RRB Technician Grade III 2024, 2025 PYQs

Concept: For a concave mirror, the centre of curvature C lies at $R=2f$ from the pole. An object placed at C gives an image at C itself.

1. Same-size image means magnification $m=-\frac{v}{u}=-1$, so $v=u$.
2. Mirror formula: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{2}{u}=\frac{1}{f}$.
3. So $u=2f$ (object at C); image is real, inverted and of the same size.

Trick: Object at C → image at C (same size); beyond C → diminished; between F and C → enlarged.

Answer: (B) equal to 2f

Concept: A neuron (nerve cell) is the structural and functional unit of nervous tissue. It has a cell body (cyton), dendrites and an axon.

1. Each neuron has one long axon that carries impulses away from the cell body.
2. It has many short branched dendrites that receive impulses.
3. The cell body contains a single nucleus, not many nuclei – so statement (D) is incorrect.

Yaad rakhein: Neuron = one nucleus, one axon, many dendrites. (Many nuclei is a feature of skeletal muscle fibres.)

Answer: (D) The cell body of a neuron has many nuclei.

Concept: In a food chain or food web, energy flows from the sun → producers → herbivores → carnivores, and never returns to the previous level.

1. Energy captured by autotrophs does not go back to the sun.
2. Energy passed to herbivores does not come back to autotrophs.
3. At each level some energy is lost as heat, so the flow is one-way (unidirectional).

Yaad rakhein: Energy flows (one-way); nutrients cycle (circular).

Answer: (A) Unidirectional

Concept: Evaporation causes cooling. A volatile liquid takes the latent heat of vaporisation from the surface it is on.

1. Acetone is highly volatile and evaporates quickly at room temperature.
2. The energy needed for evaporation is absorbed from the palm.
3. The palm loses heat, so it feels cool.

Yaad rakhein: Acetone, petrol, perfume, spirit on skin → quick evaporation → cooling.

Answer: (B) It will cause the palm to feel cool.

Concept: In respiration, glucose (6-carbon) is broken in the cytoplasm into two molecules of pyruvate (3-carbon). Pyruvate then forms different products.

1. Glucose ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ – 6 carbons ✔
2. Pyruvate – 3 carbons ✔
3. Anaerobic (yeast): ethanol ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{O}\mathrm{H}$ (2 C) + $\mathrm{C}{\mathrm{O}}_2$ ✔
4. Lack of oxygen in muscles: lactic acid ${\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_3$ – 3 carbons, not 4 ✘

Yaad rakhein: Glucose 6 → pyruvate 3 → lactic acid 3 / ethanol 2 + $\mathrm{C}{\mathrm{O}}_2$.

Answer: (A) Lactic acid – 4 carbon molecules

Concept: When copper powder is heated in air, it combines with oxygen to form copper(II) oxide, which is black.

1. Reaction: $2\mathrm{C}\mathrm{u}+{\mathrm{O}}_2\stackrel{\text{heat}}{\to }2\mathrm{C}\mathrm{u}\mathrm{O}$.
2. Copper is brown/reddish; the surface turns black due to $\mathrm{C}\mathrm{u}\mathrm{O}$.
3. Copper(I) oxide $\mathrm{C}{\mathrm{u}}_2\mathrm{O}$ is red, not black.

Yaad rakhein: Heated copper turns black ($\mathrm{C}\mathrm{u}\mathrm{O}$); passing ${\mathrm{H}}_2$ over hot $\mathrm{C}\mathrm{u}\mathrm{O}$ turns it brown again (reduction).

Answer: (A) Black-coloured copper (II) oxide

Concept: Electric current is the rate of flow of charge: $I=\frac{Q}{t}$, so $Q=I\times t$ (time in seconds).

1. $I=1\text{ A}$, $t=5\text{ min}=5\times 60=300\text{ s}$.
2. $Q=I\times t=1\times 300$.
3. $Q=300\text{ C}$.

Trick: Always convert minutes to seconds before using $Q=It$.

Answer: (C) 300 C

Concept: Kinetic energy is the energy possessed by a body due to its motion. It equals the work done to bring the body from rest to velocity $v$.

1. Using $v^2-u^2=2as$ with $u=0$: $s=\frac{v^2}{2a}$.
2. Work $W=F\times s=ma\times \frac{v^2}{2a}$.
3. So $E_k=\frac{1}{2}m{v}^2$.

Yaad rakhein: Double the speed → KE becomes 4 times (since $E_k\propto v^2$).

Answer: (B) $\frac{1}{2}m{v}^2$

Concept: Soaps are sodium or potassium salts of long-chain carboxylic acids (fatty acids), e.g. sodium stearate ${\mathrm{C}}_{17}{\mathrm{H}}_{35}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{N}\mathrm{a}$.

1. Soaps are made by saponification: oil/fat + $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ → soap + glycerol.
2. The ionic end ($-\mathrm{C}\mathrm{O}{\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+}$) is water-loving; the long hydrocarbon tail is oil-loving.
3. Detergents, in contrast, are sodium salts of sulphonic acids.

Yaad rakhein: Soap → carboxylic acid salt; Detergent → sulphonic acid salt.

Answer: (C) Carboxylic acid

Concept: Acids show acidic behaviour only in water, because water helps them release ${\mathrm{H}}^{+}$ ions.

1. $\mathrm{H}\mathrm{C}\mathrm{l}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_3{\mathrm{O}}^{+}+\mathrm{C}{\mathrm{l}}^{-}$ – without water, ${\mathrm{H}}^{+}$ does not separate from HCl. Statement a is correct.
2. ${\mathrm{H}}^{+}$ cannot exist alone; it combines with water: ${\mathrm{H}}^{+}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_3{\mathrm{O}}^{+}$.
3. So hydrogen ions are always written as ${\mathrm{H}}^{+}(\mathrm{a}\mathrm{q})$ or ${\mathrm{H}}_3{\mathrm{O}}^{+}$. Statement b is correct.

