JEST 2013PPART A: THREE MARK QUESTIONS Q1. In an observer’s rest frame, a particle is moving towards the observer with an energy E and momentum P . If c denotes the velocity of light in vacuum, the energy of the particle in another frame moving in the same direction as particle with a constant velocity v is (a) (E + vp ) 2 1 − (v / c ) (b) (E − vp ) 2 1 − (v / c ) (c) (E + vp ) [1 − (v / c )2 ]2 (E − vp ) [1 − (v / c )2 ]2 (d) Ans.: (a) vx x v v x+ x + 2x 2 ′ x c ⇒ = c c ⇒ x′ = c ∵ x = ct , x′ = ct ′ Solution: t ′ = 2 2 c v v v2 1− 2 1− 2 1− 2 c c c t+ E v E E + Pv c Now x′ = E ′, x = E ⇒ E ′ = ⇒ E = mc 2 , E = Pc ⇒ P = ⇒ E ′ = c v2 v2 1− 2 1− 2 c c E+ Q2. Consider a system of two particles A and B. each particle can occupy one of three possible quantum states 1 , 2 and 3 . The ratio of the probability that the two particles are in the same state to the probability that the two particles are in different states is calculated for bosons and classical (Maxwell- Boltzman) particles. They are respectively (a) 1,0 (b) 1/2,1 (c) 1,1/2 (d) 0,1/2 Ans.: (c) Solution: For two particle in same state: AB 3 3 2 AB AB 1 3 3 2 2 2 1 1 Boson Probability ratio: 1/ 3 =1 1/ 3 AB 1 3 AB AB 3 2 2 1 1 Classical (Maxwell - Boltzman) For two particle in different states 3 B B 2 A 1 3 2 A B 3 A 2 B 1 A 1 Boson Probability ratio: Q3. 3 2 1 3 A B B 3 2 1 A 2 A 1 B 3 A 2 B 1 3 2 1 Classical (Maxwell-Boltzmann) 1/ 3 1 = 2/3 2 At ‘equilibrium there can not be any free charge inside a metal. However, if you forcibly put charge in the interior then it takes some finite time to ‘disappear’ i.e. move to the surface. If the conductivity, σ , of a metal is 106 (Ωm ) −1 and the dielectric constant ε 0 = 8.85 × 10 −12 Farad/m, this time will be approximately: (a) 10-5 sec (b) 10-11 sec (c) 10-9 sec (d) 10-17 sec Ans.: (d) Solution: Characteristic time: τ = Q4. ∈ σ = 8.85 × 10 −12 = 8.85 × 10 −18 10 6 The free fall time of a test mass on an object of mass M from a height 2R to R is R3 (a) (π / 2 + 1) GM (b) R3 GM (c) (π / 2) R3 GM 2R 3 (d) π GM Ans.: (a) Solution: Equation of motion v md 2 r GMm d 2r GM d 2r A = − ⇒ = − ⇒ =− 2 2 2 2 2 2 dt dt dt r r r dv A dr d ⎛ v2 =− 2 ⇒ ⎜⎜ dt dt ⎝ 2 r dt ∵ GM = A ⎞ d ⎛ A⎞ v2 A ⎟⎟ = ⎜ ⎟ ⇒ = +C r 2 ⎠ dt ⎝ r ⎠ when r = 2 R, v = 0 0 A A v2 A A 2A 2A dr 2A = +C ⇒ C = − ⇒ = − ⇒v= − ⇒ = 2 2R 2R 2 r 2R r 2R dt 2R R r 2R 2R − r ∫ dr = − A t dt R ∫0 put r = u 2 , dr = 2udu when r = 2 R, r = R, u = 2 R , u = R 2R − r r A B 3 2 1 The particle is in its ground state given by ϕ 0 ( x ) = 2 / L cos(πx / L ) . what is the probability that the particle will be in the ground state after this sudden expansion? (a) (8 / 3π ) 2 (b) 0 (c) (16 / 3π ) 2 (d) (4 / 3π ) 2 . the coordinates and momenta of any particle/ system transform as: t ' = t . It turns out that the results of all 3 tosses are heads. Suppose you choose a coin at random and toss it 3 times.99 (b) 0. A particle of mass m is contained in a one-dimensional infinite well extending from x = -L/2 to x = L/2. Under a Galilean transformation. X / e −i t v .P / e i M v 2 t / (2 ) (d) e i t v . P / e −i M v 2 t / (2 2 t / (2 ) ) Ans.