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