Yaad rakhein: Dry HCl gas does not change the colour of dry blue litmus – no water, no ${\mathrm{H}}^{+}$.

Answer: (C) Both a and b

Concept: Rutherford bombarded a thin gold foil with fast α-particles and studied how they scattered. This led to the discovery of the nucleus (1911).

1. Most α-particles passed straight → most of the atom is empty space.
2. A few were deflected → positive charge occupies a very small volume.
3. Very few (about 1 in 20,000) bounced back → a tiny, dense, positively charged nucleus at the centre.

Yaad rakhein: Electron – J.J. Thomson; Proton – Goldstein; Nucleus – Rutherford; Neutron – Chadwick.

Answer: (C) Nucleus

Concept: The sky looks blue because tiny air molecules scatter blue light (shorter wavelength) much more than red light.

1. Scattering needs particles of the atmosphere.
2. With no atmosphere, there would be no scattering of sunlight.
3. No scattered light reaches the eye from the sky, so it appears dark (black).

Yaad rakhein: Astronauts and the Moon's sky see a black sky – no atmosphere, no scattering.

Answer: (A) black

Concept: Speed of light in a medium $v=\frac{c}{n}$; a greater refractive index means a smaller speed.

1. Refractive indices: air $\approx 1.0003$, water $=1.33$, glass $\approx 1.5$.
2. So speed: air $>$ water $>$ glass.
3. Hence light is faster in water than in glass.

Trick: Optically denser medium → light slows down (higher $n$, lower $v$).

Answer: (D) greater in water than in glass

Concept: Alkanes are saturated hydrocarbons with only single C–C bonds. Their general formula is ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$.

1. Methane ($n=1$): $\mathrm{C}{\mathrm{H}}_4$; ethane ($n=2$): ${\mathrm{C}}_2{\mathrm{H}}_6$.
2. Alkenes (one double bond): ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}}$.
3. Alkynes (one triple bond): ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}-2}$.

Trick: Alkane +2, alkene 0, alkyne −2 (H count relative to $2n$).

Answer: (D) ${\text{C}}_n{\text{H}}_{2n+2}$

Concept: Non-biodegradable substances (plastics, glass, metals) are not broken down by microbes and stay in the environment for a long time.

1. Reduce – use less of such materials.
2. Reuse – use items again; Recycle – convert them into new products.
3. Throwing them openly pollutes soil and water, blocks drains and harms animals – not a suitable method.

Yaad rakhein: 3 R's – Reduce, Reuse, Recycle. Open dumping is never a solution.

Answer: (C) Throw them openly

Concept: Work done by a constant force: $W=F\times s$, where $s$ is the displacement in the direction of the force.

1. $F=2\text{ N}$, $s=0$.
2. $W=2\times 0=0$.
3. No displacement along the force → no work, however large the force.

Yaad rakhein: Pushing a wall or holding a load still → work done $=0$.

Answer: (D) 0

Concept: Heat produced per second (power) in a resistor: $P=I^{2}R$.

1. $P=100\text{ J/s}=100\text{ W}$, $R=4\Omega$.
2. $I^2=\frac{P}{R}=\frac{100}{4}=25$.
3. $I=\sqrt{25}=5\text{ A}$.

Yaad rakhein: Heat per second $=I^{2}R$; total heat $H=I^{2}Rt$.

Answer: (A) 5 A

Concept: A homogeneous mixture has a uniform composition throughout and no visible boundaries between its components.

1. Air is a mixture of gases – mainly ${\mathrm{N}}_2$ (~78%) and ${\mathrm{O}}_2$ (~21%), plus Ar, $\mathrm{C}{\mathrm{O}}_2$, water vapour.
2. The gases mix uniformly, so any sample has the same composition.
3. Hence air is a homogeneous mixture (a gaseous solution).

Yaad rakhein: Air, sugar solution, brass – homogeneous; muddy water, oil + water – heterogeneous.

Answer: (D) Homogeneous mixture

Concept: Ozone (${\mathrm{O}}_3$) in the upper atmosphere shields the earth from harmful ultraviolet (UV) radiation of the sun.

1. (i) UV radiation is highly damaging (e.g. skin cancer) – incorrect.
2. (ii) UV splits ${\mathrm{O}}_2$ into O atoms, which combine with ${\mathrm{O}}_2$ to form ${\mathrm{O}}_3$ – correct.
3. (iii) Ozone levels began to drop sharply in the 1980s, linked to CFCs – correct.

Yaad rakhein: ${\mathrm{O}}_2\stackrel{\mathrm{U}\mathrm{V}}{\to }\mathrm{O}+\mathrm{O}$, $\mathrm{O}+{\mathrm{O}}_2\to {\mathrm{O}}_3$; CFCs deplete ozone (Montreal Protocol, 1987).

Answer: (B) Only (ii) and (iii)

Concept: Each part of the female reproductive system has a specific role from making eggs to nourishing the embryo.

1. Ovary → makes female gametes (eggs): (i)-(d).
2. Oviduct (fallopian tube) → site of fertilisation: (ii)-(a).
3. Uterus → implantation of the embryo: (iii)-(b).
4. Placenta → nourishes the growing embryo: (iv)-(c).

Trick: Egg made in ovary → meets sperm in oviduct → sticks in uterus → fed by placenta.

Answer: (C) (i)-(d), (ii)-(a), (iii)-(b), (iv)-(c)