JEST 2013 Question n Solution

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 (E + vp ) (E − vp ) (E + vp ) (E − vp ) [1 − (v / c )2 ]2 [1 − (v / c )2 ]2 (a) (b) (c) (d) 1 − (v / c ) 1 − (v / c ) 2 2 Ans.: (a) vx x v v t+ ′ + 2x x+ x 2 x c ⇒ = c c ⇒ x′ = c ∵ x = ct , x′ = ct ′ Solution: t ′ = v 2 c v 2 v2 1− 2 1− 2 1− 2 c c c E v E+ c E E + Pv Now x′ = E ′, x = E ⇒ E ′ = ⇒ E = mc 2 , E = Pc ⇒ P = ⇒ E ′ = v2 c v2 1− 2 1− 2 c c 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 AB 3 3 3 3 3 AB 2 AB 2 2 2 2 2 AB AB 1 1 1 1 1 1 Boson Classical (Maxwell - Boltzman) 1/ 3 Probability ratio: =1 1/ 3 For two particle in different states B B B A B A 3 3 3 3 3 3 3 3 3 B A B A A B 2 2 2 2 2 2 2 2 2 A A 1 A B A B 1 1 1 1 1 1 1 1 Boson Classical (Maxwell-Boltzmann) 1/ 3 1 Probability ratio: = 2/3 2 Q3. 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) ∈ 8.85 × 10 −12 Solution: Characteristic time: τ = = = 8.85 × 10 −18 σ 10 6 Q4. The free fall time of a test mass on an object of mass M from a height 2R to R is R3 R3 R3 2R 3 (a) (π / 2 + 1) (b) (c) (π / 2) (d) π GM GM GM GM Ans.: (a) md 2 r GMm d 2r GM d 2r A Solution: Equation of motion 2 = − 2 ⇒ 2 = − 2 ⇒ 2 =− 2 ∵ GM = A dt r dt r dt r dv A dr d ⎛ v2 ⎞ d ⎛ A⎞ v2 A v =− 2 ⇒ ⎜⎜ ⎟⎟ = ⎜ ⎟ ⇒ = +C dt r dt dt ⎝ 2 ⎠ dt ⎝ r ⎠ 2 r when r = 2 R, v = 0 0 A A v2 A A 2A 2A dr 2A 2R − r = +C ⇒ C = − ⇒ = − ⇒v= − ⇒ = 2 2R 2R 2 r 2R r 2R dt 2R r R r A t ∫2R 2R − r dr = − R ∫0 dt put r = u 2 , dr = 2udu when r = 2 R, r = R, u = 2 R , u = R A unitary operator carrying out these transformations for a system having total mass M.75 (d) 0. A particle of mass m is contained in a one-dimensional infinite well extending from x = -L/2 to x = L/2.99 (b) 0. X / e i t v . Suppose you choose a coin at random and toss it 3 times. X / e i t v . A box contains 100 coins out of which 99 are fair coins and 1 is a double-headed coin. what is the probability that the particle will be in the ground state after this sudden expansion? (a) (8 / 3π ) (c) (16 / 3π ) (d) (4 / 3π ) 2 2 2 (b) 0 . the coordinates and momenta of any particle/ system transform as: t ' = t . R u A t A R u2 ∫ 2R 2R − u 2 × 2udu = − ∫ R 0 dt ⇒ − R t = 2∫ 2R 2R − u 2 du R A ⎡ u 2 R −1 u ⎤ ⇒− t = 2 ⎢− 2R − u 2 + sin ⎥ R ⎣ 2 2 2R ⎦ 2R A ⎡− R 2 R −1 R 2R 2R ⎤ ⇒− t = 2⎢ 2R − R + sin + 2 R − 2 R − R sin −1 ⎥ R ⎣ 2 2 2R 2 2R ⎦ ⎡ − R Rπ Rπ ⎤ R R ⎛π ⎞ ⎛π ⎞ R 3 A ⇒− t = 2⎢ + − ⇒ t = ⎜ + 1⎟ ⇒ t = ⎜ + 1⎟ ∵ A = GM R ⎣ 2 4 2 ⎥⎦ A ⎝2 ⎠ ⎝ 2 ⎠ GM Q5. X / e −i t v .: (b) Q7.P / e i M v Ans. P / e −i M v 2 2 (c) e i M v .01 Ans. r ' = r + v t and p ' = p + mv where v is the velocity of the boosted frame with respect to the original frame. The walls of the box are moved suddenly to form a box extending from x = -L to x = L. What is the probability that the coin you