JKSSB JE Civil Mock Test - 2 Free Online Test 2026

Cost means original cost of construction of property or purchase whereas value means present value of property which may be higher or lower than former.

Parameters and their units-

• Turbidity unit-NTU
• Odour-expressed in terms of threshold odour number
• Colour- cobalt units
• Solids-mg/l

Non-destructive testing of concrete:

Ultrasonic test-

Energy loss in the hydraulic jump

Depth of flow before hydraulic jump (Y₁) = 2 m
Depth of flow after hydraulic jump (Y₂) = 5 m

• The minimum flow of sewage passing through the sewer is an important factor in the design of a sewer.
• At low flow, velocity will be considerably reduced causing silting
• To avoid this, slope at which the sewer has to be laid is decided in accordance with the requirement of Minimum permissible velocity at minimum hourly discharge.

Most economical rectangular channel:
Wetted perimeter (P)= b +2d = A/d +2d
For most economical section, dP/d(d) = 0
dP/d(d) = -A/d²+2 = 0
A = 2d²
bd = 2d²
b = 2d

As per IS 456:2000 Cl.38.1(c),Fig.21-

Stress strain curve for concrete is given below. For design purposes, the compressive strength of concrete in the structure shall be assumed to be 0.67 times the characteristic strength. The partial safety factor of 1.5 shall be applied in addition to this. Therefore,
maximum compressive strength of concrete = 0.67f_ck  
design maximum compressive strength of concrete 

Earthwork- cu m
Damp proof course-sq m
Plastering- sq m

Viscosity = 0.2 poise = 0.2/10 Ns/m² = 0.02 Ns/m²

As it is less than 2000,the flow is laminar.

Kinematic viscosity = 0.02 stoke = 0.02 x 10-4 m²/s

A closed contour line represents either depression or hill. A set of ring contours with higher values inside, represents a hill whereas the lower value inside, represents a depression (without an outlet).

centre line method-
Thickness of the wall = 30 cm
centre to centre length of long wall= 6 + 0.15 + 0.15 = 6.3 m
centre to centre length of short wall= 5 + 0.15 + 0.15 = 5.3 m
Total centre length = (2 x centre to centre length of long wall) + (2 x centre to centre length of short wall)
= 2 x 6.3 + 2 x 5.3
= 23.2 m

Back bearing = fore bearing ± 180^o
Fore bearing = 95 + 180 = 275^o

Temperature strain = coefficient of linear expansion x temperature difference
= 18 x 12 x10^-6
= 0.000216
If the temperature is decreased, temperature stress and strain will be tensile in nature and when temperature is increased, temperature stress and strain will be compressive in nature.

Falling weight test/tup test- As per Indian railways standards a rail specimen should withstand the blow without fracture. One Test per cast shall be carried out. Sample for 20% (minimum) of the fresh casts rolled per day shall be selected at random from straightened rails and the remaining samples shall be hot sawn. No retest shall be permitted on account of sample from straightened rails failed in Falling weight test. However, present provision of retests shall be applicable to rest 80% of samples taken from un-straightened rails. Choice of the test sample location within cast and strand shall normally lie with the manufacturer. The test sample position within bloom/ rail shall be selected at the discretion of the inspecting agency.

Centre of pressure- point of application of the total pressure on the surface of a submerged body.
Unit: meter
Dimensional formula

As per IS 456:2000, Cl G-1.1:

For a BG track, as per Indian Railways, minimum depth of ballast should be 25 cm.

Cast iron sleepers- these are widely used in Indian railways. They are of two types, pot type sleepers and plate type sleepers. Pot type sleepers cannot be used for curves sharper than 4 degrees.

Cast iron plate sleepers-

• DO plate sleeper
• Laisy pedestral
• NWR type, LK type etc.
• CST-9
• Rail free duplex

Out of the above types, CST- 9 type sleepers are most extensively used in Indian railways.

Darcy-Weisbach Equation for loss of head in a turbulent flow-

Straight line method of calculating depreciation-
Annual depreciation (D) = (original cost - scrap value)/life of property

Drive ways- connects the highway with commercial establishments like fuel stations, service stations etc. These should be located fairly away from intersections. The radius of the drive way curve should be kept as large as possible. Width of the drive way should be minimized to reduce the length of cross walks.

MANDATORY/REGULATORY SIGNS

• These signs are used to inform the road users certain laws, regulations, and prohibitions.
• Violation of these signs are legal offense.
• Examples: stop and give way signs, prohibitory signs, no parking, and no stopping signs, speed limit and vehicle control signs, restriction end sign, compulsory direction control and other signs.

WARNING/CAUTIONARY SIGNS

• These are to warn road users about certain hazardous conditions that exist on or adjacent to the roadway.
• The signs are equilateral triangle in shape with its apex pointing upward.
• Examples: right-hand/left-hand curve, right/left hair pin bend, steep ascent/descent, narrow road ahead, narrow bridge ahead, pedestrian crossing etc.

INFORMATION SIGNS

• For information and guidance of road users.
• Examples: direction, hospital, etc.

As per IS 456:2000, Cl 40.2.2:
Shear Strength of Members under Axial Compression- For members subjected to axial compression P^_u, the design shear strength of concrete, given in Table 19, shall be multiplied by the following factor;

A very thin layer next to the wall in which velocity variations are affected only by viscous effects. This layer is known as laminar sublayer and no turbulence exists in it.

Stability of bituminous mixes- defined as the resistance of the paving mix to deformation under traffic load. Two kinds of failure are;
(i) shoving - a transverse rigid deformation that occurs at areas subject to severe acceleration (ii) grooving - longitudinal ridging due to channelization of traffic

As per IS 456:2000, Cl. B-2.2 (Table 22)
For high yield strength deformed bars of Grade Fe 500, the permissible stress in direct tension and flexural tension shall be 0.55 fy (0.55 fy = 0.55 x 500 = 275 N/mm²).
The permissible stresses for shear and compression reinforcement shall be as for Grade Fe 415.

Annual depreciation = (original cost-salvage value)/useful life
Annual depreciation = (500000-380000)/15= Rs.8000
Book value at the end of second year = 500000- 2 x 8000 = Rs.484000

Here, A₁ = 27 sq m, A₂ = 60 sq m, A₃ = 54 sq m, A₄ = 33 sq m, A₅ = 36 sq m, A₆ = 48 sq m,
A₇ = 51 sq m
Volume by Prismoidal formula:
Volume (V) = D/3 X [A₁ + A₇ +4(A₂ +A₄ +A₆) +2(A₃ +A₅)]
V = 30/3 X [27 + 51 +4(60+33+48) +2 (54+36)]= 10 x (78 + 564 + 180)= 8220
V = 8220 m³