RRB Technician Grade I (Signal) 2024, 2025 PYQs

Concept: Electric field at a point is the force experienced by a unit positive test charge placed there: $\overrightarrow{E}=\frac{\overrightarrow{F}}{q}$, unit N/C (or V/m).

1. 'Energy per unit charge' is electric potential (volt), not field — option B is the tempting wrong one.
2. Force per unit mass is a gravitational field, not electric.

Trick: Field = force ÷ charge; potential = energy ÷ charge.

Answer: (C) Force acting per unit charge

Concept: For a p-type semiconductor $\sigma \approx pe{\mu }_p$ (holes are the majority carriers).

1. $\sigma$ is directly proportional to hole mobility ${\mu }_p$.
2. ${\mu }_p\to \frac{{\mu }_p}{2}$ with $p$ constant ⇒ $\sigma \to \frac{\sigma }{2}$.

Trick: $\sigma =ne\mu$: halve the mobility ⇒ halve the conductivity.

Answer: (C) Reduced to half

Concept: Charge on one electron $e=1.6\times {10}^{-19}\text{C}$; number of electrons in 1 C $=\frac{1}{e}$.

1. $n=\frac{1}{1.6\times {10}^{-19}}=6.25\times {10}^{18}\approx 6\times {10}^{18}$ electrons.

Trick: $1\text{C}\approx 6.25\times {10}^{18}$ electrons — memorise this number.

Answer: (B) $6\times {10}^{18}$ electrons

Concept: $F=BIL\sin \theta$, where $\theta$ is the angle between current and field.

1. $\sin \theta$ is maximum (= 1) at $\theta ={90}^{\circ }$ ⇒ current perpendicular to $\overrightarrow{B}$.
2. Parallel or antiparallel ($0^{\circ }$ or ${180}^{\circ }$) gives $\sin \theta =0$ ⇒ no force; zero field also gives no force.

Trick: Perpendicular ⇒ maximum force; parallel ⇒ zero force.

Answer: (B) The current is perpendicular to the magnetic field

Concept: Mass is the amount of matter in a body — an intrinsic property that does not change with location.

1. Weight $=mg$ changes because $g$ differs on the Moon, other planets or in space.
2. Gravitational force and apparent weight (e.g. in a lift) also vary with place and motion.
3. Only mass stays the same everywhere in the universe.

Trick: Mass → constant everywhere; weight → depends on $g$.

Answer: (C) Mass

Concept: By KVL, source voltage $v_s=v_D+v_L$ at every instant. The average of a sinusoid over a full cycle is zero.

1. Averaging: $⟨v_s⟩=⟨v_D⟩+⟨v_L⟩=0$ ⇒ $⟨v_D⟩=-⟨v_L⟩$.
2. $⟨v_L⟩=\frac{V_m}{\pi }$, so $⟨v_D⟩=-\frac{V_m}{\pi }$: equal magnitude, opposite sign.
3. Option D is wrong — $v_D$ is not zero on average; it carries the whole negative half cycle. Option C: rms of $v_D$ is $\frac{V_m}{2}$, not $\frac{V_m}{\sqrt{2}}$.

Trick: Ideal diode: load takes the positive half, diode 'takes' the negative half — their averages cancel.

Answer: (B) The average values of $v_L(t)$ and $v_D(t)$ are equal in magnitude and opposite in sign.

Concept: A generator (dynamo) rotates a coil in a magnetic field; the changing flux induces an emf — mechanical → electrical energy.

1. An electric motor does the reverse: electrical → mechanical.
2. A battery converts chemical energy; a capacitor only stores charge.

Trick: Generator: mechanical → electrical. Motor: electrical → mechanical.

Answer: (C) Electric generator

Concept: $E=\frac{F}{q}$.

1. $E=\frac{10}{2}=5\text{N/C}$.

Trick: Field strength is force per coulomb — divide force by charge.

Answer: (C) $5\frac{\text{N}}{\text{C}}$

Concept: $\overline{\text{PSEN}}$ (Program Store Enable) is an active-low output of the 8051 that acts as the read strobe for external program memory (ROM).

1. The bar over PSEN indicates it is active low; it is connected to the OE pin of the external ROM.
2. Internal RAM needs no external strobe; external data memory uses $\overline{\text{RD}}$/$\overline{\text{WR}}$, not PSEN.

Trick: PSEN → external Program memory; RD/WR → external Data memory; all active low.

Answer: (B) active low, external program

Concept: In circuit diagrams a resistor is drawn as a zig-zag line (US style) or a small rectangle (IEC style).

1. Two parallel plates represent a capacitor.
2. A circle with a pointer is a meter/galvanometer; an arrow is used for variable elements or current direction.

Trick: Zig-zag = resistor; parallel plates = capacitor; coil = inductor.

Answer: (C) Zig-zag line

Concept: For two resistors in parallel $R_p=\frac{R_1{R}_2}{R_1+R_2}$.

1. $R_p=\frac{6\times 3}{6+3}=\frac{18}{9}=2\Omega$.