have drawn is the double- headed one? (a) 0.: (c) Q6.P / t / (2 ) t / (2 ) (d) e i t v .925 (c) 0.P / e −i M v 2 (a) e i M v . Under a Galilean transformation. The particle is in its ground state given by ϕ 0 ( x ) = 2 / L cos(πx / L ) . It turns out that the results of all 3 tosses are heads. total momentum P and centre of mass coordinate X is t / (2 ) (b) e i M v . φ0 = cos .Ans. while the electric field outside a spherical shell with a uniform surface 1 q charge density is given by E r < R = rˆ . The correct ratio of 4πε 0 r 2 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.: (a) 2 2 πx 2 πx Solution: Probability φ 0 φ1 .: (d) ∈0 R 2 ∈0 ∞ 2 Solution: Electrostatic energy in spherical shell wsp = 2 ∫0 E1 4π r 2 dr + 2 ∫R E 2 4π r 2 dr ∞ ∈0 ∞ q2 q2 ⎛ 1 ⎞ q2 1 ⇒ 2 ∫ (4π ∈ ) R 2 r4 4π r 2 dr = ⎜− ⎟ = 8π ∈0 ⎝ r ⎠ R 8π ∈0 R 0 ∈0 R ∈0 ∞ Electrostatic energy in solid sphere ws = ∫ E1 4π r 2 dr + ∫ E 2 4π r 2 dr 2 2 2 0 2 R R ∞ q2 1 ⎡r5 ⎤ q2 ⎡ 1⎤ ⇒ × ⎢ ⎥ + ⎢− r ⎥ 8π ∈0 R 6 ⎣ 5 ⎦ 0 8π ∈0 ⎣ ⎦R . φ1 cos L L 2L 2L Since the wall of box are moved suddenly then 2 2 L/2 2 1 cos π x cos π x 2 1 L / 2 2 cos π x cos π x Probability = ∫ −L / 2 L ⋅ L L ⋅ 2L dx = L 2 ∫− L / 2 L ⋅ 2L dx 2 2 2 1 L / 2 ⎡ ⎛ 3π x ⎞ ⎛ π x ⎞⎤ 3π x 2 L π x⎤ L/2 2 1 ⎡ 2L ⇒ ⋅ ∫ ⎢cos ⎜ ⎟ + cos ⎜ ⎟ ⎥ dx ⇒ ⋅ ⎢ sin + sin L 2 − L / 2 ⎣ ⎝ 2L ⎠ ⎝ 2L ⎠⎦ L 2 ⎣ 3π 2L π 2 L ⎦⎥ − L / 2 2 2 1 ⎡ 2 L ⎛ 3π 3π ⎞ 2 L ⎛ π π ⎞⎤ ⇒ ⋅ ⎢ ⎜ sin + sin ⎟+ ⎜ sin + sin ⎟ ⎥ L 2 ⎣ 3π ⎝ 4 4 ⎠ π ⎝ 4 4 ⎠⎦ 2 2 2 2 8 ⇒ + = 3π π 3π Q8. The electric fields outside (r > R) and inside (r < R) a solid sphere with a uniform 1 q 1 q volume charge density are given by Er >R = rˆ and Er<R = rrˆ 4πε 0 r 2 4πε 0 R 3 respectively. q being the total charge. For convenience we take m= = ω = 1 for the oscillator. A spherical planet of radius R has a uniform density ρ and does not rotate. t ) = ψ * ( x. t ) = ψ 0 ( x ) + ψ 1 ( x ) + 2 Reψ 0* ( x )ψ 1 ( x ) cos ( E1 − E0 ) 2 2 2 putting t = π ψ ( x. A quantum mechanical particle in a harmonic oscillator potential has the initial wave function ψ/ 0 ( x ) + ψ/ 1 ( x ). If the planet is made up of some liquid. t ) = ψ 0 ( x ) e − i +ψ 1 ( x ) e−i Now probability density at time t t ψ ( x. 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. 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 2 2 (c) (ψ/ 1 ( x ) + ψ/ 0 ( x )) (d) (ψ/ 1 ( x )) + (ψ/ 0 ( x )) 2 2 2 Ans. q2 1 q2 6q 2 ws = ⋅ + = 5 × 8π ∈0 R 8π ∈0 R 40π ∈0 R q2 Wspherical 8π ∈0 5 Now = 2 = Wsphere 6q 6 40π ∈0 R Q9. where ψ/ 0 and ψ/ 1 are the real wavefunctions in the ground and first excited state of the harmonic oscillator Hamiltonian.: (a) Solution: ψ (x ) = ψ 0 (x ) + ψ 1 ( x ) E0t E1t ψ ( x. t )ψ ( x. t ) = ψ 0 ( x ) + ψ 1 ( x ) − 2 Reψ 0* ( x )ψ 1 ( x ) = ⎡⎣ψ 1 ( x ) −ψ 0 ( x ) ⎤⎦ 2 2 2 2 Q10. the pressure