RRB Technician Grade I (Signal) 2024, 2025 PYQs

Concept: Intel 8085 is an 8-bit microprocessor that generates its own bus-control signals; no external bus controller is needed.

1. 8085 provides $\mathrm{I}\mathrm{O}/\bar{\mathrm{M}}$, $\overline{\mathrm{R}\mathrm{D}}$ and $\overline{\mathrm{W}\mathrm{R}}$ (plus $S_0,S_1$, ALE) to control bus cycles — so statement D is false.
2. It has serial I/O via SID/SOD pins (A true), interrupts RST 5.5, 6.5, 7.5, TRAP, INTR which can be expanded with 8259 (B true).
3. With 8155 and 8355 (or 8156, 8755) it forms a minimum three-chip microcomputer (C true).

Trick: 8085 = 'self-sufficient' bus control; it is the 8086 that needs the 8288 bus controller in maximum mode.

Answer: (D) It does not provide control signals, hence external bus controller is required — this is NOT correct.

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

1. $\cos \varphi =1\Rightarrow \varphi =0^{\circ }$: voltage and current are in phase.
2. This happens only in a purely resistive circuit (or at resonance when $X_L=X_C$).
3. Purely inductive or capacitive circuits have $\varphi ={90}^{\circ }$, power factor 0, and all power is reactive.

Trick: PF = 1 → all power is real (active) power, $P=VI$.

Answer: (B) the circuit is purely resistive

Concept: In a DC motor the commutator is a split ring that reverses the direction of current in the armature coil every half rotation, so the torque always acts in the same direction and the coil keeps rotating.

1. Without reversal, the coil would stop after a half turn.
2. Brushes (not the commutator) make the sliding contact that supplies current; speed is controlled by voltage/field, not by the commutator.

Trick: Commutator = 'current reverser' (motor) / 'AC-to-DC converter' (generator).

Answer: (B) To reverse the direction of current in the armature

Concept: $P=\frac{W}{t}=\frac{F\times s}{t}$; convert to SI units first.

1. $s=1.2\text{km}=1200\text{m}$, $t=2\text{min}=120\text{s}$
2. $W=10000\times 1200=1.2\times {10}^7\text{J}$
3. $P=\frac{1.2\times {10}^7}{120}={10}^5\text{W}=100\text{kW}$

Trick: $P=F\times v$; $v=\frac{1200}{120}=10\text{m/s}$, so $P=10000\times 10={10}^5\text{W}$.

Answer: (B) 100 kW

Concept: $U=\frac{1}{2}k{x}^2$ — potential energy is proportional to the square of displacement.

1. New displacement $x^{\prime }=\frac{x}{2}$
2. $U^{\prime }=\frac{1}{2}k{\left(\frac{x}{2}\right)}^2=\frac{1}{4}·\frac{1}{2}k{x}^2=\frac{U}{4}$

Trick: Halve $x$ → energy becomes one-fourth; double $x$ → four times.

Answer: (B) It decreases to $\frac{1}{4}$ of its original value

Concept: Resolution is the smallest change in input that can be read: (value of one scale division) × (fraction of a division that can be estimated).

1. One division $=\frac{300}{100}=3\text{V}$
2. $\frac{1}{10}$ of a division can be estimated: resolution $=\frac{3}{10}=0.3\text{V}$

Trick: Resolution $=\frac{\text{full scale}}{\text{divisions}}\times \frac{1}{10}$ when a tenth of a division is readable.

Answer: (D) 0.3 V

Concept: For a conservative force (gravity, spring, electrostatic) the work done is path independent, so a potential energy function $U$ can be defined uniquely for every configuration of the system.

1. $W_{\text{conservative}}=-\Delta U$; it depends only on the initial and final positions.
2. Friction is non-conservative: work depends on the path, and no potential energy can be defined for it (options B, D wrong).
3. Potential energy is not zero in general; only its reference point is arbitrary (option C wrong).

