Assistant Engineer (Kerala Water Authority) 2022 — Kerala PSC PYQ Practice with Answers

Correct answer (Option A):\nFormula: E = (P × L) / (A × ΔL) and Bulk Modulus K = E / (3 × (1 - 2μ))\nGiven: D = 20 mm, P = 50000 N, ΔL = 0.20 mm, L = 200 mm, ΔD = 0.004 mm\nStep 1: Cross-sectional Area A = π × (20 / 2)² = 314.16 mm²\nStep 2: Young's Modulus E = (50000 × 200) / (314.16 × 0.20) = 159155 MPa\nStep 3: Linear strain = 0.20 / 200 = 0.001\nStep 4: Lateral strain = 0.004 / 20 = 0.0002\nStep 5: Poisson's ratio μ = 0.0002 / 0.001 = 0.2\nStep 6: K = 159155 / (3 × (1 - 2 × 0.2)) = 159155 / 1.8 = 88419 MPa\nAnswer: The closest value is 88384 MPa.\nOption A is correct.\n\nWhy others are wrong:\nOption B represents the approximate value for Young's Modulus, not the Bulk Modulus. Option C and Option D reflect miscalculations or the use of an incorrect relationship between elastic constants, such as using the formula for Shear Modulus instead.\n\nStudy tip:\nThe fundamental relationships between elastic constants are critical in solid mechanics. Always memorize: E = 3K(1 - 2μ), E = 2G(1 + μ), and E = (9KG) / (3K + G).

Correct answer (Option C):\nFormula: M = (w × L²) / 8 and Z = (b × d²) / 6 and σ = M / Z\nGiven: b = 0.07 m, d = 0.20 m, L = 4.5 m, load = 10 kN/m², σ = 8 MPa = 8 × 10⁶ N/m²\nStep 1: Let spacing be 's' meters. Uniformly distributed load on beam w = 10 × s kN/m = 10000 × s N/m.\nStep 2: Max bending moment M = (10000s × 4.5²) / 8 = 25312.5s Nm.\nStep 3: Section modulus Z = (0.07 × 0.20²) / 6 = 0.0004667 m³.\nStep 4: Using M = σ × Z, 25312.5s = 8 × 10⁶ × 0.0004667.\nStep 5: 25312.5s = 3733.6 → s = 0.1475 m = 14.75 cm.\nAnswer: The closest standard permissible spacing is 15 cm.\nOption C is correct.\n\nWhy others are wrong:\nOption A (120 cm) and Option B (70 cm) would lead to a bending stress far exceeding the 8 MPa limit, causing the beams to fail. Option D (10 cm) is overly conservative and less economical given the permissible stress limit.\n\nStudy tip:\nFor uniformly loaded simple beams, the maximum bending moment always occurs at the mid-span. Ensure units are consistent (convert cm to m and MPa to N/m²) before equating moments and stresses.

Correct answer (Option A):\nFormula: Polar Moment of Inertia J = (π / 32) × (d_o⁴ - d_i⁴) and Torsion Equation τ / r = T / J\nGiven: Inner diameter d_i = 90 mm, thickness t = 25 mm, T = 12 kNm = 12 × 10⁶ Nmm\nStep 1: Outer diameter d_o = 90 + (2 × 25) = 140 mm\nStep 2: J = (π / 32) × (140⁴ - 90⁴) = 31.27 × 10⁶ mm⁴\nStep 3: Inner radius r_i = d_i / 2 = 45 mm\nStep 4: Shear stress at inner face τ = (T × r_i) / J\nStep 5: τ = (12 × 10⁶ × 45) / (31.27 × 10⁶) = 17.26 MPa\nAnswer: 17.26 MPa.\nOption A is correct.\n\nWhy others are wrong:\nOption B is incorrect because it likely calculates the stress at the outer radius instead of the inner face. Option C and Option D result from arithmetic errors in calculating the polar moment of inertia or improperly converting kilonewton-meters to newton-millimeters.\n\nStudy tip:\nFor hollow circular shafts under torsion, the shear stress varies linearly from zero at the central axis to a maximum at the outer surface. Use the exact radial distance to find stress at any specific point.

Correct answer (Option D):\nFormula: Euler's Crippling Load P_cr = (π² × E × I) / L_eff²\nGiven: P_A = 0.5 × P_B. Both columns have the same actual length (L).\nStep 1: For a column fixed at one end and hinged at the other, L_eff = L / √2. Therefore, P_cr = 2 × (π²EI / L²).\nStep 2: For a column fixed at both ends, L_eff = L / 2. Therefore, P_cr = 4 × (π²EI / L²).\nStep 3: Comparing the two: 2 × (π²EI / L²) is exactly half of 4 × (π²EI / L²).\nAnswer: Column A must be fixed-hinged, and Column B must be fixed-fixed.\nOption D is correct.\n\nWhy others are wrong:\nOption A incorrectly pairs the hinged-hinged (factor 1) and fixed-fixed (factor 4) conditions, giving a 1:4 ratio. Option B yields a 1:4 ratio (factor 0.25 vs 1). Option C reverses the relationship, making Column A much stronger than Column B.\n\nStudy tip:\nAlways remember the effective length (L_eff) multipliers for standard columns: Both hinged (L), Fixed-Free (2L), Fixed-Fixed (L/2), and Fixed-Hinged (L/√2).

Correct answer (Option C):\nFormula: Discharge per unit width Q = |ψ₂ - ψ₁|\nGiven: Stream function ψ = 2(y² - x²)\nStep 1: Calculate the value of the stream function at the first point (2, 4).\nStep 2: ψ₁ = 2 × (4² - 2²) = 2 × (16 - 4) = 2 × 12 = 24\nStep 3: Calculate the value of the stream function at the second point (1, 3).\nStep 4: ψ₂ = 2 × (3² - 1²) = 2 × (9 - 1) = 2 × 8 = 16\nStep 5: Discharge Q = |24 - 16| = 8 m³/s.\nAnswer: 8 m³/s.\nOption C is correct.\n\nWhy others are wrong:\nOption A and Option D result from improperly adding the stream function values instead of finding their difference. Option B is a random mathematical distractor caused by squaring the coordinates incorrectly before subtraction.\n\nStudy tip:\nThe difference between the values of a stream function at any two points represents the volumetric flow rate (discharge per unit width) occurring between the streamlines passing through those two points.