at point r from the center is 4πρ 2 G 2 4πρG 2 (a) 3 ( R − r2 ) (b) 3 R − r2( ) 2πρ 2 G 2 ρG (c) 3 ( R − r2 ) (d) 2 (R 2 − r2 ) . : (c) r ρ ⋅ 4πr 2 drGM dm ⋅ g dm ⋅ g R3 Solution: Pressure dp = ⇒ dp = ⇒ dp = A 4πr 2 4πr 2 dr r dm (mass of elementary part ) R 4π 3 r ρ ⋅ 4πr 2 drG ⋅ ρ ⋅ R 3 ⇒ dp = 3 R ⇒ dp = 4π ρ 2 Grdr 4πr 2 3 R R 4π 2 4π 2 ⎛ r 2 ⎞ 4π 2 ⎛ R 2 r 2 ⎞ ∫ dp = ∫ r 3 ρ Grdr ⇒ p = 3 ρ G⎜⎜ ⎝ 2 ⎟⎟ ⇒ p = ⎠r 3 ρ G⎜⎜ ⎝ 2 − ⎟⎟ 2⎠ 4π ρ 2 G 2 2π 2 ⇒ p= 3 2 (R − r2 )⇒ p = 3 ρ G (R 2 − r 2 ) Re(z 2 ) + Im(z 2 ) Q11. Compute lim z →0 z2 (a) The limit does not exist. (b) 1 (c) –i (d) -1 Ans. If the particle attains a maximum height after time t.Ans. 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. and g is the gravitational acceleration.: (a) Solution: lim ( ) Re z 2 + Im z 2 ( ) = lim x 2 − y 2 + 2 xy ⇒ lim x 2 − y 2 + 2 xy =1 z →0 z2 z →0 x 2 − y 2 + 2ixy y = 0 x 2 − y 2 + 2ixy x →0 x 2 − y 2 + 2 xy lim 2 = −1 x = 0 x − y 2 + 2ixy y →0 Q12. what is the velocity? ⎛ g ⎞ k ⎛ g ⎞ gk tan⎜⎜ t ⎟⎟ (a) tan⎜⎜ t ⎟⎟ (b) k g ⎝ ⎠ ⎝ k ⎠ . : (d) Solution: It is isobaric process (constant pressure) Then δθ = nC p ΔT ⇒ ΔW = nRΔT In this process δθ is heat exchange during process. 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. 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 (d) 2/7 Ans. Function of heat supplied δW nRΔT R γ −1 1 = = = = = 1− ΔQ nC p ΔT R γ γ γ γ −1 1 Cp γR ⇒ 1− γ= ⇒ Cp = ⎛ 2⎞ CV γ −1 ⎜1 + f ⎟ ⎝ ⎠ .: (c) Q14. Consider a uniform distribution of particles with volume density n in a box. (c) g k tan ( gk t ) (d) gk tan ( gk t ) Ans.: (c) mdv dv k dv Solution: Equation of motion = mg + kv 2 ⇒ = g + v2 ⇒ = dt dt dt m k 2 g+ v m dv dv m 1 v ⇒∫ = ∫ dt ⇒ ∫ = ∫ dt ⇒ × tan −1 =t k k ⎛ gm ⎞ k gm gm g + v2 ⎜ + v2 ⎟ m m⎝ k ⎠ k k v gk gm gk ⇒ tan −1 = t⇒v= tan t gm m k m k Q13. For a diatomic ideal gas near room temperature. The particles have an isotropic velocity distribution with constant magnitude v. A metal suffers a structural phase transition from face-centered cubic (FCC) to the simple cubic (SC) structure. for diatomic molecule f = 5 f +2 5 2 ⇒ 1− ⇒ 5+2 7 Q15.e.: (d) 1 1 1 Solution: neff = nC + n f + 1× ni = × 6 + 1 = 3 a = 2r 3 2 3 neff × A 3 × πr 2 π Now largest fraction of area i. 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.: (b) 1 i ⎡0 1 ⎤ ⎡0 0⎤ Solution: J x = ( J+ + J− ) . packing fraction = = = × (2r ) 3 3 2 3 6× × a2 6× 2 4 4 Q16. A flat surface is covered with non-overlapping disks of same size. Jz are angular momentum operators. J− = ⎢ ⎥ 2 2 ⎣0 0 ⎦ ⎣1 0 ⎦ ⎡0 1 ⎤ i ⎡0 −1⎤ J + J y 1 ⎡ 0 1− i⎤ Jx = ⎢ ⎥ . What is the largest fraction of the area that can be covered? 3 5π 6 π (a) (b) (c) (d) π 6 7 2 3 Ans. J y = ( J− − J+ ) ⇒ J+ = ⎢ ⎥ .26] . If Jx. Jy = ⎢ ⎥ ⇒ x = ⎢ 2 ⎣1 0 ⎦ 2 ⎣1 0 ⎦ 2 ⎣1 + i 0 ⎥⎦ 1 ⎛ − λ 1− i⎞ eigen value ⎜⎜ ⎟⎟ ⇒ λ2 − 2 = 0 ⇒ λ = ± 2 2 ⎝1 + i − λ ⎠ Q17. Jy. 