UPHESC Assistant Professor Physics 2025 PYQs

Concept: First check whether the Zener is ON; if ON, the load voltage is clamped at $V_Z$ and $I_Z=I_S-I_L$.

1. Without Zener: $V_L=120\times \frac{10}{15}=80$ V > 50 V, so the Zener conducts and $V_L=50$ V.
2. Series current $I_S=\frac{120-50}{5\mathrm{k}\mathrm{\Omega }}=14$ mA.
3. Load current $I_L=\frac{50}{10\mathrm{k}\mathrm{\Omega }}=5$ mA.
4. $I_Z=14-5=9$ mA.

Short trick: $I_Z=I_S-I_L$; 14 mA (option D) is the trap – that is the series current.

Concept: CE amplifier gives high current gain ($\beta$) and high voltage gain, hence the highest power gain; it is the standard voltage amplifier.

1. CE: $A_v=-\beta R_C/r_{in}$ – large (hundreds).
2. CB: current gain $\alpha <1$, very low input impedance – used for impedance matching / high frequency.
3. CC (emitter follower): $A_v\approx 1$, so it is the lowest.

Short trick: CE = ‘everything high’ (voltage, current, power gain); CC = voltage gain ≈ 1.

Concept: In GaAs (III–V), a dopant with one electron fewer than the atom it replaces is an acceptor.

1. Cd (group II, 2 valence e⁻) replaces Ga (group III) → one electron short → acceptor → p-type.
2. Se, Te (group VI) replace As (group V) → extra electron → n-type.
3. Si (group IV) is amphoteric; in GaAs it mostly sits on Ga sites → n-type.

Short trick: Group II on Ga site = p (Zn, Cd, Be, Mg); Group VI on As site = n (S, Se, Te).

Concept: CMRR $=20{\log }_{10}(A_d/A_{cm})$; ideal op-amp has infinite CMRR, practical good op-amps have about 90–120 dB.

1. 20–60 dB is poor (a simple discrete differential stage).
2. IC 741 ≈ 90 dB; precision op-amps ≈ 120 dB ($A_d/A_{cm}\approx {10}^6$).
3. 200 dB (${10}^{10}$) is unrealistic.

Short trick: Good op-amp CMRR ≈ 100–120 dB.

Concept: Basic gates implement the three Boolean operations.

1. AND: $Y=A\cdot B$ → Boolean multiplication (3).
2. OR: $Y=A+B$ → Boolean addition (2).
3. NOT: $Y=\overline{A}$ → complementation (1).

Short trick: AND = ‘·’, OR = ‘+’, NOT = ‘bar’ → 3-2-1.

Concept: A photodetector converts light into an electrical signal by the photoelectric (internal or external) effect.

1. A photon with $h\nu \ge E_g$ is absorbed and lifts an electron from valence to conduction band (electron–hole pair).
2. These carriers give photocurrent (photodiode, photoconductor, PMT).
3. Thermal emission is noise (dark current), not the working principle; tunneling and magnetic interaction are irrelevant.

Short trick: Photodetector = absorb photon → excite electron → current.

Concept: A 1-to-$2^n$ DEMUX with its data input held at logic 1 behaves exactly as an $n$-to-$2^n$ decoder.

1. So (A) is true.
2. A DEMUX needs NOT gates (inverters) on the select lines in addition to AND gates; it can also be built with NAND/NOR gates alone. ‘AND gates only’ is wrong, so (R) is false.

Short trick: DEMUX + data = 1 → decoder; circuit = AND + NOT.

Concept: $n=\frac{\rho N_A}{M}$ with $M$ in kg/mol.

1. $M=197$ g/mol $=0.197$ kg/mol.
2. $n=\frac{19.32\times {10}^3\times 6.022\times {10}^{23}}{0.197}=5.9\times {10}^{28}{\mathrm{m}}^{-3}$.
3. (In per cm³ it would be $5.9\times {10}^{22}$.) None of the printed options equals $5.9\times {10}^{28}$ per m³; option A has the right mantissa but the power is short by 10, so the question is treated as a bonus.

Short trick: Metals have ~${10}^{28}$–${10}^{29}$ atoms per m³.

Concept: In an odd-A nucleus, the ground-state $J^{\pi }$ is that of the last unpaired nucleon.

1. ${}^{25}$Mg: Z = 12 (even, paired), N = 13 (odd).
2. Filling: $1s_{1/2}$(2), $1p_{3/2}$(4), $1p_{1/2}$(2) = 8; then $1d_{5/2}$ holds neutrons 9–14.
3. 13th neutron is in $1d_{5/2}$: $j=5/2$, $l=2$ → parity $(-1{)}^2=+$. So $5/2^{+}$ (also the observed value).

Short trick: Nucleon number 9–14 → $d_{5/2}$ → $5/2^{+}$.

Concept: $R=R_0{A}^{1/3}$, so $R\propto A^{1/3}$.

1. $\frac{R}{R_{Fe}}=\frac{1}{2}\Rightarrow \frac{A}{56}=\frac{1}{8}$.
2. $A=7$ → ${\mathrm{Li}}^7$.

Short trick: Half radius = one-eighth mass number.

Concept: Fusion fuel (deuterium from sea water) is practically inexhaustible, so controlled fusion can be an almost unlimited power source – (A) true.

