Title YCT - Assistant Professor Physics 2025-26.pdf ✅

3 YCT Uttar Pradesh Higher Education Service Commission Assistant Professor Exam-2021 PHYSICS Solved Paper 1. The vector potential in a region is given as: Skeâ #es$e ceW meefoMe efJeYeJe keâes efoMe ieÙee nw~ The associated magnetic induction B is: Fmemes mecyeæ ÛegbyekeâerÙe ØesjCe ( ) B nesiee– ˆ ˆ A = (x,y, z) = y + 2xj − i (a) ˆ ˆ i k + (b) ˆ 3k (c) ˆ ˆ − +i 2j (d) ˆ ˆ ˆ − + + i j k Ans. (b) : Given, A yi 2x j = − + ɵ ɵ Magnetic Induction B A = ∇× ɵ i j k d d d B dx dy dz y 2x 0 = − ɵ ɵ ( ) ( ) ɵ B i 0 j 0 k 2 1 = − + + ( ) ɵ ɵ B 3k = 2. F.E.T. is:/F.E.T. nw: (a) Current controlled device/Oeeje efveÙebef$ele Ùeb$e (b) Power controlled device/Meefòeâ (heeJej) efveÙebef$ele Ùeb$e (c) Voltage controlled device/efJeYeJeevlej efveÙebef$ele Ùeb$e (d) Resistance controlled device/ØeeflejesOe efveÙebef$ele Ùeb$e Ans. (c) : F.E.T. is a voltage controle device. A field-Effect transistor (FET) is a three-terminal semiconductor device that controle the flow of current using an electric field applied to one of its terminal and acting as voltage controlled device. 3. The average energy of Planck's oscillator of frequency v at an obsolute temperature T is: T leehe hej v DeeJe=efòe kesâ hueebkeâ oesefue$e keâer Deewmele Tpee& nesleer nw– (a) hv (b) nhv (c) hv/kT hv e 1       − (d) kT Ans. (c) : Average energy of plank oscillator it is total Energy (E) of oscillator and number of an oscillator (N). it is given by Eavg h 1 KT h e ν − ν = Where h = Plank constant ν = Frequency of oscillation K = Boltz man constant T = Absolute temperature 4. Fermi level for extrinsic semiconductor depends on: yee¢e (Skeämš^sveefpekeâ) DeOe&Ûeeuekeâ kesâ efueS Heâceea mlej efveYe&j keâjlee nw– (a) Donor element/oelee lelJe hej (b) Impurity concentration/DeMegælee meevõlee hej (c) Temperature/leehe hej (d) All above/GheÙeg&òeâ meYeer hej Ans. (d) : The Fermi level for extrinsic semiconductor depends on the donor element concentration of impurity and temperature Hence it depend upon all of these. 5. The transformation Q = f (q, p) and P = Q (q, p) is canonical if: Q = f (q, p) SJeb P = Q (q, p) keâe TMheevlejCe ‘‘kewâveesefvekeâue’’ neslee nw Ùeefo: (a) [Q P] = 0 (b) [Q Q] = 1 (c) [Q P] = 1 (d) [P P ] = 1 Ans. (c) : Q = f (q, p) P = g (q, p) {Q, Q} = {P, P} = 0 Poisson bracket, { } A B A B A,B q p p q ∂ ∂ ∂ ∂ = − ∂ ∂ ∂ ∂ So, {Q, p} = 1 6. A particle is constrained to move on the circumference of a circle. The number of degree of freedom of the particle is: Skeâ keâCe Je=òe keâer heefjefOe hej Ietceves keâes Øeefleyeæ nw~ keâCe kesâ ‘ef[«eer Dee@Heâ øeâer[ce’ keâer mebKÙee nesieer: Exam Date: 28.11.2021 [email protected]
4 YCT (a) One/Skeâ (b) Two/oes (c) Three/leerve (d) Four/Ûeej Ans. (a) : A particle constrained to move on the circumference of circle has one degree of freedom. This is because the particle position is fully described by a single variable Degree of freedom (D.O.F.)=Total dimension– constraint D.O.F. = 2–1 = 1 D.O.F. = 1 7. The value of − ∫ x + 2 2 2 e [H (x)] dx − ∞ ∞ is: meceerkeâjCe − ∫ x + 2 2 2 e [H (x)] dx − ∞ ∞ keâe ceeve nesiee: (a) π (b) 2 π (c) 4 π (d) 8 π Ans. (d) : ( ) ( )n 2 2 x x n n d H x 1 e e dx − = − H0(x) = 1 H1(x) = 2x H2(x) = 4x2 –2 H3(x) = 8x3 – 12x So ( ) 2 2 x 2 e 4x x dx ∞ − ∞ = − ∫ ( ) 2 x 4 2 e 16x 16x 4 dx ∞ − −∞ = − + ∫ By solving 3 1 1 16 16 4 2 2 2 = × × π − × π + π = π − π + π 12 8 4 = π 8 8. If φ is a cyclic coordinate, then: Ùeefo φ Skeâ Ûe›eâerÙe efveÙeecekeâ nw, lees– (a) L = 0 ∂ ∂φ i (b) L = 0 ∂ ∂φ (c) L = 0 ∂ ∂φ i i (d) L = 0 ∂ ∂φ i Ans. (b) : If φ is a cyclic co-ordinate, then dL 0 d = φ because the generalized momentum conjugate to a cyclic co-ordinate is conserved Generalized momentum- dL P dq dt =   ∂     If co-ordinate is cyclic, the corresponding generalized momentum is a constant of motion and does not depend upon the time, So rate of change of lagrangian with respect to cyclic co-ordinate will be zero dL 0 d   =     φ 9. A reversible Carnot engine is operted between temperature T1 and T2 (T2 > T1) with a photon gas as the working substance. The efficiency of the engine will be: Heâesšesve iewme keâes ef›eâÙeeMeerue heoeLe& keâer lejn GheÙeesie ceW ueskeâj Skeâ ØeefleJeleea keâeveex (Carnot) Fbpeve leehe T1 leLee T2 (T2 > T1) kesâ yeerÛe ÛeueeÙee peelee nw, Fmekeâer #ecelee: (a) 1 2 T 1 T − (b) 2 1 2 T 1 T   −       (c) 0 (d) ∞ Ans. (a) : Output Input η = 2 2 2 Q Q Q − η = 1 2 Q 1 Q η = − 1 2 T 1 T η = − 10. In a canonical ensemble the quantities same for each system are: efJeefnle (kesâveesefvekeâue) mecegoeÙe ceW ØelÙeskeâ efvekeâeÙe kesâ efueS meceeve neslee nw: (a) Temperature, volume and the number of particles/leehe, DeeÙeleve leLee keâCeeW keâer mebKÙee (b) Energy, volume and the number of particles/ Tpee&, DeeÙeleve leLee keâCeeW keâer mebKÙee [email protected]