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Question 71

FCI JE 2015• Test

In a mortar, the binding material is:

  1. AOption (a)

    Cement

  2. BOption (b)

    Sand

  3. COption (c)

    Cinder

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (a)A.

Cement

Why this answer is correct

Explanation:

The paste obtained by the addition of water in suitable quantity in the mixture of binding material like cement or lime and aggregates like sand is known as Mortar.

This binding material is also termed as Matrix and the aggregate used is also known as Adulterant.

Admixture is added in the concrete to modify its property such as strength, setting time etc.

Question 72

FCI JE 2015• Test

Minimum size of the particles of silt soil, is:

  1. AOption (a)

    0.002 mm

  2. BOption (b)

    0.04 mm

  3. COption (c)

    0.06 mm

  4. DOption (d)

    0.08 mm

View answer & explanation

Correct answer

Answer: (a)A.

0.002 mm

Why this answer is correct

Explanation:

Silt size varies from 0.002 mm - 0.075 mm.

In the Indian Standard Soil Classification System (ISSCS) or BIS, soils are classified according to their grain size as boulder, cobble, gravel, sand, silt, or clay, as shown below in the tabulated form.

Soil GroupType of SoilSub - GroupSize Range
Very Coarse SoilsBoulder> 300 mm
Cobble80 – 300 mm
Coarse SoilsGravelCoarse20 – 80 mm
Fine4.75 – 20 mm
SandCoarse2- 4.75 mm
Medium0.425- 2 mm
Fine0.075 – 0.425 mm
Fine SoilsSilt0.002 – 0.075 mm
Clay< 0.002 mm

Question 73

FCI JE 2015• Test

A steel cable of 2 cm diameter is used to lift a board 500π kg. Given that the length of the cable is 10 m and the value of E = 2 × 106 kg/cm2, the elongation of the cable will be:

  1. AOption (a)

    0.25 cm

  2. BOption (b)

    1.00 cm

  3. COption (c)

    0.50 cm

  4. DOption (d)

    1/π cm

View answer & explanation

Correct answer

Answer: (a)A.

0.25 cm

Why this answer is correct

Concept:

Elongation of the cable is given by,

δL = \(\frac{PL}{AE}\)

E = Elastic Modulus, P = Applied Load, L = Length of cable

A = Area of the cross-section, δL = Elongation

Explanation:

Given:

E =  2 × 106 kg/cm2 

P = 500π kg, L = 10 m = 1000 cm

A = \(\frac{π}{4}\times (2)^2\) cm2 = π cm2

∴ δL = \(\frac{PL}{AE}\)

δL = \(\frac{500π ;\times; 1000}{\pi ;\times; 2;\times; 10^6}\)

δL  = 0.25 cm

Question 74

FCI JE 2015• Test

The azimuth of a line is the:

  1. AOption (a)

    Clockwise angle that a line makes with the south end of the selected material

  2. BOption (b)

    Anti-clockwise angle that a line makes with the south end of the selected material

  3. COption (c)

    Clockwise angle that a line makes with the north end of the selected material

  4. DOption (d)

    None of these

View answer & explanation

Correct answer

Answer: (c)C.

Clockwise angle that a line makes with the north end of the selected material

Why this answer is correct

Explanation-

Surveyors express angles in several ways. When specifying directions, as is done in the preparation of a property survey, angles may be specified as bearings or azimuths.

A bearing is an angle less than 90° within a quadrant defined by the cardinal directions.

Azimuth is an angle between 0° and 360° measured clockwise from the North.

"South 45° East" and "135°" are the same direction expressed as a bearing and as an azimuth.

Question 75

FCI JE 2015• Test

The tie bars in a concrete pavement are provided in:

  1. AOption (a)

    Contraction joints

  2. BOption (b)

    Expansion Joints

  3. COption (c)

    Longitudinal joints

  4. DOption (d)

    Construction joints

View answer & explanation

Correct answer

Answer: (c)C.

Longitudinal joints

Why this answer is correct

Explanation:

Longitudinal joint

  • These are provided along the length of the pavement.
  • They are provided if the pavement width is more than 4.5 m.
  • It reduces the warping stress and uneven settlement of subgrade.
  • The tie bars in concrete pavement are provided across the Longitudinal joint.
  • Tie bars are not load transfer devices but serve as a means to tie two slabs.
  • Hence tie bars must be deformed or hooked and must be finally anchored into the concrete to function properly.
  • They are smaller than dowel bars and placed at large intervals.

Additional Information

There are mainly two types of joints provided in the cement concrete pavement:

(i) Longitudinal joint

(ii) Transverse joint

  • Expansion joint
  • Contraction joint
  • Construction joint
  • Warping joint

Question 76

FCI JE 2015• Test

In a CBR test, If the CBR value at 5 mm is greater than that at 2.5 mm:

  1. AOption (a)

    The higher value should be chosen 

  2. BOption (b)

    The test should be repeated 

  3. COption (c)

    Average value of the two should be used 

  4. DOption (d)

    None of these 

View answer & explanation

Correct answer

Answer: (b)B.

