers the origins of their reference frames coincide, therefore t0 is negative; this means that (0, 10) event (as observed by O0 ) occurs 9

ct0 3 · 108 · 13 · 10−9 |t | seconds before the other). Let α = = = 0.39, then, using units in x0 10 v2 v α or α2 = ≈ α = 0.39 (for which c = 1, one has α = γv = √ =⇒ v = √ 2 2 1−v 1−v α2 + 1 such an approximation the calculation error will be ∼ 7%). The closest choice is (C). If you calculate the square root in the above equation fairly, you will get v ≈ 0.363c. 0

95. (D) As you can check for yourself [Jx , Jy ] = i~Jz (the other two commutation relations can be obtain be cyclic permutation of the indexes x → y, y → z, z → x; note also that I do not use the signˆfor the operators). Using a sweet relation one must (or derive if needed) on the PGRE [AB, C] = A[B, C] + [A, C]B one has [Jx Jy , Jx ] = Jx [Jy , Jx ] + [Jx , Jx ]Jy = −i~Jx Jz (here I’ve used the property that any operator commutes with itself). 96. (D) If you impure into tetravalent germanium atom Ge a pentavalent element such as As, P , Sb or N , then four of the five valence electrons introduced by impurity atoms come into with four neighboring atoms of Ge and form a stable shell of eight electrons, while the fifth electron will be weakly bound to the nucleus of the impurity atom. This creates an excess of negative (n-type) electron charge carriers. The only element that cannot be used to create an excess of negative charge is B, since it has only three valence electrons. 97. (E) According to Compton scattering formula, the difference in wavelength of the scattered h (1 − cos θ), where θ is the scattering angle. In our λ0 and incident λ photon is λ0 − λ = mc hc hc h . But E = , so E 0 = 0 , where E 0 is the energy of the case, θ = 90◦ and λ0 − λ = mc λ λ hc hc h E · mc2 0 scattered photon. Thus, 0 − = , from which E = . E E mc E + mc2 98. (D) Lepton number Le is conserved only for (D). , Le = +1 for the electron, the muon, and the neutrino, and Le = −1 for the positron, the positive muon and the antineutrino (all other particles are given a lepton number of zero). 99. (E) In inertial reference frame associated with the platform, Newton’s second law reads ma = fs + mg + N, where N is a normal force perpendicular to the surface of , mg is the downward force of gravity. Acceleration a has a tangential component aτ and a radial component an : a = aτ τˆ + an n ˆ , where τˆ is a unit vector tangent to the path pointing in the ˆ is the unit (inward) normal vector direction of motion at the chosen moment in time and n to the particle’s trajectory, aτ = dv = d(ωr)/dt = rdω/dt = rα; an = v 2 /r = ω 2 r2 /r = ω 2 r. dt 2 Projecting Newton’s second law of motion onto τˆ and n ˆ gives mαr = fs sin θ and mω r = 2 −1 α . fs cos θ. From these two equations α/ω = tan θ, so θ = tan ω2 100. (E) The partition function for a single QM harmonic oscillator is: ∞ ∞ X ~ω X En n~ω Z= exp − = exp − exp − . kT 2kT n=0 kT n=0 ∞ X n~ω ~ω 2~ω exp − = 1 + exp − + exp − + . . . which is an infinite geometric kT kT kT n=0 ∞ ~ω ~ω X exp − 2kT exp 2kT n~ω 1 . Thus, Z = = series: exp − = . ~ω ~ω ~ω kT 1 − exp − 1 − exp − exp −1 kT kT kT n=0

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