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. f ⇒ 1− f = degree of freedom. It is observed that this phase transition does not involve any change of volume. 17: () Solution: Nearest neighbour in SC → a → C. l = 2rp .: (b) ⎛ 494 494 106 106 ⎞ 2 (m 2 − mμ2 ) c 2 ⎜ 2 × 2 − 2 × 2 ⎟c ⇒⎝ c c c c ⎠ Solution: k → μ + ν . in a Kepler potential. then (a) rc = 2rp (b) rc = rp (c) 2rc = rp (d) rc = 2rp Ans. which can be massless. 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. = 1. l = rC ⇒ 2rp = rC rp rC Q19. (a) 1. If. The energy of the neutrino.N = 12 2 a dFCC 1 1 = 2 = = = 0.414 Q18. is approximately (a) 120 MeV (b) 236 MeV (c) 300 MeV (d) 388 MeV Ans.374 (d) 1.707 dSC a 2 1. e = 0 . 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 .: (a) l Solution: Ionic equation = 1 + e cos θ for parabola e = 1 for circle.N = 6 a Nearest neighbour in FCC → → C. Eν = k 2mk 494 2× 2 c 11236 ⇒ 244036 − = 235. 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.029 (b) 1. θ = 0 r l l = 1+ e .6275 ≈ 236 MeV 988 Q20.122 (c) 1.130 Ans. 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.: (d) Q22. Ay = y. The probability that there are no buses in five minutes is closest to (a) 0.60 (c) 0. The probability that they meet after n steps is 1 2n! 1 2n! 1 1 (a) (b) (c) 2n! (d) n! 4 n n!2 2 n n!2 2n 4n Ans. (d) ρ (z sin 2 φ + cos 2 φ )eˆ ρ + ρ sin φ cos φ (1 − z )eˆφ Ans.: (a) Solution: Into probability of taking ' r ' steps out of N steps r N −r ⎛1⎞ ⎛1⎞ = N Cr ⎜ ⎟ ⎜ ⎟ ⎝2⎠ ⎝2⎠ total steps = N = n + n = 2n for taking probability of n steps out of N n N −n n 2n−n 2n ⎛1⎞ ⎛1⎞ N! ⎛1⎞ ⎛1⎞ 2n ! ⎛ 1 ⎞ 2n ! P = N Cn ⎜ ⎟ ⎜ ⎟ = = = ⎝2⎠ ⎝2⎠ ( N − n ) !n ! ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜ ⎟ n !n ! ⎝ 2 ⎠ ( n !) 2 4n . There are on average 20 buses per hour at a point.19 Ans. each having equal probability of making a step simultaneously to the left or right along the x axis. but at random times.36 (d) 0.: (a) Solution: A = xziˆ + yˆj Ax = xz .07 (b) 0. Two drunks start out together at the origin. 0.: (c) Q24. 1 ⎛1⎞ ⎛1⎞ ⎜ ⎟ 1 ⎜ ⎟ Now state after measurement yielding 1⇒ φ1 + φ 3 = ⎜0⎟ = ⎜0⎟ ⎜1⎟ 2 ⎜1⎟ ⎝ ⎠ ⎝ ⎠ . what is the state after the measurement? ⎛1⎞ ⎛ 1/ 3 ⎞ ⎛ 0⎞ ⎛1 / 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (a) ⎜ 0 ⎟ (b) ⎜ 0 ⎟ (c) ⎜ 0 ⎟ (d) ⎜ 0 ⎟ ⎜ 0⎟ ⎜⎜ ⎟⎟ ⎜1⎟ ⎜⎜ ⎟⎟ ⎝ ⎠ ⎝ 2/3⎠ ⎝ ⎠ ⎝1 / 2 ⎠ Ans. Consider the state ⎜⎜1 / 2 ⎟⎟ corresponding to the angular momentum l = 1in the Lz basis ⎝1 / 2 ⎠ of states with m = +1.Q23. If L2z is measured in this state yielding a result 1.: (d) ⎛1 0 0 ⎞ ⎛1 0 0⎞ ⎛1⎞ ⎛0⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Solution: L z = ⎜ 0 0 0 ⎟ . ⎜1⎟. -1.: (b) ⎛1 / 2 ⎞ Q25. eigenvector 2 ⎜ 0⎟ . 