Trick: Conservative force ⇔ path independent ⇔ potential energy exists.

Answer: (A) can be defined for any configuration of the system

Concept: Motional emf $\epsilon =Blv\sin \theta$, where $\theta$ is the angle between the velocity and the magnetic field.

1. $\epsilon =2\times 1.5\times 10\times \sin {60}^{\circ }$
2. $=30\times 0.866=25.98\approx 26\text{V}$

Trick: $\sin {60}^{\circ }=\frac{\sqrt{3}}{2}\approx 0.87$ — multiply the 'maximum' emf ($Blv=30$ V) by 0.87.

Answer: (D) 26 volts

Concept: An electric fan runs on an AC induction motor — the rotating magnetic field of the stator induces current in the rotor (electromagnetic induction), which produces torque.

1. Toaster and electric iron work on the heating effect of current ($I^{2}Rt$).
2. A refrigerator's compressor also uses a motor, but as a whole it works on the vapour-compression refrigeration cycle; the fan is the direct induction-motor appliance.

Trick: Heating appliances → Joule heating; rotating appliances → induction motor.

Answer: (C) Electric Fan

Concept: An emf (and current) is induced only when the magnetic flux $\Phi =BA\cos \theta$ linked with the coil changes.

1. Moving the coil towards a magnet or rotating it in a field changes the flux → current induced.
2. Moving the coil parallel to the field lines keeps the number of lines through it constant → no change in flux → no induced current.
3. Option C explicitly says the flux is changing, so current is induced.

Trick: No change in flux = no emf. Motion along the field never cuts field lines.

Answer: (D) Moving the coil parallel to the magnetic field lines

Concept: A BJT has two junctions; their biasing decides the mode.

1. Active mode (amplifier, most useful): emitter-base forward-biased, base-collector reverse-biased.
2. Saturation: both forward-biased (switch ON); Cut-off: both reverse-biased (switch OFF). Options C and D have these swapped.
3. Option A is reverse-active mode — rarely used.

Trick: Active = 'FR' (Forward EB, Reverse CB). Saturation = FF, Cut-off = RR.

Answer: (B) Active Mode: emitter-base forward-biased, base-collector reverse-biased

Concept: In a dual-slope ADC the input is integrated for a fixed time and then a reference is integrated back to zero; the result depends only on the ratio of times, so the values of $R$, $C$ and clock frequency cancel out.

1. Single-slope ADC accuracy depends directly on the integrator's $R$, $C$ and clock stability.
2. Dual-slope also averages out noise (good 50 Hz rejection) → higher accuracy; used in digital multimeters.

Trick: Dual-slope = slow but very accurate (DMMs); flash = fast but less accurate.

Answer: (C) A dual-slope integrating-type A/D converter has a higher accuracy than a single-slope type.

Concept: A resistance (resistive) potential divider is used to measure high DC/low-frequency voltages by scaling them down. Its range is limited to a few kilovolts because of heating ($I^{2}R$ losses), insulation and leakage/stray-capacitance errors at higher voltages.

1. For higher voltages, capacitive dividers (AC) or electrostatic voltmeters are used.
2. Micro/milli-volt and volt ranges need no divider at all.

Trick: Resistive divider → few kV; capacitive divider → hundreds of kV.

Answer: (D) Kilo Volts

Concept: Electrostatic deflection in a CRT: $D=\frac{Ll{V}_d}{2d{E}_a}$, where $V_d$ is the deflecting voltage, $E_a$ the accelerating voltage, $l$ plate length, $d$ plate spacing and $L$ distance to the screen.

1. For a given tube and fixed $E_a$, all other quantities are constant → $D\propto V_d$.
2. Deflection is inversely proportional to $E_a$ (higher anode voltage → faster electrons → less deflection).

Trick: $D\propto \frac{V_d}{E_a}$ — directly with deflecting, inversely with accelerating voltage.

Answer: (B) directly proportional