1. Energy per event: D–T fusion ≈ 17.6 MeV; fission of one ${}^{235}$U ≈ 200 MeV.
2. So a single fusion gives LESS energy than a single fission – (R) false.
3. Fusion wins only per unit mass (≈3.5 MeV/nucleon vs ≈0.85 MeV/nucleon).

Short trick: Per reaction: fission bigger; per kg: fusion bigger.

Concept: Surface nucleons have fewer neighbours; their number ∝ surface area $4\pi R^2$.

1. $R=R_0{A}^{1/3}\Rightarrow$ area ∝ $A^{2/3}$.
2. Surface term $=-a_s{A}^{2/3}$.
3. Others: volume $a_{v}A$, Coulomb $a_{c}Z(Z-1)A^{-1/3}$, asymmetry $(A-2Z{)}^2/A$, pairing ∝ $A^{-3/4}$ (or $A^{-1/2}$).

Short trick: Volume → $A$, Surface → $A^{2/3}$, Coulomb → $A^{-1/3}$.

Concept: Each lepton family number ($L_e$, $L_{\mu }$) is separately conserved.

1. Left: $L_{\mu }=-1$, $L_e=0$. Right: $L_{\mu }=0$, $L_e=-1$ → muon and electron lepton numbers both violated.
2. Charge (+1 = +1), baryon number (0 = 0), energy–momentum (muon heavier) and angular momentum ($\frac{1}{2}\to \frac{1}{2}+1$) can all be satisfied.
3. Observed decay is ${\mu }^{+}\to e^{+}{\nu }_e{\overline{\nu }}_{\mu }$.

Short trick: No neutrinos in a muon decay ⇒ lepton (flavour) number broken.

Concept: A composite particle made of an odd number of fermions has half-integer spin → fermion; even number → boson.

1. α-particle: 2p + 2n = 4 → boson.
2. ${}^7$Be nucleus: 4p + 3n = 7 → fermion.
3. H atom: p + e = 2 → boson; deuteron: p + n = 2 → boson.

Short trick: Count the constituent fermions – odd ⇒ fermion.

Concept: Face diagonals are $⟨110⟩$ directions; angle from $\cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|a||b|}$.

1. Two face diagonals meeting at a cube corner (adjacent faces): $[110]$ and $[011]$ → $\cos \theta =\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$ → $60°$.
2. Together with the third one, $[101]$, they form an equilateral triangle – all angles $60°$.
3. (The two diagonals of the same face, $[110]$ and $[1\overline{1}0]$, are at $90°$; the standard textbook answer for ‘angle between face diagonals’ is the $⟨110⟩$ pair from a common vertex, $60°$.)

Short trick: Face diagonals from one corner make an equilateral triangle → 60°.

Concept: $\mu =\sigma R_H=\frac{R_H}{\rho }$ ⇒ $R_H=\rho \mu$.

1. $\rho =7\times {10}^{-3}\mathrm{\Omega }$ m, $\mu =0.05$ m²/V s.
2. $R_H=7\times {10}^{-3}\times 0.05=3.5\times {10}^{-4}$ m³/C.

Short trick: Hall coefficient = resistivity × mobility.

Concept: ${\mu }_r=1+\chi$; the two rise and fall together.

1. Ferromagnets (Fe, Co, Ni): $\chi \sim {10}^2$–${10}^5$, so ${\mu }_r$ is also very large.
2. ‘High one, low other’ is impossible because ${\mu }_r=1+\chi$.

Short trick: Ferro: $\chi \gg 1$, ${\mu }_r\gg 1$; para: $\chi$ small +; dia: $\chi$ small −.

Concept: $E_F=\frac{h^2}{8m}{\left(\frac{3n}{\pi }\right)}^{2/3}$ ⇒ $n=\frac{8\pi }{3}\frac{(2m{E}_F{)}^{3/2}}{h^3}$.

1. $E_F=4.72\times 1.6\times {10}^{-19}=7.56\times {10}^{-19}$ J.
2. $2m{E}_F=1.377\times {10}^{-48}$; $(2m{E}_F{)}^{3/2}=1.616\times {10}^{-72}$.
3. $n=\frac{8\pi }{3}\times \frac{1.616\times {10}^{-72}}{(6.63\times {10}^{-34}{)}^3}\approx 4.64\times {10}^{28}{\mathrm{m}}^{-3}$.

Short trick: Electron density in metals is always ~${10}^{28}$ m⁻³ – only option A fits.

Concept: The reciprocal lattice is uniquely fixed by the real lattice (${\overrightarrow{b}}_1=2\pi \frac{{\overrightarrow{a}}_2\times {\overrightarrow{a}}_3}{V}$, …) but its type may or may not be the same.

1. Simple cubic → simple cubic (same type).
2. FCC → BCC and BCC → FCC (different).
3. So it ‘can be’ the same; it is strongly (exactly) related, not weakly.

Short trick: SC ↔ SC, FCC ↔ BCC.

Concept: ${\lambda }_L=\sqrt{\frac{m^{\ast }}{{\mu }_0{n}_s{e}^{\ast 2}}}$.

1. It depends on the superfluid (Cooper-pair) density $n_s$ and effective mass $m^{\ast }$.
2. $n_s$ falls with temperature, so $\lambda (T)=\frac{\lambda (0)}{\sqrt{1-(T/T_c{)}^4}}$ – temperature dependent.
3. Critical field, coherence length, lattice constant or thermal conductivity do not enter the formula.

Short trick: $\lambda \propto \sqrt{m/n_s}$, diverges at $T_c$.