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. A simple model of a helium-like atom with electron-electron interaction is replaced by Hooke’s law force is described by Hamiltonian − 2 2 λ 2m ( ) 1 ( ∇1 + ∇ 22 + mω 2 r12 + r22 − 2 ) 4 mω 2 r1 − r2 . ⎜0⎟ ⎜ 0 0 − 1⎟ ⎜0 0 1⎟ ⎜ 0⎟ ⎜0⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Corresponding eigenvalue 1. Lz = ⎜ 0 0 0 ⎟ . 0. 2 What is the exact ground state energy? (a) E = 3 2 ( ω 1+ 1+ λ ) (b) E = 3 2 ω 1+ λ ( ) (c) E = 3 2 ω 1− λ (d) E = 3 2 ( ω 1+ 1− λ ) Ans. 5° C (c) 10° C (d) 8. If 50% of the heat generated remains in the bullet. Consider the differential equation dG ( x ) + kG ( x ) = δ ( x ) . B∝ r r r r So m = n = 1 Q28.2° C Ans. dx where k is a constant. what is the increase in its temperature? (The specific heat of the bullet = 160 Joule per Kg per degree C) (a) 14° C (b) 12.: (c) qa sin θ qa sin θ 1 1 Solution: For large distance F = .: (c) Q29.: (c) Q 2 Q Qω r 2 Solution: Magnetic dipole moment M ′ = IA = πr ⇒ × 2π × π r = 2 T 2π T 2 M′ Q Angular momentum J = Mr 2ω ⇒ = J 2M Q27. m = 2 Ans. B= ⇒E∝ . (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. Ans.: (c) . A metal bullet comes to rest after hitting its target with a velocity of 80m/s. A thin uniform ring carrying charge Q and mass M rotates about its axis. m = 1 (c) n = 1. PART B: ONE MARK QUESTIONS Q26. 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. What are the value of n and m? (a) n = 1. m = 1 (d) n = 2. m = 2 (b) n = 2. 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. a z ) ⎛0 1⎞ ⎛ 0 −i ⎞ ⎛1 0 ⎞ =⎜ ⎟ ax + ⎜ ⎟ ay + ⎜ ⎟ az ⎝1 0⎠ ⎝i 0 ⎠ ⎝ 0 −1⎠ ⎛ az (a − ia y )⎞ ⎛ (a z − λ ) (a x − ia y ) ⎞ ⇒ ⎜⎜ ⎟⇒⎜ ⎟ x ⎝ (a x + ia y ) − a z ⎟⎠ ⎜⎝ (a x + ia y ) − (a z + λ )⎟⎠ ⇒ −(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) ∂ ⎞ ⎛ −∂ψ ( x ) ⎞ † * ⎛ * Solution: ⇒ ⎜ψ ( x ) − ψ ( x ) ⎟ = ⎜ ψ ( x) ⎟ ⎝ ∂x ⎠ ⎝⎜ ∂x ⎟ ⎠ ∞ ⎡ ∂ ⎤ ∞ ∞ ∂ψ ( x ) * ⇒ ∫ ψ * ( x ) ⎢ − ψ ( x ) ⎥ dx −ψ * ( x )ψ ( x ) − ∫ − ψ ( x ) dx −∞ ⎣ ∂x ⎦ −∞ −∞ ∂x ∞ ∂ψ * ( x ) ⇒∫ ψ ( x ) dx −∞ ∂x . 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 y + σ z .: (d) Solution: H = σ ⋅ a = (σ x . The hermitian conjugate of the operator ⎜ ⎟ is ⎝ ∂x ⎠ ∂ ∂ ∂ ∂ (a) (b) − (c) i (d) − i ∂x ∂x ∂x ∂x Ans. 1 2 1 Solution: Conservation of momentum mv × 50 % = mcΔT ⇒ 80 × 80 = 160 ΔT 2 2 80 × 80 1 ⇒ ΔT = × = 100 C 4 160 Q30. What are the eigenvalues of the operator H = σ ⋅ a .a x + σ y . : (b) 1 Solution: Electrons are Fermions of spin 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) . then the expectation value of momentum for the wavefunction e i k x / υ/ ( x ) is (a) k (b) p − k (c) p + k (d) p Ans.Q32. Two electrons are confined in a one dimensional box of length L. x 2 ) = sin ⎜ ⎟ sin ⎜ ⎟ L ⎝ L ⎠ ⎝ L ⎠ 2 ⎛ 2πx1 ⎞ ⎛ πx 2 ⎞ (d) υ/ ( x1 . What would be the ground state wave function υ/ ( x1 . x 2 ) if both electrons are arranged to have the same spin state? 