RRB Technician Grade III 2024, 2025 PYQs

Concept: Plaster of Paris (POP) is calcium sulphate hemihydrate, $\mathrm{C}\mathrm{a}\mathrm{S}{\mathrm{O}}_4·\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$.

1. On mixing with water it sets into a hard mass of gypsum: $\mathrm{C}\mathrm{a}\mathrm{S}{\mathrm{O}}_4·\frac{1}{2}{\mathrm{H}}_2\mathrm{O}+1\frac{1}{2}{\mathrm{H}}_2\mathrm{O}\to \mathrm{C}\mathrm{a}\mathrm{S}{\mathrm{O}}_4·2{\mathrm{H}}_2\mathrm{O}$.
2. This property is used for plaster casts supporting fractured bones, toys, decorative items and smooth surfaces.
3. Washing soda removes hardness; baking soda is used in baking and fire extinguishers.

Trick: POP + water → hard gypsum → bone cast.

Answer: (D) Creating casts for supporting fractured bones

Concept: A bisexual flower has both stamens (male) and pistil (female); a unisexual flower has only one of them.

1. Mustard and Hibiscus flowers have both stamens and pistil → bisexual.
2. Papaya and watermelon have separate male and female flowers → unisexual.
3. Pumpkin (cucurbit) flowers are also unisexual.

Yaad rakhein: Bisexual: Mustard, Hibiscus; Unisexual: Papaya, Watermelon (and cucurbits).

Answer: (B) Mustard

Concept: Meristems are growing tissues whose cells keep dividing.

1. Apical meristem – at the tips of stem and root – increases the length.
2. Lateral meristem (cambium) increases the girth (thickness).
3. Intercalary meristem – at the base of leaves or internodes – also adds to length.

Yaad rakhein: Apical = Apex = Length; Lateral = Girth.

Answer: (B) length

Concept: The number of heart chambers differs among vertebrate groups.

1. Fishes: 2 chambers (1 atrium + 1 ventricle).
2. Amphibians (and most reptiles): 3 chambers (2 atria + 1 ventricle), so oxygenated and deoxygenated blood mix slightly.
3. Birds and mammals: 4 chambers – complete separation of blood.

Yaad rakhein: Fish 2, Amphibian 3, Bird/Mammal 4.

Answer: (A) Amphibians

Concept: Vegetative propagation: new plants grow from vegetative parts – root, stem or leaf.

1. Leaves: Bryophyllum (buds on leaf margins).
2. Stem: potato, sugarcane, rose cuttings; Root: sweet potato, dahlia.
3. The flower is a reproductive part used for sexual reproduction, so it does not take part.

Yaad rakhein: Vegetative = root, stem, leaf; flower = sexual reproduction.

Answer: (D) Flower

Concept: The magnetic field at the centre of a current-carrying coil is $B\propto \frac{nI}{r}$.

1. Field increases when current $I$ is increased.
2. Field increases when number of turns $n$ is increased.
3. Field decreases as we move away from the coil.
4. So only 'current is increased' raises the field.

Trick: More current, more turns → stronger field; more distance → weaker.

Answer: (D) current through the coil is increased

Concept: Newton's law: $F=\frac{G{m}_1{m}_2}{d^2}$, so $F\propto m_1{m}_2$.

1. New mass $m_1^{\prime }=4m_1$, $m_2$ and $d$ unchanged.
2. $F^{\prime }=\frac{G(4m_1)m_2}{d^2}=4\times \frac{G{m}_1{m}_2}{d^2}$
3. $F^{\prime }=4F$

Trick: Force changes in the same ratio as the product of masses.

Answer: (D) 4F

Concept: Alkenes have general formula ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}}$; successive members differ by $\mathrm{C}{\mathrm{H}}_2$.

1. ${\mathrm{C}}_3{\mathrm{H}}_6$: $n=3$, $2n=6$ ✓; ${\mathrm{C}}_4{\mathrm{H}}_8$: $n=4$, $2n=8$ ✓ – both alkenes, differ by $\mathrm{C}{\mathrm{H}}_2$.
2. $\mathrm{C}{\mathrm{H}}_4$ and ${\mathrm{C}}_2{\mathrm{H}}_6$ are alkanes (${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$).
3. ${\mathrm{C}}_3{\mathrm{H}}_{10}$ is not a valid hydrocarbon; so only (A) fits.

Yaad rakhein: Alkane ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$, Alkene ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}}$, Alkyne ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}-2}$.

Answer: (A) ${\text{C}}_3{\text{H}}_6$ and ${\text{C}}_4{\text{H}}_8$

Concept: The alimentary canal runs from mouth to anus: oesophagus → stomach → small intestine → large intestine.

1. The small intestine is the longest part (about 6–7 m in adults), highly coiled.
2. Its length gives a large surface for complete digestion and absorption (villi).
3. The large intestine is shorter (about 1.5 m) but wider.

Yaad rakhein: Small intestine = longest; large intestine = wider, not longer.

Answer: (D) Small intestine

Concept: Standard rays for a concave mirror: a ray parallel to the principal axis passes through F after reflection, and vice versa.

