RRB Technician Grade I (Signal) 2024, 2025 PYQs

Concept: Work-energy theorem: the work done by the net force equals the gain in kinetic energy, $K=F·x$.

1. Starting from rest, $v^2=u^2+2ax=2ax$.
2. $K=\frac{1}{2}m{v}^2=\frac{1}{2}m(2ax)=max$.
3. Since $m$ and $a$ are constant, $K\propto x$ – a straight-line (linear) relation.

Trick: Constant force ⇒ $K=Fx$ ⇒ $K\propto x$ (not $\sqrt{x}$ or $x^2$).

Answer: (C) $K\propto x$

Concept: SI has seven fundamental units (m, kg, s, A, K, mol, cd); any unit formed by combining them is a derived unit.

1. Volt $={\text{kg m}}^2{\text{s}}^{-3}{\text{A}}^{-1}$, Newton $={\text{kg m s}}^{-2}$, Joule $={\text{kg m}}^2{\text{s}}^{-2}$ – all derived.
2. Options (A), (B), (C) contain only fundamental units (metre, kilogram, ampere, kelvin, mole, second).

Yaad rakhein: Units named after scientists (volt, newton, joule, watt, pascal) are always derived.

Answer: (D) Volt, Newton, Joule

Concept: In this two-diode circuit each diode conducts only when the input exceeds its battery voltage; between the two thresholds both diodes stay off and no output appears.

1. For $V_i>+E_1$, $D_1$ conducts and output follows $V_i-E_1$; for $V_i<-E_2$, $D_2$ conducts.
2. For $-E_2<V_i<+E_1$ both diodes are reverse biased, so $V_o=0$ – the transfer curve is flat.
3. This flat band where the input produces no output is called the dead zone (dead band).

Yaad rakhein: Input present but output zero = dead zone; cut-off and saturation are transistor regions, not diode-clipper terms.

Answer: (D) dead zone

Concept: An electrostatic voltmeter measures voltage by the force between charged plates. Its range is extended by a potential divider, not by a series resistor as in ordinary voltmeters.

1. With a resistance potential divider the instrument reads only a known fraction of the total voltage (used for DC and low-frequency AC); for pure AC a capacitance divider is also used.
2. A plain resistor or capacitor in series cannot give a fixed multiplying factor for an instrument that draws almost no current, so options (B) and (D) are wrong.

Yaad rakhein: Electrostatic voltmeter range ↑ → potential divider (resistance / capacitance).

Answer: (A) Resistance Potential Divider

Concept: A piezoelectric crystal (quartz, Rochelle salt) produces charge only while the applied force is changing; under a steady force the charge leaks away through the crystal and amplifier resistance.

1. Hence it responds to dynamic (varying) displacement, vibration, shock and pressure, but cannot hold a reading for a static displacement.
2. Its high natural frequency makes it ideal for vibration and accelerometer applications.

Yaad rakhein: Piezo = dynamic only; for static displacement use LVDT or strain gauge.

Answer: (B) Piezoelectric transducer can be used for the measurement of dynamic displacement only.

Concept: Power is the rate of doing work: $P=\frac{W}{t}$, so for fixed work $t=\frac{W}{P}$, i.e. $t\propto \frac{1}{P}$.

1. Original: $t=\frac{W}{P}$.
2. New power $2P$: $t^{\prime }=\frac{W}{2P}=\frac{t}{2}$.

Trick: Double the power → half the time (inverse relation).

Answer: (A) $\frac{t}{2}$

Concept: Resistivity $\rho$ is a property of the material; resistance $R=\rho \frac{l}{A}$ depends on dimensions.

1. $\rho$ depends on the material and its atomic structure (free-electron density) and on temperature.
2. Length and area change $R$, not $\rho$ – a long and a short copper wire have the same resistivity.

Yaad rakhein: Resistance → size matters; resistivity → only material and temperature.

Answer: (C) Length of the conductor

Concept: In free fall, total mechanical energy is conserved: loss in PE = gain in KE.

