RRB Technician Grade I (Signal) 2024, 2025 PYQs

Concept: For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides.

1. $\frac{\text{ar}(ABC)}{\text{ar}(LMN)}={\left(\frac{BC}{MN}\right)}^2\Rightarrow \frac{25}{36}={\left(\frac{2.5}{MN}\right)}^2$.
2. $\frac{BC}{MN}=\frac{5}{6}\Rightarrow MN=\frac{2.5\times 6}{5}=3\text{cm}$.

Trick: $25:36$ areas → $5:6$ sides; $2.5\times \frac{6}{5}=3$.

Answer: (C) 3

Concept: $\frac{1}{{\sin }^{2}A}={\csc }^{2}A$ and the identity $1+{\cot }^{2}A={\csc }^{2}A$.

1. $\frac{1}{{\sin }^{2}A}-1={\csc }^{2}A-1={\cot }^{2}A$.
2. Directly: $\frac{1-{\sin }^{2}A}{{\sin }^{2}A}=\frac{{\cos }^{2}A}{{\sin }^{2}A}={\cot }^{2}A$.
3. Check with $A={30}^{\circ }$: $\frac{1}{{\sin }^2{30}^{\circ }}-1=4-1=3={\cot }^2{30}^{\circ }$ ✓ (while ${\tan }^2{30}^{\circ }=\frac{1}{3}$, so ${\tan }^{2}A$ is wrong).

Yaad rakhein: $1+{\cot }^{2}A={\csc }^{2}A$ and $1+{\tan }^{2}A={\sec }^{2}A$ – cot pairs with cosec, tan with sec.

Answer: (A) ${\cot }^{2}A$

Concept: The centre of a circle is the midpoint of any diameter: $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

1. Let $A=(x,y)$, $B=(1,4)$, centre $=(2,-3)$.
2. $\frac{x+1}{2}=2\Rightarrow x=3$; $\frac{y+4}{2}=-3\Rightarrow y=-10$.

Trick: $A=2\times \text{centre}-B=(4-1,-6-4)=(3,-10)$.

Answer: (C) (3, −10)

Concept: $\tan \theta =\frac{8}{15}$ ⇒ perpendicular 8, base 15, hypotenuse $\sqrt{64+225}=17$, so $\sin \theta =\frac{8}{17}$.

1. $1-\sin \theta =\frac{9}{17}$, $1+\sin \theta =\frac{25}{17}$.
2. $\frac{\sqrt{1-\sin \theta }}{\sqrt{1+\sin \theta }}=\sqrt{\frac{9}{25}}=\frac{3}{5}$.

Trick: 8-15-17 is a Pythagorean triplet; $\frac{17-8}{17+8}=\frac{9}{25}$ – both perfect squares, root $=\frac{3}{5}$.

Answer: (A) $\frac{3}{5}$

Concept: Drawing one by one without replacement – multiply the successive probabilities; a pack has 4 jacks in 52 cards.

1. $P=\frac{4}{52}\times \frac{3}{51}\times \frac{2}{50}$.
2. $=\frac{24}{132600}=\frac{1}{5525}$.

Trick: Same as choosing 3 jacks together: $\frac{{}^4{C}_3}{{}^{52}{C}_3}=\frac{4}{22100}=\frac{1}{5525}$.

Answer: (B) $\frac{1}{5525}$

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%

Concept: Transmission-line conductors need low resistivity plus low cost, light weight and good strength.

1. Silver has the lowest resistivity but is too costly; gold also costly.
2. Copper ($\rho \approx 1.7\times {10}^{-8}\Omega \text{m}$) and aluminium ($\approx 2.8\times {10}^{-8}\Omega \text{m}$) are cheap and conductive — aluminium (ACSR) for overhead lines, copper for cables.
3. Tungsten, iron, lead, tin have much higher resistivity.

Yaad rakhein: Resistivity order: Ag $<$ Cu $<$ Au $<$ Al — practical choice: Cu and Al.