1. The path of light is reversible.
2. So a ray passing through the principal focus F, after reflection, emerges parallel to the principal axis.
3. A ray through C retraces its path; a ray to the pole reflects at an equal angle.

Yaad rakhein: Parallel → F, and F → Parallel.

Answer: (C) emerge parallel to the principal axis

Concept: A solution is a homogeneous mixture of two or more substances.

1. Particle size is less than 1 nm (${10}^{-9}$ m) – true.
2. Particles are too small to scatter light, so no Tyndall effect – true.
3. Solute particles pass through filter paper and cannot be separated by filtration – so (D) is NOT a characteristic.

Yaad rakhein: Solution: no Tyndall effect, no settling, not separable by filtration.

Answer: (D) The solute particles can be separated from the mixture by the process of filtration.

Concept: The pH scale runs from 0 to 14 and measures ${\mathrm{H}}^{+}$ ion concentration.

1. pH less than 7 → acidic.
2. pH = 7 → neutral (e.g. pure water).
3. pH greater than 7 → basic (alkaline).

Yaad rakhein: Pure water at room temperature has pH exactly 7.

Answer: (D) 7

Concept: Mass number $A$ = number of protons + number of neutrons.

1. Protons $=13$, neutrons $=14$.
2. $A=13+14=27$
3. This is aluminium: ${\mathrm{}}_{13}^{27}\mathrm{A}\mathrm{l}$.

Trick: Atomic number = protons; mass number = protons + neutrons.

Answer: (D) 27 u

Concept: Minerals from which metals can be extracted profitably are called ores.

1. Many metals occur in the crust as oxides (e.g. bauxite, haematite).
2. Others occur as sulphides (e.g. zinc blende, cinnabar) and carbonates (e.g. calamine, limestone).
3. Since oxygen and sulphur are reactive, ores of oxides, sulphides and carbonates are the most common.

Yaad rakhein: Common ores = oxides, sulphides, carbonates.

Answer: (A) oxides, sulphides and carbonates

Concept: Non-biodegradable substances (plastic, DDT, glass) are not broken down by biological processes such as bacteria and decomposers.

1. They harm various members of the ecosystem, e.g. through biological magnification – (i) true.
2. They do not break down biologically – (ii) false.
3. Hence they persist in the environment for a long time – (iii) false.

Yaad rakhein: Non-biodegradable = not broken by microbes + persists + harms the ecosystem.

Answer: (A) Only (i)

Concept: Metals at the bottom of the reactivity series are least reactive and occur in the free (native) state.

1. Gold, silver, platinum and copper are found free in nature.
2. Sodium and potassium are highly reactive and are never found free; they occur only as compounds.
3. So the pair with both metals free is gold and silver.

Yaad rakhein: Least reactive (Au, Ag, Pt, Cu) → free state; K, Na, Ca → always combined.

Answer: (A) Gold and silver

Concept: Acceleration = change in velocity ÷ time, $a=\frac{v-u}{t}$.

1. Unit of velocity $=\mathrm{m}{\mathrm{s}}^{-1}$, unit of time $=s$.
2. Unit of $a=\frac{\mathrm{m}{\mathrm{s}}^{-1}}{s}=\mathrm{m}{\mathrm{s}}^{-2}$.
3. $\mathrm{m}{\mathrm{s}}^{-1}$ is the unit of speed/velocity, not acceleration.

Trick: Each division by time adds one more ${\mathrm{s}}^{-1}$: distance m → velocity $\mathrm{m}{\mathrm{s}}^{-1}$ → acceleration $\mathrm{m}{\mathrm{s}}^{-2}$.

Answer: (B) m $s^{-2}$

Concept: Transpiration is the loss of water in the form of water vapour from the aerial parts of a plant, mainly through stomata of leaves.

1. It creates a suction pull that helps upward movement of water and minerals in xylem.
2. It also helps in temperature regulation (cooling).
3. Imbibition = absorption of water by dry seeds; transcription = DNA → RNA.

Trick: Transpiration = water vapour out through stomata; 'trans-port' is a distractor.

Answer: (D) Transpiration

Concept: Unicellular organisms reproduce by fission: binary (into two) or multiple (into many).

1. Leishmania: binary fission in a definite orientation (due to flagellum) – correct.
2. Amoeba: splitting can occur in any plane – correct.
3. Yeast: reproduces by budding – correct.
4. Plasmodium divides into many daughter cells simultaneously by multiple fission, not binary fission – so (B) is INCORRECT.

Yaad rakhein: Many cells at once = multiple fission (Plasmodium); two cells = binary fission (Amoeba, Leishmania).

Answer: (B) Plasmodium divides into many daughter cells simultaneously by binary fission.

Concept: Newton's first law: a body stays at rest or in uniform motion unless an external force acts on it.

1. This natural tendency to resist change in the state of rest or uniform motion is called inertia.
2. Mass is the measure of inertia – heavier bodies have more inertia.
3. Example: passengers jerk forward when a moving bus stops suddenly.

Yaad rakhein: First law = law of inertia; mass measures inertia.

Answer: (A) inertia