1. Initial PE $=mgh$; at half height PE $=mg\frac{h}{2}=\frac{1}{2}mgh$.
2. KE gained $=mgh-\frac{1}{2}mgh=\frac{1}{2}mgh$, i.e. $\frac{1}{2}$ of the initial PE.

Trick: PE ∝ height, so fraction of PE converted = fraction of height fallen $=\frac{1}{2}$.

Answer: (A) $\frac{1}{2}$

Concept: Power is the rate at which energy is transferred or work is done.

1. $P=\frac{E}{t}$; SI unit watt $=\text{J/s}$.
2. $E\times t$, $E+t$ or $\frac{E}{t^2}$ do not have the unit of watt, so they are wrong.

Yaad rakhein: Watt = joule per second → $P=\frac{E}{t}$.

Answer: (A) $P=\frac{E}{t}$

Concept: In series the same current $I$ flows through both resistors, so $P=I^{2}R$ gives $P\propto R$.

1. $\frac{P_1}{P_2}=\frac{I^2{R}_1}{I^2{R}_2}=\frac{R_1}{R_2}$.
2. $P_1=4P_2\Rightarrow R_1:R_2=4:1$.

Trick: Series → $P\propto R$; parallel → $P\propto \frac{1}{R}$ (then the answer would be $1:4$).

Answer: (D) 4 : 1

Concept: Band gap energy ($E_g$) is the energy needed to move an electron from the valence band to the conduction band.

1. Standard values at room temperature: $E_g(\mathrm{G}\mathrm{e})\approx 0.67\text{eV}$, $E_g(\mathrm{S}\mathrm{i})\approx 1.1\text{eV}$, $E_g(\mathrm{G}\mathrm{a}\mathrm{A}\mathrm{s})\approx 1.43\text{eV}$.
2. Hence $E(\mathrm{G}\mathrm{a}\mathrm{A}\mathrm{s})>E(\mathrm{S}\mathrm{i})>E(\mathrm{G}\mathrm{e})$.
3. This is why Ge devices are more temperature-sensitive (smallest gap) and GaAs is used in LEDs/high-frequency devices.

Yaad rakhein: "Ge < Si < GaAs" – 0.7, 1.1, 1.4 eV (roughly increasing by 0.35).

Answer: (A) E(GaAs) > E(Si) > E(Ge)

Concept: NAND output $=\overline{AB}$; NOR output $=\overline{X+Y}$. Use De Morgan's laws to simplify.

1. Both NAND gates have inputs A and B, so each gives $P=\overline{AB}$.
2. The NOR gate receives $P$ and $P$: $Y=\overline{P+P}=\bar{P}$.
3. $Y=\overline{\stackrel{―}{AB}}=AB$ (double negation).

Trick: A NOR gate with both inputs tied together is just a NOT gate. NOT of NAND = AND.

Answer: (B) $AB$

Concept: For an infinitely long straight wire, $B=\frac{{\mu }_{0}I}{2\pi r}$, where $\frac{{\mu }_0}{2\pi }=2\times {10}^{-7}\text{T m/A}$.

1. $r=\frac{{\mu }_{0}I}{2\pi B}=\frac{2\times {10}^{-7}\times 15}{5\times {10}^{-6}}$.
2. $=\frac{3\times {10}^{-6}}{5\times {10}^{-6}}=0.6\text{m}$.

Trick: Remember $B=\frac{2\times {10}^{-7}I}{r}$ – it avoids writing ${\mu }_0$ and $\pi$ every time.

Answer: (A) 0.6 m

Concept: The magnetic field lines around a straight current-carrying wire are concentric circles centred on the wire (right-hand thumb rule).

1. At any point the field is tangent to the circle through that point, i.e. tangential and perpendicular to the radius.
2. Reversing the current reverses the direction of circulation (thumb = current, curled fingers = field).
3. The field is neither radially outward nor inward – that would be the pattern of an electric field around a line charge.