The test should be repeated 

Why this answer is correct

Concept:

CBR Test:

  • The California bearing ratio test is penetration test meant for the evaluation of subgrade strength of road and pavements. The results obtained by these tests are used with the empirical curves to determine the thickness of pavement and its component layers.
  • This is the most widely used method for the design of flexible pavement.
  • Generally CBR value is calculated at 2.5 mm and 5 mm penetration.
  • The CBR value at 2.5 mm penetration will be greater than the CBR value at 5 mm penetration and in such a case the former will be taken as the CBR value.
  • If CBR for 5 mm exceeds that for 2.5 mm, the test should be repeated. If identical result follows, the CBR for 5 mm penetration should be taken for design.

Question 77

FCI JE 2015• Test

Salvaging is:

  1. AOption (a)

    Dumping of solid waste

  2. BOption (b)

    Sanitary land filling of solid waste

  3. COption (c)

    Composing and soil conditioning

  4. DOption (d)

    Extraction of essence from waste

View answer & explanation

Correct answer

Answer: (d)D.

Extraction of essence from waste

Why this answer is correct

Explanation:

Salvaging:

  • Salvaging is described as an observed collection of solid waste for reuse.
  • Salvaging is considered an advanced way of solid waste management.
  • Salvaged material should be treated with care and should be avoided to mix with surrounding materials.
  • Salvaged material is then reused by passing through different stages of recycling.

Question 78

FCI JE 2015• Test

A cantilever beam is one which is -

  1. AOption (a)

    Fixed at one end and free at the other end

  2. BOption (b)

    Fixed at both ends

  3. COption (c)

    Supported at its ends

  4. DOption (d)

    Supported at more than two points

View answer & explanation

Correct answer

Answer: (a)A.

Fixed at one end and free at the other end

Why this answer is correct

Explanation:

There are different types of beam:

Overhanging beam:

  • A beam having its end portion extended beyond the support is known as an overhanging beam.
  • A beam may be overhanging on one side or on both sides.

Cantilever beam:

  • A beam fixed at one end and free at the other end is known as a cantilever beam.

Simply Supported Beam:

  • A beam supported at its both ends is known as a simply supported beam.

Fixed beam:

  • A beam whose both ends are fixed is known as a fixed beam.

Continuous beam:

  • A beam supported on more than two supports is known as a continuous beam.

Question 79

FCI JE 2015• Test

Compression index on a soil helps to determine:

  1. AOption (a)

    Total time required for consolidation 

  2. BOption (b)

    Time required for 50% consolidation 

  3. COption (c)

    Total settlement of clay layer 

  4. DOption (d)

    Pre-consolidation pressure of clay 

View answer & explanation

Correct answer

Answer: (c)C.

Total settlement of clay layer 

Why this answer is correct

Concept:

The coefficient of compression or compression index:

  • The compression index (Cc) is the slope of the linear portion of the pressure–void ratio curve on a semi-log plot, with pressure on the log scale (IS: 8009 – Part 1, 1976).
  • This is a dimensionless parameter.

Compression index on a soil helps to determine total settlement of clay layer.

Mathematically consolidation settlement(Sc) is given by,

\({S_C} = H\frac{{{C_C}}}{{1 + {e_0}}}log\frac{{{p_0} + {{\rm{\Delta }}_{\rm{p}}}}}{{{p_0}}}\)

Where CC = Co-efficient of the compression index, H = Thickness of clay layer, e0 = Initial void ratio = \(wG\over S\) , w = water content, G = specific gravity, S = degree of saturation, p0 = Initial overburden pressure at the middle of clay layer, \(\Delta p\) = Extra pressure due to the new

Important Points

For undisturbed low to medium sensitivity soils,  (Terzaghi and Peck empirical index) 

→ Cc = 0.009 × (WL - 10%)

For disturbed(remolded) low to medium sensitive soil (Skempton's empirical index)

→ Cc = 0.007 × (WL - 10%)

Question 80

FCI JE 2015• Test

If levelling staff is held inclined at a staff station, the reduced level calculated from observation would be:

  1. AOption (a)

    True R.L.

  2. BOption (b)

    More than true R. L.

  3. COption (c)

    Less than true R. L.

  4. DOption (d)

    Equal to R. L. of Bench mark

View answer & explanation

Correct answer

Answer: (c)C.

Less than true R. L.

Why this answer is correct

Explanation: 

  • Staff reading is less when staff held vertical as compare to when staff is held inclined at staff station.
  • Reduced level calculated for the inclined position of the staff is less as the staff reading is more as compare to staff held vertical.
  • Reduce level of a point = Height of instrument - Staff reading at that point
  • So as the staff reading increases, the RL of a that staff station decreases as compare to R.L of instrument station.

 

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