1 ⎡ 2 ⎛ πx1 ⎞ ⎛ 2πx 2 ⎞ 2 ⎛ 2πx1 ⎞ ⎛ πx 2 ⎞⎤ (a) υ/ ( x1 .: (c) ∞ ⎛ ∂ ⎞ Solution: ∫ ψ * ( x ) ⎜ −i ⎟ψ ( x ) dx = p −∞ ⎝ ∂x ⎠ Now ∞ − ikx ⎛ ∂ ⎞ ikx ∞ − ikx ⎡ ikx ∂ ik ikx ⎤ ∫ e ψ * ( x ) ⎜ −i ⎟ e ψ ( x ) dx ⇒ ∫ e ψ * ( x )( −i ) ⎢e ψ ( x ) + e ψ ( x )⎥ −∞ ⎝ ∂x ⎠ −∞ ⎣ ∂x ⎦ ∞ − ikx ⎛ ∂ ⎞ ikx ∞ ik − ikx ⇒∫ e ψ * ( x ) ⎜ −i ψ ( x ) ⎟ e + ∫ −i . x 2 ) = sin ⎜ ⎟ sin ⎜ ⎟ L ⎝ L ⎠ ⎝ L ⎠ Ans. The one-electron states are given by υ/ n ( x ) = 2 / L sin (nπx / L ) . x 2 ) = ⎢ sin ⎜ ⎟ sin ⎜ ⎟ + sin ⎜ ⎟ sin ⎜ ⎟⎥ 2 ⎣L ⎝ L ⎠ ⎝ L ⎠ L ⎝ L ⎠ ⎝ L ⎠⎦ 1 ⎡ 2 ⎛ πx1 ⎞ ⎛ 2πx 2 ⎞ 2 ⎛ 2πx1 ⎞ ⎛ πx 2 ⎞⎤ (b) υ/ ( x1 . x 2 ) = ⎢ sin ⎜ ⎟ sin ⎜ ⎟ − sin ⎜ ⎟ sin ⎜ ⎟⎥ 2 ⎣L ⎝ L ⎠ ⎝ L ⎠ L ⎝ L ⎠ ⎝ L ⎠⎦ 2 ⎛ πx1 ⎞ ⎛ 2πx 2 ⎞ (c) υ/ ( x1 . e ψ * ( x )ψ ( x ) dx −∞ ⎝ ∂x ⎠ −∞ ∞ ⎡ ∂ ⎤ ∞ ⇒ ∫ ψ * ( x ) ⎢ −i ψ ( x )⎥ + k ∫ ψ * ( x )ψ ( x ) ⇒ P + K −∞ ⎣ ∂x ⎦ −∞ Q33. If the expectation value of the momentum is p for the wavefunction υ/ ( x ) . ... ⎟ ⎝ 2! 4! ⎠ ⎝ 3! 5! ⎠ (a) 0 (b) e (c) e2 (d) 1 Ans. Then 1 ⎡ 2 ⎛ πx1 ⎞ ⎛ 2πx 2 ⎞ 2 ⎛ 2πx1 ⎞ ⎛ πx 2 ⎞⎤ = ⎢ L sin ⎜ L ⎟ ....⎟ + ⎜1 − + − ..: (a) c v+ −1 n = ⎛v + c ⎞⎛ v ⎞ ⎛ c ⎞⎛ v v2 ⎞ Solution: now u = ⎜ ⎟⎜1 + ⎟ = ⎜ v + ⎟ ⎜ 1 − + 2 2 ⎟ v⋅c ⎝ n ⎠⎝ cn ⎠ ⎝ n ⎠ ⎝ cn c n ⎠ 1+ 2 c ⋅n v2 v3 c v cv 2 c ⎛ 1 ⎞ ⇒v− + 2 2 + − 2 + 3 ⇒ u = + v ⎜1 − 2 ⎟ cn c n n cn cn n ⎝ n ⎠ Q36. What is the value of the following series? 2 2 ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎜1 − + − ... sin ⎜ L ⎟ − L sin ⎜ L ⎟. .⎟ + ⎜1 − + ⎟ ⇒ cos12 + sin 12 = 1 ∵ sin 2 θ + cos 2 θ = 1 ⎝ 2! 4! ⎠ ⎝ 3! 5! ⎠ Q35. 2! 4! 3! 5! 2 2 ⎛ 1 1 ⎞ ⎛ 1 1⎞ ⇒ ⎜1 − + .... the mean value of x is (a) λ (b) 2λ (c) λ / 2 (d) 0 . If the distribution function of x is f ( x ) = xe − x / λ over the interval 0 < x < ∞.. A light beam is propagating through a block of glass with index of refraction n...: (d) θ2 θ4 θ3 θ5 Solution: ⇒ cos θ = 1 − + . sin θ = θ − + . the velocity of the light in the glass block as measured by an observer in the laboratory is approximately c ⎛ 1 ⎞ c ⎛ 1 ⎞ (a) u = + v ⎜1 − 2 ⎟ (b) u = − v⎜ 1 − 2 ⎟ n ⎝ n ⎠ n ⎝ n ⎠ c ⎛ 1 ⎞ c (c) u = + v⎜1 + 2 ⎟ (d) u = n ⎝ n ⎠ n Ans. sin ⎜ L ⎟⎥ 2⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ Q34. If the glass is moving at constant velocity v in the same direction as the beam. 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. There is a magnetic field h present and the energy for a spin state s is –hs. Consider a particle with three possible spin states: s = 0 and ±1. S = 0 Q38.xe λ dx ∫ x 2 e λ dx Solution: ∵ it is distribution function so x = −∞ ∞ = 0 ⇒ 0 −x = 2λ f ( x ) dx x ∫ −∞ ∫ ∞ 0 − xe dx λ ∫ ∞ 0 λ xe dx Q37. and ln 3 at high T Ans. and 3 at high T (d) S(T) = 0 at T = 0. Which of the following statements is true about