RRB Group D 27 Nov 2025 Shift 2 — Kerala PSC PYQ Practice with Answers
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Paper details
- Paper code: rrb-group-d-27-nov-2025-shift-2
- Format: Full previous year paper — PYQ practice with answers
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Preview questions (5)
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Question 1 General Science
What is the location of the cell wall in an organism?
Correct answer: C. Just outside the plasma membrane in a plant cell
Correct answer (Option C):\nIn plant cells, fungi, and some prokaryotes, a rigid outermost layer called the cell wall is present. It is located just outside the plasma membrane, providing structural support, protection, and maintaining the cell shape. Animal cells completely lack a cell wall, having only a plasma membrane as their outermost boundary.\n\nWhy others are wrong:\nOption A and D are incorrect because animal cells do not possess a cell wall. Option B is incorrect because the cell wall resides outside, not inside, the plasma membrane in plants.\n\nStudy tip:\nThe plant cell wall is primarily composed of cellulose, a complex carbohydrate that provides structural tensile strength.
Question 2 General Science
If the length of a wire is doubled and its cross-sectional area is halved, its resistance will become:
Correct answer: B. Four times the original
Correct answer (Option B):\nResistance of a conductor is determined by the formula:\nR = ρ × (l / A)\nGiven:\nNew length (l') = 2l\nNew cross-sectional area (A') = A / 2\nStep 1: Substitute the new parameters into the resistance formula:\nR' = ρ × (2l / (A / 2))\nStep 2: Simplify the fraction multiplication:\nR' = ρ × (2 × 2 × l / A) = 4 × (ρ × l / A) = 4R\nAnswer: 4 times the original resistance.\nOption B is correct.\n\nWhy others are wrong:\nOption A, C, and D do not match the mathematically derived factor of 4 resulting from concurrent elongation and narrowing of the conductor.\n\nStudy tip:\nResistance is directly proportional to length and inversely proportional to cross-sectional area.
Question 4 General Science
The increase in which of the following factor(s) will increase the resistance of a conductor?\n(i) Temperature of the conductor\n(ii) Length of the conductor\n(iii) Area of cross-section of the conductor
Correct answer: A. Both (i) and (ii)
Correct answer (Option A):\nThe electrical resistance of a metallic conductor increases with an increase in length (R ∝ l) and an increase in temperature due to increased thermal vibrations of atoms impeding electron flow. Conversely, increasing the cross-sectional area decreases resistance because it provides a wider path for electrons.\n\nWhy others are wrong:\nOption B is wrong because increasing cross-sectional area reduces resistance. Option C omits length. Option D includes area, which has an inverse relationship.\n\nStudy tip:\nRemember the formula R = ρl/A. Temperature variations shift the intrinsic resistivity (ρ) value of metals upward.
Question 5 General Science
¹⁸₄₀Ar and ²⁰₄₀Ca are a pair of:
Correct answer: B. Isobars
Correct answer (Option B):\nIsobars are atoms of different chemical elements that have the same mass number (total number of nucleons) but different atomic numbers (number of protons). Here, both Argon (Ar) and Calcium (Ca) possess an identical mass number of 40, matching the strict definition of isobars.\n\nWhy others are wrong:\nOption A is incorrect because isotones have equal neutron numbers (40-18=22 vs 40-20=20). Option C is incorrect as they have different electron counts (18 vs 20). Option D is incorrect because isotopes share the same atomic number.\n\nStudy tip:\nIsotopes = Same Atomic Number; Isobars = Same Mass Number.
Question 6 General Science
A heavy wooden block is pushed across a rough floor. The applied force is gradually increased:\nAt first, the block does not move.\nThen, at a certain force, the block just begins to slide.\nFinally, when the block is sliding, it moves with constant velocity.\nWhich of the following correctly describes the force situation in these three stages?
Correct answer: A. Stage 1 - Balanced, Stage 2 - Unbalanced, Stage 3 - Balanced
Correct answer (Option A):\nIn Stage 1, the block remains stationary because the applied force is completely balanced by static friction (net force = 0). In Stage 2, the block just starts moving, indicating the forces have become unbalanced to break the static equilibrium. In Stage 3, the block moves at a constant velocity, meaning there is no acceleration and net force is zero again, indicating balanced kinetic friction and applied forces.\n\nWhy others are wrong:\nOptions B, C, and D misidentify the net equilibrium acceleration states across Newton's laws of motion.\n\nStudy tip:\nConstant velocity always implies that net acceleration equals zero, which means the external forces acting on the system are completely balanced.
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