Yaad rakhein: Electric field of a line charge = radial; magnetic field of a line current = circular/tangential.

Answer: (A) tangential to the wire and depends on the direction of the current

Concept: Hysteresis is the lag between cause and effect – the output depends on whether the input is increasing or decreasing.

1. In a magnetic material, flux density $B$ for the same current (same $H$) is different on the rising and falling curves – this forms the B–H loop.
2. The difference in readings for the same input value is called hysteresis error.
3. Frequency, friction and temperature errors do not depend on the direction of change of input.

Yaad rakhein: "Increasing vs decreasing gives different reading" = Hysteresis (the loop).

Answer: (A) Hysteresis

Concept: Repeatability is the closeness of repeated readings of the same input by the same instrument, same observer, same conditions, in a short time.

1. So the instrument should give the same output for repeated applications of the same input value.
2. Any option with "different" breaks the definition – an instrument giving different outputs for the same input is not repeatable.

Yaad rakhein: Repeatability = same–same–same (output, input, conditions); Reproducibility = same output, same input, but different conditions/observer/location.

Answer: (A) same; same

Concept: The 8051 has five interrupt sources plus RESET; the 8052 adds a Timer 2 interrupt.

1. 8051: INT0, Timer 0, INT1, Timer 1, Serial (TI/RI) = 5 sources; with RESET counted, 6 interrupts (vectors 0000H, 0003H, 000BH, 0013H, 001BH, 0023H).
2. 8052: the same plus Timer 2 (vector 002BH) = 6 sources, i.e. 7 with RESET.

Yaad rakhein: Vector addresses jump by 8 bytes: 03, 0B, 13, 1B, 23 (8051), 2B (8052 Timer 2).

Answer: (A) 6; 7

Concept: Assume $T\propto m^a{L}^b{g}^c$ and equate dimensions on both sides.

1. $[T]=[M{]}^a[L{]}^b[L{T}^{-2}{]}^c=M^a{L}^{b+c}{T}^{-2c}$.
2. Compare with $M^0{L}^0{T}^1$: $a=0$, $b+c=0$, $-2c=1$ ⇒ $c=-\frac{1}{2}$, $b=\frac{1}{2}$.
3. So $T\propto L^{1/2}{g}^{-1/2}=\sqrt{\frac{L}{g}}$; the period does not depend on mass.

Trick: Only $\sqrt{L/g}$ has the dimension of time ($\sqrt{\text{m}/({\text{m/s}}^2)}=\text{s}$); $L/g$ gives ${\text{s}}^2$ and $L/\sqrt{g}$ does not give seconds – eliminate by units.

Answer: (C) $\sqrt{\frac{L}{g}}$

Concept: Reproducibility is the degree of closeness with which a given value may be repeatedly measured – the same output for the same input under the same conditions, over a longer period.

1. The definition requires same output for same input under same environmental conditions.
2. Any "different" in the blanks would describe drift or poor reproducibility, not the property itself.

Yaad rakhein: Repeatability = short time; Reproducibility = long time – both: same–same–same.

Answer: (B) same; same; same

Concept: Write the output of each gate and simplify with Boolean laws: $\overline{X+Y}=\overline{X}\overline{Y}$, $\overline{\stackrel{―}{X}}=X$.

1. First stage: NAND gives $P=\overline{AB}$; NOR gives $Q=\overline{A+B}$.
2. Upper NOR: $\overline{P+Q}=\overline{P}\overline{Q}=AB(A+B)=AB$.
3. Lower NOR (inputs $Q$ and $A$): $\overline{Q+A}=\overline{Q}\overline{A}=(A+B)\overline{A}=\overline{A}B$.
4. Final OR: $Y=AB+\overline{A}B=B(A+\overline{A})=B$.

Trick: Quick check with A = 0, B = 1 and A = 1, B = 0: output follows B both times, so $Y=B$.

Answer: (A) B