50 Important Numerical MCQs for JKSSB JE Civil (Solved)
1. Strength of Materials
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A bar of length 2 m is subjected to axial tensile force of 50 kN. Area = 500 mm². Find the stress. - 
A. 50 MPa 
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B. 100 MPa ✅ 
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C. 25 MPa 
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D. 75 MPa 
 Solution: Stress = Force / Area = 50000 / 500 = 100 MPa 
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- 
A mild steel rod of 1 m elongates by 1 mm under a load. What is the strain? - 
A. 0.001 
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B. 0.01 
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C. 0.0001 ✅ 
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D. 0.1 
 Strain = ΔL / L = 1 / 1000 = 0.001 = 0.001 
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- 
If Young's Modulus = 200 GPa, and strain = 0.0005, find stress. - 
A. 10 MPa 
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B. 200 MPa ✅ 
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C. 100 MPa 
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D. 250 MPa 
 Stress = E × Strain = 200 × 10⁹ × 0.0005 = 100 × 10⁶ = 200 MPa 
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A cantilever of 2 m has a point load of 5 kN at free end. Find maximum bending moment. - 
A. 5 kNm 
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B. 10 kNm 
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C. 2.5 kNm 
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D. 10 kNm ✅ 
 M = W × L = 5 × 2 = 10 kNm 
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A simply supported beam has UDL of 10 kN/m over 4 m. Max BM? - 
A. 20 kNm 
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B. 10 kNm 
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C. 8 kNm 
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D. 20 kNm ✅ 
 M = wL² / 8 = 10 × 4² / 8 = 20 kNm 
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2. RCC and Steel
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For a singly reinforced RCC beam, d = 500 mm, Mu = 250 kNm. Find lever arm if moment of resistance = Mu. - 
A. 400 mm 
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B. 450 mm 
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C. 500 mm 
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D. 555.56 mm ✅ 
 M = 0.87 × fy × Ast × z ⇒ z = M / (0.87 × fy × Ast) 
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- 
Modular ratio (m) = E_steel / E_concrete = ? - 
A. 10 
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B. 12 
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C. 13.33 ✅ 
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D. 15 
 E_steel = 2×10⁵ MPa, E_concrete = 15×10³ MPa ⇒ m = 200000/15000 = 13.33 
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- 
A steel rod of 20 mm dia carries 20 kN. Find stress. - 
A. 50 MPa 
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B. 63.7 MPa ✅ 
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C. 80 MPa 
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D. 90 MPa 
 A = π/4 × d² = 314.16 mm² ⇒ Stress = 20000 / 314.16 = 63.7 MPa 
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- 
Fe415 means: - 
A. Yield strength = 415 MPa ✅ 
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B. UTS = 415 MPa 
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C. Elongation = 41.5% 
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D. None 
 
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- 
Concrete mix 1:2:4 means: 
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A. Cement:Sand:Aggregate ✅ 
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B. Cement:Aggregate:Sand 
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C. Sand:Cement:Aggregate 
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D. None 
3. Soil Mechanics
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A soil has G = 2.7, and e = 0.6. Find dry density. 
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A. 1.2 g/cc 
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B. 1.6 g/cc ✅ 
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C. 2.0 g/cc 
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D. 2.2 g/cc 
γ_d = G × γ_w / (1 + e) = 2.7 × 1 / 1.6 = 1.6875 g/cc
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A soil sample has void ratio = 0.5, porosity = ? 
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A. 25% 
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B. 33.33% 
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C. 50% ✅ 
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D. 60% 
n = e / (1 + e) = 0.5 / 1.5 = 0.3333 = 33.33%
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Flow net has 5 flow channels and 10 equipotentials. H = 5 m. Find seepage. 
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A. 0.25 k 
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B. 2.5 k ✅ 
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C. 5.0 k 
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D. 10 k 
Q = k × H × Nf/Nd = k × 5 × 5/10 = 2.5k
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Unconfined compression test gives failure load = 500 N on 50 mm dia. Find cohesion. 
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A. 50 kPa 
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B. 127.3 kPa 
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C. 254.6 kPa ✅ 
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D. 100 kPa 
A = πd²/4 = 1963.5 mm² ⇒ q_u = 500 / 1963.5 = 254.6 kPa ⇒ c = q_u / 2
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A clay soil has liquid limit = 60%, plastic limit = 30%, water content = 40%. Find plasticity index. 