∫ u R 2R 2R − u 2 A t A dt ⇒ − t = 2∫ ∫ R 0 R × 2udu = − A 2 R −1 u ⎤ ⎡ u ⇒− t = 2 ⎢− 2R − u 2 + sin ⎥ R 2 2R ⎦ ⎣ 2 Q5. r ' = r + v t and p ' = p + mv where v is the velocity of the boosted frame with respect to the original frame. What is the probability that the coin you have drawn is the doubleheaded one? (a) 0.P / e −i M v (a) e i M v .925 (c) 0. X / e i t v . The walls of the box are moved suddenly to form a box extending from x = -L to x = L.P / (c) e i M v .: (c) Q6.: (b) Q7. X / e i t v . R 2R u2 2R − u 2 du R 2R ⇒− ⎡− R A 2 R −1 R 2R 2R ⎤ + t = 2⎢ 2R − R + sin 2 R − 2 R − R sin −1 ⎥ R 2 2 2R 2R ⎦ ⎣ 2 ⇒− 3 A R R ⎛π ⎡ − R Rπ Rπ ⎤ ⎞ ⎛π ⎞ R + − ∵ A = GM ⇒ = + ⇒ = + t = 2⎢ t t 1 1 ⎜ ⎟ ⎜ ⎟ R 4 2 ⎥⎦ A ⎝2 ⎠ ⎣ 2 ⎝ 2 ⎠ GM A box contains 100 coins out of which 99 are fair coins and 1 is a double-headed coin. A unitary operator carrying out these transformations for a system having total mass M. total momentum P and centre of mass coordinate X is (b) e i M v .75 (d) 0.01 Ans. q being the total charge. 2 2 + 3π π 2 = 2 2 2 8 3π The electric fields outside (r > R) and inside (r < R) a solid sphere with a uniform Er >R = volume charge density are given by q rˆ and 4πε 0 r 2 1 Er<R = q rrˆ 4πε 0 R 3 1 respectively. The correct ratio of 4πε 0 r 2 1 the electrostatic energies for the second case to the first case is (a) 1:3 (b) 9:16 (c) 3:8 (d) 5:6 Ans. while the electric field outside a spherical shell with a uniform surface charge density is given by E r < R = q rˆ .: (a) 2 Solution: Probability φ 0 φ1 πx πx 2 2 cos . φ1 cos L L 2L 2L .: (d) Solution: Electrostatic energy in spherical shell wsp = ⇒ ∈0 2 ∞ R 0 ∫ R 0 2 E1 4π r 2 dr + ∈0 2 ∫ ∞ R 2 E 2 4π r 2 dr ∞ q2 ∫ (4π ∈ ) ∈0 2 2 r4 4π r 2 dr = q2 ⎛ 1 ⎞ q2 1 ⎜− ⎟ = 8π ∈0 ⎝ r ⎠ R 8π ∈0 R Electrostatic energy in solid sphere ws = R q2 1 ⎡r5 ⎤ q2 ⇒ × ⎢ ⎥ + 8π ∈0 R 6 ⎣ 5 ⎦ 0 8π ∈0 ∞ ⎡ 1⎤ ⎢− r ⎥ ⎣ ⎦R ∈0 2 ∫ R 0 E1 4π r 2 dr + 2 ∈0 2 ∫ ∞ R E 2 4π r 2 dr 2 .Ans.φ0 = Since the wall of box are moved suddenly then 2 Probability = ∫ 2 1 cos π x cos π x 2 1 L / 2 2 cos π x cos π x ⋅ ⋅ dx = ⋅ dx L L L L 2 ∫− L / 2 L 2L 2L L/2 −L / 2 2 2 2 1 L / 2 ⎡ ⎛ 3π x ⎞ 2 1 ⎡ 2L 3π x 2 L π x⎤ ⎛ π x ⎞⎤ ⇒ ⋅ ∫ ⎢cos ⎜ + cos ⎜ ⋅ ⎢ sin + dx ⇒ sin ⎟ ⎟ ⎥ L 2 − L / 2 ⎣ ⎝ 2L ⎠ L 2 ⎣ 3π 2L π 2 L ⎦⎥ − L / 2 ⎝ 2L ⎠⎦ L/2 2 1 ⎡ 2 L ⎛ 3π 3π ⎞ 2 L ⎛ π π ⎞⎤ ⇒ ⋅ ⎢ ⎜ sin + sin ⎟+ ⎜ sin + sin ⎟ ⎥ 4 4 ⎠ π ⎝ 4 4 ⎠⎦ L 2 ⎣ 3π ⎝ ⇒ Q8. where ψ/ 0 and ψ/ 1 are the real wavefunctions in the ground and first excited state of the harmonic oscillator Hamiltonian. Wspherical Wsphere q2 8π ∈0 5 = = 2 6q 6 40π ∈0 R A quantum mechanical particle in a harmonic oscillator potential has the initial wave function ψ/ 0 ( x ) + ψ/ 1 ( x ). t ) = ψ 0 ( x ) + ψ 1 ( x ) + 2 Reψ 0* ( x )ψ 1 ( x ) cos ( E1 − E0 ) 2 2 2 t putting t = π ψ ( x. 2 2 2 A spherical planet of radius R has a uniform density ρ and does not rotate. For convenience we take m= = ω = 1 for the oscillator. t ) = ψ 0 ( x ) + ψ 1 ( x ) + 2 Reψ 0* ( x )ψ 1 ( x ) cos π 2 2 2 ∵ E1 − E0 = ω = 1 ψ ( x. the pressure at point r from the center is (a) 4πρ 2 G 2 R − r2 3 ( ) (b) (c) 2πρ 2 G 2 R − r2 3 ) (d) ( 4πρG 2 R − r2 3 ( ρG 2 (R 2 − r2 ) ) .: (a) Solution: ψ (x ) = ψ 0 (x ) + ψ 1 ( x ) ψ ( x. What is the probability density of finding the particle at x at time t = π ? (a) (ψ/ 1 ( x ) − ψ/ 0 ( x )) (b) (ψ 1 ( x )) − (ψ/ 0 ( x )) 2 (c) (ψ/ 1 ( x ) + ψ/ 0 ( x )) 2 2 (d) (ψ/ 1 ( x )) + (ψ/ 0 ( x )) 2 2 2 Ans. t )ψ ( x. t ) = ψ 0 ( x ) e − i E0t +ψ 1 ( x ) e−i E1t Now probability density at time t ψ ( x. t ) = ψ * ( x.q2 1 q2 6q 2 ws = ⋅ + = 5 × 8π ∈0 R 8π ∈0 R 40π ∈0 R Now Q9. If the planet is made up of some liquid. t ) = ψ 0 ( x ) + ψ 1 ( x ) − 2 Reψ 0* ( x )ψ 1 ( x ) = ⎡⎣ψ 1 ( x ) −ψ 0 ( x ) ⎤⎦ 2 Q10. what is the velocity? (a) ⎛ g ⎞ k tan⎜⎜ t ⎟⎟ g ⎝ k ⎠ (b) ⎛ g ⎞ gk tan⎜⎜ t ⎟⎟ k ⎝ ⎠ . (b) 1 (c) –i (d) -1 Ans. 4π 2 4π 2 ⎛ r 2 ρ Grdr ⇒ p = ρ G⎜⎜ 3 3 ⎝ 2 R ⎞ 4π 2 ⎛ R 2 r 2 ⎞ ⎟⎟ ⇒ p = ρ G⎜⎜ − ⎟⎟ 3 2⎠ ⎠r ⎝ 2 4π ρ 2 G 2 2π 2 R − r2 )⇒ p = ( ρ G (R 2 − r 2 ) 3 2 3 Re(z 2 ) + Im(z 2 ) z →0 z2 Compute lim (a) The limit does not exist. A particle of mass m is thrown upward with velocity v and there is retarding air resistance proportional to the square of the velocity with proportionality constant k.: (c) dm ⋅ g dm ⋅ g ⇒ dp = ⇒ dp = Solution: Pressure dp = A 4πr 2 ρ ⋅ 4πr 2 drGM 4πr 2 r R3 dr dm (mass of elementary part ) r R 4π 3 r R 3 3 R ⇒ dp = 4π ρ 2 Grdr 2 3 4πr ρ ⋅ 4πr 2 drG ⋅ ρ ⋅ ⇒ dp = ∫ dp = ∫ R r ⇒ p= Q11. and g is the gravitational acceleration.: (a) Solution: lim ( ) ( ) = lim x Re z 2 + Im z 2 z →0 z2 z →0 − y 2 + 2 xy x 2 − y 2 + 2 xy ⇒ lim =1 y = 0 x 2 − y 2 + 2ixy x 2 − y 2 + 2ixy x →0 2 x 2 − y 2 + 2 xy lim 2 = −1 x = 0 x − y 2 + 2ixy y →0 Q12.Ans. If the particle attains a maximum height after time t. For a diatomic ideal gas near room temperature. The particles have an isotropic velocity distribution with constant magnitude v.: (c) Solution: Equation of motion ⇒∫ dv dv m = ∫ dt ⇒ × = ∫ dt ⇒ ∫ k k ⎛ gm k ⎞ g + v2 + v2 ⎟ ⎜ m m⎝ k ⎠ v ⇒ tan −1 Q13.: (c) Q14.(c) g tan k ( gk t ) (d) gk tan ( gk t ) Ans. what fraction of the heat supplied is available for external work if the gas is expanded at constant pressure? (a) 1/7 (b) 5/7 (c) 3/4 Ans. The rate at which the particles will be emitted from a hole of area A on one side of this box is (a) nvA (b) nvA/2 (c) nvA/4 (d) none of the above Ans.: (d) Solution: It is isobaric process (constant pressure) Then δθ = nC p ΔT ⇒ ΔW = nRΔT In this process δθ is heat exchange during process. mdv dv k dv = mg + kv 2 ⇒ = g + v2 ⇒ = dt k 2 dt dt m g+ v m gm k = gk t⇒v= m 1 gm k tan −1 v gm k =t gm gk tan t k m Consider a uniform distribution of particles with volume density n in a box. Function of heat supplied = ⇒ 1− δW ΔQ 1 ⎛ 2⎞ ⎜1 + f ⎟ ⎝ ⎠ = nRΔT R γ −1 1 = = = 1− γ γ nC p ΔT R γ γ −1 γ= Cp CV ⇒ Cp = γR γ −1 (d) 2/7 . The nearest neighbor distances dfc and dsc for the FCC and the SC structures respectively are in the ratio (dfc/dsc) [Given 21/3 = 1. Jy. What is the largest fraction of the area that can be covered? (a) 3 (b) π 5π 6 (c) 6 7 (d) π 2 3 Ans. ⇒ 1− f f +2 ⇒ 1− 5 2 ⇒ 5+2 7 f = degree of freedom. J y = ( J− − J+ ) ⇒ J+ = ⎢ ⎥ ⎥ 2 2 ⎣0 0 ⎦ ⎣1 0 ⎦ J + J y 1 ⎡ 0 1− i⎤ ⎡0 1 ⎤ i ⎡0 −1⎤ . the eigenvalues of the operator (J x + j y ) / are (a) real and discrete with rational spacing (b) real and discrete with irrational spacing (c) real and continuous (d) not all real Ans. It is observed that this phase transition does not involve any change of volume. Jz are angular momentum operators.26] . J− = ⎢ ( J+ + J− ) . for diatomic molecule f = 5 A flat surface is covered with non-overlapping disks of same size. Jy = ⎢ ⇒ x = ⎢ ⎢ ⎥ ⎥ 2 ⎣1 0 ⎦ 2 ⎣1 0 ⎦ 2 ⎣1 + i 0 ⎥⎦ eigen value Q17.Q15.: (d) 1 1 1 Solution: neff = nC + n f + 1× ni = × 6 + 1 = 3 3 2 3 a = 2r neff × A Now largest fraction of area i.e. 3 × a2 4 3 × πr 2 = 6× 3 2 × (2r ) 4 = π 2 3 If Jx. 1 ⎛ − λ 1− i⎞ ⎜⎜ ⎟⎟ ⇒ λ2 − 2 = 0 ⇒ λ = ± 2 2 ⎝1 + i − λ ⎠ A metal suffers a structural phase transition from face-centered cubic (FCC) to the simple cubic (SC) structure. packing fraction = 6× Q16.: (b) Solution: J x = Jx = ⎡0 1 ⎤ ⎡0 0⎤ i 1 . in a Kepler potential. The energy of the neutrino.029 (b) 1. which can be massless. e = 0 . rC l = rC ⇒ 2rp = rC A K meson (with a rest mass of 494 MeV) at rest decays into a muon (with a rest mass of 106 MeV) and a neutrino. (m = 2 k − mμ2 ) c 2 2mk ⎛ 494 494 106 106 ⎞ 2 ⎜ 2 × 2 − 2 × 2 ⎟c c c c c ⎠ ⇒⎝ 494 2× 2 c 11236 = 235. θ = 0 r l = 1. then (a) rc = 2rp (b) rc = rp (c) 2rc = rp (d) rc = 2rp Ans. is approximately (a) 120 MeV (b) 236 MeV (c) 300 MeV Ans. Eν ⇒ 244036 − Q20. l = 2rp . If.130 Ans. l = 1 + e cos θ for parabola e = 1 for circle.707 dSC a 2 1. 17: () Solution: Nearest neighbour in SC → a → C. the pericentre distance of particle in a parabolic orbit is rp while the radius of the circular orbit with the same angular momentum is rc.(a) 1.122 (c) 1.: (a) Solution: Ionic equation l = 1+ e .N = 12 a dFCC 1 1 = 2 = = = 0.6275 ≈ 236 MeV 988 The vector field xziˆ + yˆj in cylindrical polar coordinates is (a) ρ (z cos 2 φ + sin 2 φ )eˆ ρ + ρ sin φ cos φ (1 − z )eˆφ (b) ρ (z cos 2 φ + sin 2 φ )eˆ ρ + ρ sin φ cos φ (1 + z )eˆφ (c) ρ (z sin 2 φ + cos 2 φ )eˆ ρ + ρ sin φ cos φ (1 + z )eˆφ (d) 388 MeV .: (b) Solution: k → μ + ν .N = 6 Nearest neighbour in FCC → a 2 → C. rp Q19.414 Q18.374 (d) 1. : (d) Q22. each having equal probability of making a step simultaneously to the left or right along the x axis. but at random times.07 (b) 0. Az = 0 Aρ = A ⋅ eˆρ = Ax ( xˆ ⋅ eˆρ ) + Ay ( yˆ ⋅ eˆρ ) + Az ( zˆ ⋅ eˆρ ) = ρ cos φ z ⋅ cos φ + ρ sin φ ⋅ sin φ + 0 ⇒ Aρ = ( ρ cos φ 2 z + ρ sin 2 φ ) eˆρ Aφ = A ⋅ eˆφ = Ax ( xˆ ⋅ eˆφ ) + Ay ( yˆ ⋅ eˆφ ) + Az ( zˆ ⋅ eˆφ ) = ρ cos φ ( − sin φ ) z + ρ sin φ ⋅ cos φ Aφ = ρ cos φ ⋅ sin φ (1 − z ) eˆφ ( ) A = Aρ eˆρ + Aφ eˆφ + Az eˆz = ρ cos 2 φ z + sin 2 φ eˆρ + ρ cos φ sin φ (1 − z ) eˆφ Q21. The probability that they meet after n steps is (a) 1 2n! 4 n n!2 (b) 1 2n! 2 n n!2 (c) 1 2n! 2n (d) 1 n! 4n Ans.: (a) Solution: A = xziˆ + yˆj Ax = xz . Ay = y.: (a) Solution: Into probability of taking ' r ' steps out of N steps r ⎛1⎞ ⎛1⎞ = N Cr ⎜ ⎟ ⎜ ⎟ ⎝2⎠ ⎝2⎠ N −r total steps = N = n + n = 2n for taking probability of n steps out of N n ⎛1⎞ ⎛1⎞ P = N Cn ⎜ ⎟ ⎜ ⎟ ⎝2⎠ ⎝2⎠ N −n n N! ⎛1⎞ ⎛1⎞ = ( N − n ) !n ! ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ 2n−n 2n ! ⎛ 1 ⎞ = ⎜ ⎟ n !n ! ⎝ 2 ⎠ 2n = 2n ! ( n !) 2 4n .19 Ans.60 (c) 0.36 (d) 0.