the entropy S(T)? (a) S(T) = ln 3 at T = 0.e.: (b) ⎛d ⎞⎛ d ⎞ ⎛d ⎞⎡ d ⎤ Solution: ⇒ ⎜ − x ⎟⎜ + x ⎟ f ( x ) ⇒ ⎜ − x ⎟ ⎢ f ( x ) + xf ( x )⎥ ⎝ dx ⎠⎝ dx ⎠ ⎝ dx ⎠ ⎣ dx ⎦ d ⎡d ⎤ ⇒ ⎢ f ( x ) + xf ( x )⎥ − x d f (x ) − x 2 f (x ) dx ⎣ dx ⎦ dx d2 df ( x ) d ⇒ 2 f ( x) + f ( x) + x −x f ( x ) − x2 f ( x ) dx dx dx .: (b) x −x ∞ ∞ − ∞ ∫ xf ( x )dx ∫ x. The operator ⎛d ⎞⎛ d ⎞ ⎜ − x ⎟⎜ + x ⎟ ⎝ dx ⎠⎝ dx ⎠ is equivalent to d2 d2 (a) 2 − x2 (b) 2 − x2 +1 dx dx d2 d d2 d (c) 2 − x x2 +1 (d) 2 − 2x − x 2 dx dx dx dx Ans. The system is at a temperature T.: (d) Solution: S = k ln ω where ω = number of microstates S = k ln 3 at high T ω =3 and at T = 0 it is perfect ordered i. 99995 me Solution: En = × ⇒ −13. EB and Ecl respectively. which of the following is true? (a) EF = EB = Ecl (b) EF > EB = Ecl (c) EF < EB < Ecl (d) EF > EB > Ecl Ans. the ground state energy of the electron in a hydrogen atom would be (a) less (b) more (c) the same (d) less.: (b) −13.99995me n2 me Q41. 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. Consider three situations of 4 particles in one dimensional box of width L with hard walls. and in case (iii) they are classical. more or equal depending on the electron mass Ans. If the total ground state energy of the four particles in these three cases are EF. the particles are fermions.: (d) Solution: For electrostatic field ∇ × E = 0 . For Maxwell = 4×∈0 EF > EB = Ecl Q40. In case (i). If a proton were ten times.: (b) π2 2 Solution: For fermions =∈0 2ml 2 1×∈0 +1× 4 ∈0 +1× 9 ∈0 +1× 16 ∈0 = 30 ∈0 For Boson = 4×∈0 . d2 ⎛ d2 ⎞ ⇒ 2 f ( x ) − x 2 f ( x ) + f ( x ) = ⎜⎜ 2 − x 2 + 1⎟⎟ f (x ) dx ⎝ dx ⎠ Q39.59932 ∵ μ = 0. in case (ii) they are bosons.6 0. The period of a simple pendulum inside a stationary lift is T. 238 U decays with a half life of 4. A rock sample analysis shows that the ratio of the numbers of 206 238 206 atoms of Pb to U is 0. the period of the pendulum will be (a) T (b) T / 4 (c) 2T / 3 (d) 2T / 5 Ans. If the lift accelerates downwards with an acceleration g / 4.51 × 109 years.0058. the speed of light in vacuum) . the decay series eventually ending at 206 Pb. Assuming that all the Pb has been produced by the 238 decay of U and that all other half-lives in the chain are negligible. The velocity of a particle at which the kinetic energy is equal to its rest energy is (in terms of c. which is stable.: (a) 1 ⎛ N pb + N u ⎞ Solution: t = ln ⎜ ⎟ λu ⎝ N u ⎠ 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.: (c) l Solution: T = 2π ⇒ lift accelerates down wards then g l l 4l l T = 2π ⇒ T = 2π = 2π ⇒ 2π × 2 g − g′ g 3g 3g g− 4 2T T′ = 3 Q44. iˆ ˆj kˆ ∂ ∂ ∂ ∇ × E2 = ∂x ∂y ∂z y2 2 xy + z 2 2 yz (2 z − 2 z )iˆ + 0 + (2 y − 2 y )zˆ = 0 iˆ ˆj kˆ ∂ ∂ ∂ ∇ × E1 = = ( 0 − 2 y ) iˆ + 0 + xjˆ ≠ 0 ∂x ∂y ∂z xy ∂yz yxz Q42. E. rest