RRB Group D — 2025 — 28 Nov 2025 — Shift III — Previous Year Paper — Kerala PSC PYQ Practice with Answers

Correct answer (Option A):\nWhen solid lead(II) nitrate is heated strongly in a test tube, it undergoes thermal decomposition to produce a yellow residue of lead(II) oxide (PbO), along with the evolution of two distinct gases. The reddish-brown gas is nitrogen dioxide (NO₂), and the colorless gas that supports combustion is oxygen (O₂).\n\nFormula:\n2Pb(NO₃)₂ (s) → 2PbO (s) + 4NO₂ (g) + O₂ (g)\n\nWhy others are wrong:\nOption B is incorrect because nitric oxide (NO) is a colorless gas, not reddish-brown. Option C is incorrect because carbon dioxide and nitrogen gas are not formed during this reaction. Option D is incorrect because hydrogen gas is not evolved.\n\nStudy tip:\nThermal decomposition of nitrates of heavy metals like lead and copper typically yields the corresponding metal oxide, nitrogen dioxide gas, and oxygen gas.

Correct answer (Option A):\nThis question is based on a unit analogy. Newton (N) is the SI unit of weight (and force). Similarly, the SI unit of pressure is the Pascal (Pa).\n\nFormula:\nPressure = Force / Area\nUnit = N/m² = Pascal\n\nWhy others are wrong:\nOption B (Kilogram) is the SI unit of mass, not pressure. Option C (Kelvin) is the SI unit of thermodynamic temperature. Option D (Kilogram per cubic metre) is the SI unit of density.\n\nStudy tip:\nAlways memorize core SI units and their corresponding physical quantities as they are frequently asked in both science and reasoning sections of railway exams.

Correct answer (Option C):\nIn Gregor Mendel's monohybrid cross experiment, crossing a homozygous tall pea plant (TT) with a homozygous short pea plant (tt) results in all offspring in the F1 (first filial) generation being heterozygous tall (Tt). Because the trait for tallness is completely dominant over shortness, all plants express the tall phenotype.\n\nWhy others are wrong:\nOption A is incorrect because shortness is a recessive trait and does not appear in the F1 generation. Option B describes the expected outcome of a test cross (Tt x tt), not an F1 generation cross. Option D is incorrect because pea plant height shows complete dominance, not incomplete dominance or blending.\n\nStudy tip:\nRemember Mendel's Law of Dominance: in a heterozygote, one trait will conceal the presence of another trait for the same characteristic.

Correct answer (Option B):\nSodium (Na) is an alkali metal located in Group 1 of the periodic table. It is highly reactive and reacts vigorously and exothermically with cold water to produce sodium hydroxide and highly inflammable hydrogen gas, often catching fire.\n\nFormula:\n2Na + 2H₂O → 2NaOH + H₂\n\nWhy others are wrong:\nOption A (Iron) reacts very slowly with steam, not cold water. Option C (Lead) does not react with water under normal conditions. Option D (Zinc) reacts only with hot water or steam, much less vigorously than sodium.\n\nStudy tip:\nReactivity series order for these metals is Na > Zn > Fe > Pb. Highly reactive metals like Sodium and Potassium are stored under kerosene to prevent accidental reactions with moisture in the air.

Correct answer (Option D):\nWork done ($W$) is defined as the dot product of force ($F$) and displacement ($S$), calculated using the formula:\nWork = F × S × cos(θ)\nWork done is positive when the angle θ between the force vector and the displacement vector is acute (0° ≤ θ < 90°), meaning the force has a component working in the direction of the displacement.\n\nWhy others are wrong:\nOption A results in zero work done since displacement is zero. Option B results in zero work done because cos(90°) = 0. Option C results in negative work done because the opposing force creates an obtuse angle (θ = 180°).\n\nStudy tip:\nRemember that work can be positive, negative, or zero depending entirely on the direction of the force relative to the object's movement.