(d) ρ (z sin 2 φ + cos 2 φ )eˆ ρ + ρ sin φ cos φ (1 − z )eˆφ Ans. Two drunks start out together at the origin. There are on average 20 buses per hour at a point. The probability that there are no buses in five minutes is closest to (a) 0. ⎛1 / 2 ⎞ Consider the state ⎜⎜1 / 2 ⎟⎟ corresponding to the angular momentum l = 1in the Lz basis ⎝1 / 2 ⎠ of states with m = +1. The equation describing the shape of curved mirror with the property that the light from a point source at the origin will be reflected in a beam of rays parallel to the x-axis is (with a as some constant) (a) y2 = ax + a2 (b) 2y = x2 + a2 (c) y2 = 2ax + a2 (d) y2 = ax3 + 2a2 Ans. If L2z is measured in this state yielding a result 1. 2m 2 4 ( ) ( ) What is the exact ground state energy? ( (a) E = 3 ω 1+ 1+ λ 2 (c) E = 3 ω 1− λ 2 ) ( ) (b) E = 3 ω 1+ λ 2 (d) E = 3 ω 1+ 1− λ 2 ( ) Ans. ⎜ 0⎟ ⎝ ⎠ ⎛0⎞ ⎜ ⎟ ⎜1⎟. -1.: (c) Q24.: (b) Q25. Lz = ⎜ 0 0 0 ⎟ . ⎜0⎟ ⎝ ⎠ ⎛0⎞ ⎜ ⎟ ⎜0⎟ ⎜1⎟ ⎝ ⎠ Corresponding eigenvalue 1. 1 Now state after measurement yielding 1⇒ φ1 + φ 3 ⎛1⎞ ⎛1⎞ ⎜ ⎟ 1 ⎜ ⎟ = ⎜0⎟ = ⎜0⎟ 2 ⎜1⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ . A simple model of a helium-like atom with electron-electron interaction is replaced by Hooke’s law force is described by Hamiltonian λ 1 − 2 2 2 ∇1 + ∇ 22 + mω 2 r12 + r22 − mω 2 r1 − r2 .Q23. what is the state after the measurement? ⎛1⎞ ⎜ ⎟ (a) ⎜ 0 ⎟ ⎜ 0⎟ ⎝ ⎠ ⎛ 1/ 3 ⎞ ⎟ ⎜ (b) ⎜ 0 ⎟ ⎟⎟ ⎜⎜ ⎝ 2/3⎠ ⎛1 / 2 ⎞ ⎟ ⎜ (d) ⎜ 0 ⎟ ⎟⎟ ⎜⎜ ⎝1 / 2 ⎠ ⎛ 0⎞ ⎜ ⎟ (c) ⎜ 0 ⎟ ⎜1⎟ ⎝ ⎠ Ans. eigenvector ⎜ 0 0 − 1⎟ ⎜0 0 1⎟ ⎝ ⎠ ⎝ ⎠ ⎛1⎞ ⎜ ⎟ ⎜ 0⎟ . 0. 0.: (d) ⎛1 0 0 ⎞ ⎛1 0 0⎞ ⎜ ⎟ ⎜ ⎟ 2 Solution: L z = ⎜ 0 0 0 ⎟ . : (c) Solution: For large distance F = qa sin θ qa sin θ 1 1 . A thin uniform ring carrying charge Q and mass M rotates about its axis.5° C (c) 10° C (d) 8. What are the value of n and m? (a) n = 1.: (c) Q29. (c) G(x) is discontinuous at x = 0 (d) The continuity properties of G(x) and G’(x) at x = 0 depend on the value of k. dx where k is a constant.: (c) (b) 12.2° C . B∝ r r r r So m = n = 1 Q28. what is the increase in its temperature? (The specific heat of the bullet = 160 Joule per Kg per degree C) (a) 14° C Ans.: (c) Solution: Magnetic dipole moment M ′ = IA = Angular momentum J = Mr 2ω ⇒ Q27. Q 2 Q Qω r 2 2 πr ⇒ × 2π × π r = T 2π T 2 M′ Q = J 2M The electric and magnetic field caused by an accelerated charged particle are found to scale as E ∝ r − n and B ∝ r − m at large distances. What is the gyromagnetic ratio (defined as ratio of magnetic dipole moment to the angular momentum) of this ring? (a) Q / (2πM ) (b) Q/M (c) Q/(2M) (d) Q / (πM ) Ans. m = 1 (c) n = 1. Which of the following statements is true? (a) Both G(x) and G’(x) are continuous at x = 0 (b) G(x) is continuous at x = 0 but G’(x) is not. m = 2 (b) n = 2. m = 1 (d) n = 2. A metal bullet comes to rest after hitting its target with a velocity of 80m/s.PART B: ONE MARK QUESTIONS Q26. Ans. m = 2 Ans. If 50% of the heat generated remains in the bullet. Consider the differential equation dG ( x ) + kG ( x ) = δ ( x ) . B= ⇒E∝ . where σ are the three Pauli matrices and a is a vector? (a) a x + a y and a z (b) a x + a z ± ia y (c) ± (a x + a y + a z ) (d) ± a Ans.a x + σ y .Solution: Conservation of momentum ⇒ ΔT = Q30.: (a) † * ⎞ ∂ ⎛ * ⎞ ⎛ −∂ψ ( x ) ψ x) ⎟ Solution: ⇒ ⎜ψ ( x ) − ψ ( x ) ⎟ = ⎜ ( ⎟ ∂x ∂x ⎝ ⎠ ⎝⎜ ⎠ ∞ ∞ ∂ψ ( x ) ∞ ⎡ ∂ ⎤ ⇒ ∫ ψ * ( x ) ⎢ − ψ ( x ) ⎥ dx −ψ * ( x )ψ ( x ) − ∫ − ψ ( x ) dx −∞ −∞ −∞ ∂x ⎣ ∂x ⎦ * ∞ ∂ψ * ( x ) −∞ ∂x ⇒∫ ψ ( x ) dx ∂ ∂x . 