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A. 30% ✅ 
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B. 20% 
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C. 10% 
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D. 15% 
4. Surveying
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Staff reading at benchmark = 2.3 m, at unknown point = 0.8 m. RL of BM = 100 m. RL of point = ? 
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A. 98.5 
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B. 101.5 ✅ 
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C. 99.2 
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D. 100.8 
Rise = 2.3 - 0.8 = 1.5 ⇒ RL = 100 + 1.5 = 101.5 m
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A 20 m tape was 0.1 m too long. What is correction for 100 m? 
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A. +0.5 m 
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B. -0.5 m ✅ 
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C. +1 m 
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D. 0 
Correction = L × (Δl / l) = 100 × (-0.1/20) = -0.5 m
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An angle measured is 59°30′, true angle = 60°. Find error. 
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A. -0°30′ ✅ 
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B. +0°30′ 
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C. 0 
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D. 1° 
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Sensitivity of bubble tube is 20″/2 mm. If movement is 4 mm, angle = ? 
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A. 10″ 
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B. 20″ 
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C. 40″ ✅ 
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D. 80″ 
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Error due to slope in chaining 100 m at 1 in 10: 
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A. 0.5 m 
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B. 0.45 m ✅ 
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C. 1 m 
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D. 2 m 
Correction = L - √(L² - h²) ≈ h² / 2L = (10)² / (2 × 100) = 0.5 m
5. Hydraulics & Fluid Mechanics
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A pipe of diameter 0.3 m carries water at velocity of 2 m/s. Find discharge. 
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A. 0.141 m³/s ✅ 
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B. 0.282 m³/s 
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C. 0.3 m³/s 
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D. 0.06 m³/s 
A = πd²/4 = 0.0707 m² ⇒ Q = A × v = 0.0707 × 2 = 0.141 m³/s
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Pressure head of 300 kPa = ? in meters 
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A. 10 
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B. 20 
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C. 30 
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D. 30.6 ✅ 
h = P / (ρg) = 300000 / (1000 × 9.81) = 30.6 m
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Flow is laminar if Reynolds number is: 
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A. < 2000 ✅ 
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B. > 4000 
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C. 2500–5000 
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D. > 6000 
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A rectangular channel 3 m wide, flow depth = 2 m, velocity = 3 m/s. Find discharge. 
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A. 9 m³/s 
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B. 12 m³/s 
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C. 15 m³/s ✅ 
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D. 18 m³/s 
Q = A × V = (3×2) × 3 = 18 m³/s
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For a venturimeter, inlet area = 0.1 m², throat area = 0.05 m². If inlet velocity is 2 m/s, throat velocity? 
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A. 4 m/s ✅ 
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B. 2 m/s 
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C. 6 m/s 
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D. 8 m/s 
A1V1 = A2V2 ⇒ V2 = A1V1 / A2 = (0.1 × 2) / 0.05 = 4 m/s
6. Transportation Engineering
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Design speed = 60 kmph. Find stopping sight distance (SSD) if reaction time = 2.5 sec, f = 0.35. 
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A. 54 m 
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B. 70.1 m ✅ 
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C. 90 m 
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D. 100 m 
SSD = Vt + V² / (2gf) = 16.67×2.5 + (16.67)² / (2×9.81×0.35) = 70.1 m
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Radius of curve = 100 m, superelevation (e) = 0.07. Find design speed. 
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A. 45 kmph 
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B. 52 kmph ✅ 
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C. 60 kmph 
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D. 70 kmph 
v = √(gRe) = √(9.81 × 100 × 0.07) = 8.3 m/s ≈ 30 kmph
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Calculate number of repetitions for 20-year design if daily vehicles = 4500, growth rate = 7% 
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A. 30 million 
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B. 55 million 
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C. 65 million ✅ 
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D. 90 million 
N = 365×AADT × [(1+r)^n - 1]/r
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Width of 2-lane road = ? 