Trick: Product ÷ sum; the result is always less than the smaller resistor.

Answer: (A) $2\Omega$

Concept: NAND output $=\overline{A·B}$. With both inputs tied to $A$: $\overline{A·A}=\bar{A}$.

1. Input 0 ⇒ output 1; input 1 ⇒ output 0 — exactly a NOT gate.
2. This is why NAND is a universal gate: NOT, AND and OR can all be built from it.

Trick: Shorted NAND (or NOR) inputs = inverter.

Answer: (A) NOT gate (Inverter)

Concept: With $V$ fixed, $I\propto \frac{1}{R}$. Halving the current means the total resistance has doubled.

1. Initially $I=\frac{V}{10}$. New current $\frac{I}{2}=\frac{V}{10+R_2}$.
2. $10+R_2=20\Rightarrow R_2=10\Omega$.

Trick: Current halved ⇒ resistance doubled ⇒ added resistor equals the original.

Answer: (B) $10\Omega$

Concept: $\varphi =BA\cos \theta$ with $\theta$ measured from the normal to the surface.

1. Field parallel to the plane ⇒ perpendicular to the normal ⇒ $\theta ={90}^{\circ }$.
2. $\cos {90}^{\circ }=0\Rightarrow \varphi =0$ — no field lines pass through the surface.

Trick: Field along the surface ⇒ zero flux; field perpendicular to the surface ⇒ maximum flux.

Answer: (B) zero

Concept: The SI system has seven base quantities: length, mass, time, electric current, temperature, amount of substance and luminous intensity.

1. Mass (kilogram) is one of these seven.
2. Work (joule), pressure (pascal) and force (newton) are derived quantities built from base units.

Trick: Seven base quantities — m, kg, s, A, K, mol, cd.

Answer: (D) Mass

Concept: An electron accelerated from rest through a potential difference $V$ gains kinetic energy $eV$, so $\frac{1}{2}m{v}^2=eV\Rightarrow v=\sqrt{\frac{2eV}{m}}$.

1. $2eV=2\times 1.6\times {10}^{-19}\times 182=5.824\times {10}^{-17}\text{J}$.
2. $v^2=\frac{5.824\times {10}^{-17}}{9.1\times {10}^{-31}}=6.4\times {10}^{13}=64\times {10}^{12}$.
3. $v=\sqrt{64\times {10}^{12}}=8\times {10}^6\text{m/s}$.

Trick: $\frac{2\times 1.6}{9.1}\approx 0.35$ and $0.35\times 182\approx 64$ — a perfect square, so the root comes out clean.

Answer: (B) $8\times {10}^6\text{m/s}$

Concept: In parallel, $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\dots$ — every branch adds a path for current, so $R_{eq}$ is smaller than the smallest branch.

1. Removing $R_1$ removes the term $\frac{1}{R_1}$, so $\frac{1}{R_{eq}}$ becomes smaller.
2. A smaller reciprocal means a larger $R_{eq}$ — the total resistance increases.
3. Example: $6\Omega \parallel 6\Omega =3\Omega$; remove one and it becomes $6\Omega$.

Yaad rakhein: Parallel = more paths = less resistance; fewer paths = more resistance.

Answer: (C) increase

Concept: Power factor $=\cos \varphi$, where $\varphi$ is the phase angle between voltage and current.

1. In a pure inductor the current lags V by ${90}^{\circ }$; in a pure capacitor it leads by ${90}^{\circ }$.
2. $\cos {90}^{\circ }=0$, so the average (true) power is zero — energy only oscillates between source and element.
3. Pure resistor: $\varphi =0$, pf $=1$ (option B is for resistors).

Yaad rakhein: Pure L or C → "wattless" current → pf $=0$.

Answer: (A) 0

Concept: The 8085 has a multiplexed address/data bus $\mathrm{A}{\mathrm{D}}_0$–$\mathrm{A}{\mathrm{D}}_7$ (8 lines) and a separate higher address bus ${\mathrm{A}}_8$–${\mathrm{A}}_{15}$.

1. The Address/Data buffer drives $\mathrm{A}{\mathrm{D}}_0$–$\mathrm{A}{\mathrm{D}}_7$, so it is 8-bit.
2. Data must flow both into and out of the CPU, so this buffer is bi-directional.
3. The Address buffer (${\mathrm{A}}_8$–${\mathrm{A}}_{15}$) is 8-bit and unidirectional (output only).

Yaad rakhein: Address buffer = 8-bit, one-way; Address/Data buffer = 8-bit, two-way.

Answer: (B) 8 bit and bi-directional

Concept: The 8085 has an on-chip clock generator: the crystal frequency at X1/X2 is divided by 2 internally.

1. Dividing by 2 with a flip-flop gives a symmetrical square wave — equal ON and OFF times.
2. Hence the internal clock (CLK OUT) has a 50% duty cycle.
3. Example: 6 MHz crystal → 3 MHz operating clock.

Yaad rakhein: 8085: crystal $÷2$, duty cycle 50%.

Answer: (B) 50%