1 2 1 mv × 50 % = mcΔT ⇒ 80 × 80 = 160 ΔT 2 2 80 × 80 1 × = 100 C 4 160 What are the eigenvalues of the operator H = σ ⋅ a .a z ) ⎛0 1⎞ ⎛ 0 −i ⎞ ⎛1 0 ⎞ =⎜ ⎟ ax + ⎜ ⎟ ay + ⎜ ⎟ az ⎝1 0⎠ ⎝i 0 ⎠ ⎝ 0 −1⎠ ⎛ az ⇒ ⎜⎜ ⎝ (a x + ia y ) (a − ia y )⎞ ⎛ (a z − λ ) (a x − ia y ) ⎞ ⎟⇒⎜ ⎟ − a z ⎟⎠ ⎜⎝ (a x + ia y ) − (a z + λ )⎟⎠ x ⇒ −(a z − λ )(a z + λ ) − (a x − ia y )(a x + ia y ) − az2 + λ 2 − ax2 − a y2 = 0 λ 2 = ax2 + a y2 + az2 ⇒λ =± a Q31.a y + σ z . ⎛−∂⎞ The hermitian conjugate of the operator ⎜ ⎟ is ⎝ ∂x ⎠ (a) ∂ ∂x (b) − ∂ ∂x (c) i ∂ ∂x (d) − i Ans.: (d) Solution: H = σ ⋅ a = (σ x . : (c) Solution: ∫ ∞ −∞ ∂ ⎞ ⎟ψ ( x ) dx = p ∂x ⎠ ⎛ ⎝ ψ * ( x ) ⎜ −i Now ∫ ∞ −∞ e − ⎛ ⎝ ikx ∞ ψ * ( x ) ⎜ −i ⇒∫ e −∞ − ⎛ ⎝ ikx − ikx ∞ ∂ ⎞ ikx e ψ x dx ⇒ e ψ * ( x )( −i ( ) ⎟ ∫ −∞ ∂x ⎠ ψ * ( x ) ⎜ −i ⎡ ) ⎢e ⎣ ikx ⎤ ∂ ik ikx ψ ( x ) + e ψ ( x )⎥ ∂x ⎦ ikx ∞ ∂ ik − ikx ⎞ ψ ( x ) ⎟ e + ∫ −i . x 2 ) = 1 ⎡ 2 ⎛ πx1 ⎞ ⎛ 2πx 2 ⎞ 2 ⎛ 2πx1 ⎞ ⎛ πx 2 ⎟ sin ⎜ ⎟ + sin ⎜ ⎟ sin ⎜ ⎢ sin ⎜ 2 ⎣L ⎝ L ⎠ ⎝ L ⎠ L ⎝ L ⎠ ⎝ L ⎞⎤ ⎟⎥ ⎠⎦ (b) υ/ ( x1 . x 2 ) = 2 ⎛ 2πx1 ⎞ ⎛ πx 2 ⎞ sin ⎜ ⎟ sin ⎜ ⎟ L ⎝ L ⎠ ⎝ L ⎠ Ans. e ψ * ( x )ψ ( x ) dx −∞ ∂x ⎠ ∞ ∞ ∂ ⎡ ⎤ ⇒ ∫ ψ * ( x ) ⎢ −i ψ ( x )⎥ + k ∫ ψ * ( x )ψ ( x ) ⇒ P + K −∞ −∞ ∂x ⎣ ⎦ Q33.: (b) Solution: Electrons are Fermions of spin 1 and it wave functions are anti symmetric 2 Spin part is symmetric and space part will be anti symmetric (since total wave function is anti symmetric) . If the expectation value of the momentum is p for the wavefunction υ/ ( x ) . The one-electron states are given by υ/ n ( x ) = 2 / L sin (nπx / L ) . x 2 ) = 2 ⎛ πx1 ⎞ ⎛ 2πx 2 ⎞ sin ⎜ ⎟ sin ⎜ ⎟ L ⎝ L ⎠ ⎝ L ⎠ (d) υ/ ( x1 . Two electrons are confined in a one dimensional box of length L. x 2 ) if both electrons are arranged to have the same spin state? (a) υ/ ( x1 .Q32. x 2 ) = 1 ⎡ 2 ⎛ πx1 ⎞ ⎛ 2πx 2 ⎞ 2 ⎛ 2πx1 ⎞ ⎛ πx 2 ⎟ sin ⎜ ⎟ − sin ⎜ ⎟ sin ⎜ ⎢ sin ⎜ 2 ⎣L ⎝ L ⎠ ⎝ L ⎠ L ⎝ L ⎠ ⎝ L ⎞⎤ ⎟⎥ ⎠⎦ (c) υ/ ( x1 . then the expectation value of momentum for the wavefunction e i k x / υ/ ( x ) is (b) p − k (a) k (c) p + k (d) p Ans. What would be the ground state wave function υ/ ( x1 . −1 c ⎞⎛ v v2 ⎞ c ⎞⎛ v ⎞ ⎛ ⎟⎜1 + ⎟ = ⎜ v + ⎟ ⎜ 1 − + 2 2 ⎟ n ⎠ ⎝ cn c n ⎠ n ⎠⎝ cn ⎠ ⎝ c 1 ⎞ v2 v3 c v cv 2 ⎛ + 2 2 + − 2 + 3 ⇒ u = + v ⎜1 − 2 ⎟ n cn c n n cn cn ⎝ n ⎠ If the distribution function of x is f ( x ) = xe − x / λ over the interval 0 < x < ∞.: (d) Solution: ⇒ cos θ = 1 − θ2 2! + θ4 4! sin θ = θ − ..⎟ + ⎜1 − + ⎟ ⇒ cos12 + sin 12 = 1 ⎝ 2! 4! ⎠ ⎝ 3! 5! ⎠ Q35. .Then = Q34. 1 ⎡ 2 ⎛ πx1 ⎞ ⎛ 2πx 2 ⎞ 2 ⎛ 2πx1 ⎞ ⎛ πx 2 ⎢ L sin ⎜ L ⎟ ... ∵ sin 2 θ + cos 2 θ = 1 A light beam is propagating through a block of glass with index of refraction n.... the velocity of the light in the glass block as measured by an observer in the laboratory is approximately (a) u = c 1 ⎞ ⎛ + v ⎜1 − 2 ⎟ n ⎝ n ⎠ (b) u = c 1 ⎞ ⎛ − v⎜ 1 − 2 ⎟ n ⎝ n ⎠ (c) u = c 1 ⎞ ⎛ + v⎜1 + 2 ⎟ n ⎝ n ⎠ (d) u = c n Ans. sin ⎜ L 2⎣ ⎠ ⎝ ⎠ ⎝ ⎝ ⎠ ⎝ ⎞⎤ ⎟⎥ ⎠⎦ What is the value of the following series? 2 1 1 ⎛ ⎞ ⎛ 1 1 ⎞ ⎜1 − + − . 2 1 1 ⎞ ⎛ 1 1⎞ ⎛ ⇒ ⎜1 − + .. the mean value of x is (a) λ (b) 2λ (c) λ / 2 (d) 0 . 2 θ3 θ5 3! + 5! .: (a) c n = ⎛v + Solution: now u = ⎜ v⋅c ⎝ 1+ 2 c ⋅n v+ ⇒v− Q36. ⎟ ⎝ 2! 4! ⎠ ⎝ 3! 5! ⎠ (a) 0 2 (c) e2 (b) e (d) 1 Ans.. If the glass is moving at constant velocity v in the same direction as the beam. sin ⎜ L ⎟ − L sin ⎜ L ⎟...