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A. 5.5 m ✅ 
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B. 7.5 m 
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C. 9 m 
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D. 3.5 m 
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Camber is provided for: 
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A. Drainage ✅ 
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B. Speed 
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C. Strength 
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D. Friction 
7. Estimation, Costing & Valuation
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Area of trapezoid with bases 4 m and 6 m, height 3 m: 
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A. 10 m² 
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B. 15 m² ✅ 
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C. 20 m² 
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D. 18 m² 
A = ½ × (a + b) × h = ½ × (4+6) × 3 = 15
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Earthwork in a pit 20 m × 10 m × 2 m = ? 
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A. 200 m³ ✅ 
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B. 300 m³ 
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C. 400 m³ 
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D. 500 m³ 
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Volume of a cylinder of 3 m diameter and 4 m height? 
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A. 28.27 m³ 
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B. 56.55 m³ ✅ 
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C. 60 m³ 
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D. 100 m³ 
V = πr²h = 3.14 × (1.5)² × 4 = 28.27 × 2 = 56.55
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No. of bricks in 1 m³ (size = 190×90×90 mm)? 
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A. 500 
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B. 525 
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C. 560 ✅ 
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D. 580 
Volume of brick with mortar ≈ 0.002 m³ ⇒ 1/0.002 ≈ 500–560 bricks
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For 12 mm thick plaster over 10 m², find quantity of mortar (1:6). 
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A. 0.12 m³ 
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B. 0.15 m³ ✅ 
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C. 0.2 m³ 
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D. 0.25 m³ 
8. CPM & Project Management
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If an activity has optimistic time = 4, most likely = 6, pessimistic = 10, find expected time. 
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A. 6 
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B. 7 
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C. 6.5 ✅ 
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D. 8 
Te = (4 + 4×6 + 10)/6 = 6.5
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Critical path is the: 
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A. Longest duration path ✅ 
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B. Shortest path 
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C. Fastest way 
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D. Least cost 
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Activity float = ? 
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A. LS - ES ✅ 
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B. LF - EF 
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C. LF - LS 
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D. EF - ES 
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If project duration = 30 days, penalty is Rs. 1000/day. Delay = 5 days. Total penalty? 
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A. Rs. 3000 
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B. Rs. 5000 ✅ 
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C. Rs. 7000 
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D. Rs. 10000 
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Time cost trade-off is done to: 
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A. Delay work 
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B. Maximize cost 
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C. Minimize time ✅ 
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D. None 
9. Environmental Engineering
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Daily water demand for 5,000 people @ 135 LPCD = ? 
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A. 0.5 MLD 
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B. 0.675 MLD ✅ 
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C. 1.0 MLD 
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D. 2.0 MLD 
Demand = 5000 × 135 = 675000 L/day = 0.675 MLD
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BOD removal in primary treatment = ? 
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A. 20–30% ✅ 
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B. 50% 
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C. 70% 
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D. 90% 
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A sedimentation tank removes particles of specific gravity 2.65. What helps settling? 
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A. Gravity ✅ 
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B. Pressure 
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C. Viscosity 
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D. Temperature 
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COD = 250 mg/L, BOD = 150 mg/L. Waste is: 
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A. Easily biodegradable ✅ 
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B. Not treatable 
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C. Toxic 
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D. None 
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Quantity of bleaching powder for 1000 L at 2 ppm? 
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A. 1 g 
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B. 2 g ✅ 
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C. 5 g 
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D. 10 g 
2 mg/L × 1000 = 2000 mg = 2 g
10. Miscellaneous (Construction, Building Materials)
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1 bag of cement = ? 
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A. 40 kg 
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B. 50 kg ✅ 
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C. 60 kg 
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D. 25 kg 
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Compressive strength of M20 concrete = ? 
- 
A. 20 N/mm² ✅ 
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B. 25 N/mm² 
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C. 30 N/mm² 
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D. 15 N/mm² 
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Fineness modulus of sand is: 
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A. < 2.2 
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B. 2.2–3.2 ✅ 
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C. 4.0 
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D. > 5 
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Initial setting time of OPC = ? 
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A. 10 min 
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B. 30 min ✅ 
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C. 1 hour 
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D. 3 hours 
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Water-cement ratio for M20 = ? 
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A. 0.4 
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B. 0.45 ✅ 
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C. 0.5 
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D. 0.6 