mass energy = m0 c 2 K . When the car is at rest. p}x + 0 ⇒ x ( −1) + ( −1) x ⇒ −2 x Q46.: (a) Solution: K .: (a) { } { } Solution: x 2 + p.8 y represents (a) a translation (b) a proper rotation (c) a reflection (d) none of the above Ans.8 x + 0.6 y. p = x 2 . decreases due to (ii) (c) increases due to (i).: (b) Q48. = rest mass energy mc 2 − m0 c 2 = m0 c 2 mc 2 = 2m0 c 2 m0 1 ⎛ v2 ⎞ v2 3 c 2 = 2m0 c 2 ⇒ = 2 ⇒ 4⎜⎜1 − 2 ⎟⎟ = 1 ⇒ 4 2 = 3 ⇒ v = c v2 v2 ⎝ c ⎠ c 2 1− 1− c2 c2 Q45. remains unchanged due to (ii) (b) decreases due to (i). It is attached above the window of a car.E = mc 2 − m0 c 2 . p} ⇒ x{x. p}is (a) -2x (b) 2x (c) 1 (d) -1 Ans. p} = -1. p + {p. then the Poisson bracket {x2 + p. The binding energy of the k-shell electron in a Uranium atom (Z = 92. the string hangs vertically. y ′ = 0. remains unchanged due to (ii) Ans. p} + {x.: (b) Q47. (a) 3c / 2 (b) 3c / 4 (c) 3 / 5c (d) c / 2 Ans. The angle . The coordinate transformation x ′ = 0.6 x − 0. 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). If the Poisson bracket {x. A small mass M hangs from a thin string and can swing like a pendulum. increases due to (ii) (d) decreases due to (i). Ans. It is polarized in the y-direction and the amplitude of the electric field is E0.: (b) a a ⎛ 1. Which of the following statements is false? (a) The electric field inside the sphere. tan θ = ⇒ θ = tan −1 = tan −1 ⎜ ⎟ = 6. A charge q is placed at the centre of an otherwise neutral dielectric sphere of radius a and relative permittivity ε r .98 ≈ 7 0 0 g g ⎝ 9.da = Qenc ⇒ E × 4π r 2 = q ⇒ E = 4π ∈0 r 2 rˆ r > a Q50. is given by E (r ) / ε r (b) The field outside the sphere. An electromagnetic wave of frequency ω travels in the x-direction through vacuum. With k = ω/c where c is the speed of light in vacuum.: (b) Solution: E = E0 cos ( kx − ω t ) yˆ .2 m/s2 is approximately (a) 1° (b) 7° (c) 15° (d) 90° Ans. made by the string with the vertical when the car has a constant acceleration a = 1. T cos θ = mg . r > a. We denote the expression q / 4πε 0 r 2 by E(r). is given by E(r) (c) The total charge inside a sphere of radius r > a is given by q.2 ⎞ Solution: T sin θ = ma .: (d) Solution: E r>a q ⇒ ∫ E. the electric and the magnetic fields are then conventionally given by (a) E = E 0 cos(ky − ω t )xˆ and B = cos(ky − ω t ) zˆ E0 c E0 (b) E = E 0 cos(kx − ω t ) yˆ and B = cos(kx − ω t ) zˆ c E0 (c) E = E 0 cos(kx − ω t )zˆ and B = cos(ky − ω t ) yˆ c E0 (d) E = E 0 cos(kx − ω t )xˆ and B = cos(ky − ω t ) yˆ c Ans.8 ⎠ Q49. (d) The total charge inside a sphere of radius r < a is given by q. r < a. B= c ( 1 ˆ ) 1 k × E ⇒ B = ⎡⎣ xˆ × E0 cos ( kx − ωt ) yˆ ⎤⎦ c E0 E ⇒B= cos ( kx − ω t )( xˆ × yˆ ) ⇒ B = 0 cos ( kx − ω t )( zˆ ) c c . 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