⎟ + ⎜1 − + − ...... and ln 3 at high T Ans. xf ( x )dx f ( x ) dx = ∫ ∞ 0 ∫ ∞ 0 − x x. Which of the following statements is true about the entropy S(T)? (a) S(T) = ln 3 at T = 0. The system is at a temperature T. There is a magnetic field h present and the energy for a spin state s is –hs.: (b) ⎛d ⎞⎛ d ⎞ ⎤ ⎛d ⎞⎡ d Solution: ⇒ ⎜ − x ⎟⎜ + x ⎟ f ( x ) ⇒ ⎜ − x ⎟ ⎢ f ( x ) + xf ( x )⎥ ⎠⎝ dx ⎠ ⎝ dx ⎝ dx ⎠ ⎣ dx ⎦ ⇒ d ⎡d d ⎤ ( ) ( ) f x + xf x − x f (x ) − x 2 f (x ) ⎢ ⎥ dx ⎣ dx dx ⎦ ⇒ df ( x ) d2 d f ( x) + f ( x) + x −x f ( x ) − x2 f ( x ) 2 dx dx dx .: (b) Solution: ∵ it is distribution function so x = ∫ ∫ ∞ −∞ ∞ −∞ Q37. and 3 at high T (d) S(T) = 0 at T = 0. The operator ⎛d ⎞⎛ d ⎞ ⎜ − x ⎟⎜ + x ⎟ ⎝ dx ⎠⎝ dx ⎠ is equivalent to (a) d2 − x2 2 dx (b) d2 − x2 +1 2 dx (c) d2 d − x x2 +1 2 dx dx (d) d2 d − 2x − x 2 2 dx dx Ans.xe λ dx − x λ xe dx ⇒ ∫ ∞ 0 ∫ ∞ 0 −x x 2 e λ dx −x = 2λ λ xe dx Consider a particle with three possible spin states: s = 0 and ±1. and 3 at high T (b) S(T) = ln 3 at T = 0.Ans. and zero at high T (c) S(T) = 0 at T = 0. S = 0 Q38.e.: (d) Solution: S = k ln ω S = k ln 3 where ω = number of microstates at high T ω =3 and at T = 0 it is perfect ordered i. −13. in case (ii) they are bosons. more or equal depending on the electron mass Ans. the particles are fermions. which of the following is true? (a) EF = EB = Ecl (b) EF > EB = Ecl (c) EF < EB < Ecl (d) EF > EB > Ecl Ans. and in case (iii) they are classical.99995 me × ⇒ −13.: (b) Solution: For fermions π2 2 2ml 2 =∈0 1×∈0 +1× 4 ∈0 +1× 9 ∈0 +1× 16 ∈0 = 30 ∈0 For Boson = 4×∈0 .59932 n2 me ∵ μ = 0. In case (i).⇒ Q39. If a proton were ten times. For Maxwell = 4×∈0 EF > EB = Ecl Q40. ⎞ ⎛ d2 d2 2 ⎜⎜ 2 − x 2 + 1⎟⎟ f (x ) ( ) ( ) ( ) f x x f x f x − + = 2 dx ⎠ ⎝ dx Consider three situations of 4 particles in one dimensional box of width L with hard walls.6 0.: (b) Solution: En = Q41.: (d) Solution: For electrostatic field ∇ × E = 0 .99995me If E1 = xyiˆ + 2 yzˆj + 3xzkˆ and E 2 = y 2 iˆ + (2 xy + z 2 ) ˆj + 2 yzkˆ then (a) Both are impossible electrostatic fields (b) Both are possible electrostatic fields (c) Only E1 is a possible electrostatic field (d) Only E 2 is a possible electrostatic field Ans. If the total ground state energy of the four particles in these three cases are EF. EB and Ecl respectively. the ground state energy of the electron in a hydrogen atom would be (a) less (b) more (c) the same (d) less. the decay series eventually ending at 206 Pb.0058. A rock sample analysis shows that the ratio of the numbers of atoms of 206 decay of 238 Pb to 238 U is 0. the speed of light in vacuum) . the period of the pendulum will be (a) T (c) 2T / 3 (b) T / 4 (d) 2T / 5 Ans.: (a) Solution: t = Q43. the age of the rock sample is (a) 38 × 106 years (b) 48 × 106 years (c) 38 × 107 years (d) 48 × 107 years Ans. Assuming that all the 206 Pb has been produced by the U and that all other half-lives in the chain are negligible. If the lift accelerates downwards with an acceleration g / 4. l ⇒ lift accelerates down wards then g l ⇒ T = 2π g − g′ l g− g 4 = 2π 4l l ⇒ 2π × 2 3g 3g 2T 3 The velocity of a particle at which the kinetic energy is equal to its rest energy is (in terms of c. ⎛ N pb + N u ⎞ ln ⎜ ⎟ λu ⎝ N u ⎠ 1 The period of a simple pendulum inside a stationary lift is T.51 × 109 years.iˆ ∂ ∇ × E2 = ∂x y2 ˆj ∂ ∂y 2 xy + z 2 kˆ ∂ ∂z 2 yz (2 z − 2 z )iˆ + 0 + (2 y − 2 y )zˆ = 0 iˆ ∂ ∇ × E1 = ∂x xy Q42.: (c) Solution: T = 2π T = 2π T′ = Q44. which is stable. ˆj ∂ ∂y ∂yz kˆ ∂ = ( 0 − 2 y ) iˆ + 0 + xjˆ ≠ 0 ∂z yxz 238 U decays with a half life of 4. 