Answer: (D) Copper and aluminium

Concept: Inside a long current-carrying solenoid the magnetic field is uniform.

1. Uniform field ⇒ field lines are parallel straight lines, equally spaced, along the axis.
2. Outside the solenoid the field is weak and the lines spread out like a bar magnet's.
3. Circular lines are for a straight wire, not inside a solenoid.

Yaad rakhein: Solenoid inside = uniform = parallel straight lines ($B={\mu }_{0}nI$).

Answer: (C) They are in the form of parallel straight lines

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

1. $\sin \theta$ is maximum ($=1$) at $\theta ={90}^{\circ }$ → $F_{max}=BIL$.
2. Parallel or anti-parallel ($0^{\circ }$ or ${180}^{\circ }$): $\sin \theta =0$, no force.
3. The conductor need not move — the force is on the current, not on motion.

Yaad rakhein: Perpendicular → maximum force (Fleming's left-hand rule).

Answer: (C) When the direction of the current is at right angles to the magnetic field

Concept: If $Q\propto m^n$, then $\frac{\Delta Q}{Q}=n\frac{\Delta m}{m}$.

1. $\frac{\Delta m}{m}=\frac{0.0001}{0.005}=0.02=2\%$.
2. For $Q\propto m^2$: $\frac{\Delta Q}{Q}=2\times 2\%=4\%$.

Trick: Power $n$ multiplies the percentage error by $n$.

Answer: (D) 4%

Concept: The depletion layer width depends on the net barrier potential across the junction.

1. Reverse bias adds to the built-in potential → more ions uncovered → depletion layer widens.
2. Forward bias opposes the barrier → depletion layer narrows and current flows.
3. The width changes in both cases, so options saying "remains constant" are wrong.

Yaad rakhein: Reverse = wider, Forward = narrower.

Answer: (A) Depletion layer gets wider when it is reverse biased.

Concept: Work done by the net force equals the change in kinetic energy: $W=\Delta KE$.

1. Constant force $F$ over distance $d$: $W=F\times d$.
2. Car starts from rest, so $K{E}_{initial}=0$ → $KE=F\times d$.
3. Check: $v^2=2ad=\frac{2Fd}{m}$, so $\frac{1}{2}m{v}^2=Fd$.

Yaad rakhein: From rest, KE gained = work done = $Fd$ (mass cancels out).

Answer: (B) $F\times d$

Concept: In an intrinsic (pure) semiconductor the number of electrons in the conduction band equals the number of holes in the valence band.

1. Equal electron and hole concentrations place the Fermi level at the centre of the forbidden gap.
2. n-type doping shifts it up towards the conduction band; p-type shifts it down towards the valence band.

Yaad rakhein: Intrinsic → middle; n-type → near CB; p-type → near VB.

Answer: (D) lies mid-way between the conduction band and valance band

Concept: PCB = Printed Circuit Board — an insulating board (usually glass-epoxy FR-4) with copper tracks etched on it to connect components.

1. Components are soldered onto the board; the copper "printed" tracks replace loose wiring.
2. Other expansions (Processor Control Board etc.) are not standard electronics terms.

Yaad rakhein: The green board inside every gadget = Printed Circuit Board.

Answer: (B) Printed Circuit Board

Concept: Random errors vary unpredictably between readings; systematic errors shift every reading the same way.

1. Random errors are reduced by repeating observations, averaging and controlling conditions (temperature, vibration).
2. Calibration corrects a systematic (instrumental) error — it does not reduce the scatter of random errors.

Yaad rakhein: Random → average it out; Systematic → calibrate it out.

Answer: (A) Calibrating the instrument

Concept: Accuracy = closeness to the true value; precision = closeness of repeated readings to each other.

1. Hitting the same point repeatedly = high precision (readings agree).
2. That point being far from the centre = low accuracy (systematic offset).
3. Option B is high precision and high accuracy; option C is low in both.

Yaad rakhein: Tight group off-centre = precise but not accurate.

Answer: (D) Hitting the same point far from the centre repeatedly