8 x + 0. y ′ = 0. It is attached above the window of a car. increases due to (ii) (d) decreases due to (i). the string hangs vertically.: (b) Q47. = rest mass energy mc 2 − m0 c 2 = m0 c 2 mc 2 = 2m0 c 2 m0 1− Q45. p + {p. v2 c2 c 2 = 2m0 c 2 ⇒ 1 1− v2 c2 ⎛ v2 ⎞ v2 3 = 2 ⇒ 4⎜⎜1 − 2 ⎟⎟ = 1 ⇒ 4 2 = 3 ⇒ v = c 2 c ⎝ c ⎠ If the Poisson bracket {x. remains unchanged due to (ii) Ans.: (b) Q48. The binding energy of the k-shell electron in a Uranium atom (Z = 92. p} ⇒ x{x. A small mass M hangs from a thin string and can swing like a pendulum. p} + {x.8 y represents (a) a translation (b) a proper rotation (c) a reflection (d) none of the above Ans. A = 238) will be modified due to (i) screening caused by other electrons and (ii) the finite extent of the nucleus as follows: (a) increases due to (i). p}x + 0 ⇒ x ( −1) + ( −1) x ⇒ −2 x Q46.E. p = x 2 .: (a) { } { } Solution: x 2 + p.: (a) Solution: K . then the Poisson bracket {x2 + p. p} = -1.6 x − 0. rest mass energy = m0 c 2 K .(b) 3c / 4 3c / 2 (a) (c) 3 / 5c (d) c / 2 Ans.6 y. When the car is at rest. decreases due to (ii) (c) increases due to (i). The coordinate transformation x ′ = 0. remains unchanged due to (ii) (b) decreases due to (i). The angle . p}is (a) -2x (b) 2x (c) 1 (d) -1 Ans.E = mc 2 − m0 c 2 . r>a ∫ E. It is polarized in the y-direction and the amplitude of the electric field is E0. Ans. With k = ω/c where c is the speed of light in vacuum.8 ⎠ A charge q is placed at the centre of an otherwise neutral dielectric sphere of radius a and relative permittivity ε r .: (d) Solution: E ⇒ Q50.: (b) Solution: T sin θ = ma .: (b) Solution: E = E0 cos ( kx − ω t ) yˆ . We denote the expression q / 4πε 0 r 2 by E(r).2 m/s2 is approximately (a) 1° (b) 7° (c) 15° (d) 90° Ans. the electric and the magnetic fields are then conventionally given by (a) E = E 0 cos(ky − ω t )xˆ and B = E0 cos(ky − ω t ) zˆ c (b) E = E 0 cos(kx − ω t ) yˆ and B = E0 cos(kx − ω t ) zˆ c (c) E = E 0 cos(kx − ω t )zˆ and B = E0 cos(ky − ω t ) yˆ c (d) E = E 0 cos(kx − ω t )xˆ and B = E0 cos(ky − ω t ) yˆ c Ans. r < a. r > a. a a ⎛ 1. is given by E (r ) / ε r (b) The field outside the sphere.da = Qenc ⇒ E × 4π r 2 = q ⇒ E = q rˆ r > a 4π ∈0 r 2 An electromagnetic wave of frequency ω travels in the x-direction through vacuum.98 ≈ 7 g g ⎝ 9. T cos θ = mg .made by the string with the vertical when the car has a constant acceleration a = 1. (d) The total charge inside a sphere of radius r < a is given by q.2 ⎞ 0 0 ⇒ θ = tan −1 = tan −1 ⎜ ⎟ = 6. is given by E(r) (c) The total charge inside a sphere of radius r > a is given by q. tan θ = Q49. Which of the following statements is false? (a) The electric field inside the sphere. B= ( ) 1 ˆ 1 k × E ⇒ B = ⎡⎣ xˆ × E0 cos ( kx − ωt ) yˆ ⎤⎦ c c ⇒B= E0 E cos ( kx − ω t )( xˆ × yˆ ) ⇒ B = 0 cos ( kx − ω t )( zˆ ) c c .