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Tamil Nadu PSC Soil Mechanics: practice questions

Tamil Nadu PSC — Agricultural Engineering (Code 164, degree standard) · India · Tamil Nadu Public Service Commission

Original practice questions written to the published syllabus · not past papers
10worked here, free
26written to this exam in this subject
8syllabus units

About Tamil Nadu PSC

Tamil Nadu PSC — Agricultural Engineering (Code 164, degree standard) is a government recruitment exam run by Tamil Nadu Public Service Commission, for candidates in India. Degree-standard subject paper (for posts such as Assistant Engineer (Agri)).

Source: tnpsc.gov.in.

Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.

Soil Mechanics in Tamil Nadu PSC

Earthen dams and farm structures: seepage, filters, slope protection and compaction of embankments, with standard definitions.

Soil as an engineering material: its phases, flow of water through it, compaction, consolidation, shear strength and stability.

Our 26 Tamil Nadu PSC questions in this subject: 14 easy, 9 medium, 3 hard, across 8 units.

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

These 10 were written to Tamil Nadu PSC and are not shown on any other exam's page.

Question 1 · easy

Which of the following soil parameters can take a value greater than 100% (or greater than 1)?

  1. Porosity
  2. Degree of saturation
  3. Water content
  4. Air content
Answer and working

C. Water content

Water content is water mass divided by dry solids mass, so peats and very soft clays can exceed 100%. Porosity, degree of saturation and air content are all fractions of a whole volume and cannot exceed 1. Void ratio can also exceed 1, but it is not offered here.

Question 2 · medium

A flow net under a concrete weir on soil with k = 5 × 10⁻⁶ m/s has 4 flow channels and 12 equipotential drops. The head difference across the weir is 6 m. What is the seepage per metre length of weir per day?

  1. 0.864 m³/day
  2. 7.78 m³/day
  3. 0.216 m³/day
  4. 8.64 m³/day
Answer and working

A. 0.864 m³/day

q = kH·Nf/Nd = 5 × 10⁻⁶ × 6 × 4/12 = 1.0e-05 m³/s per m = 0.864 m³/day per m. Inverting the shape factor (Nd/Nf) gives 7.78; leaving out Nf gives 0.216. 8.64 is a factor-of-ten slip in k (5 × 10⁻⁵ m/s).

Question 3 · medium

During the first filling of a farm pond, sand boils carrying fine soil appear at the downstream toe. Which immediate action is correct?

  1. Seal the boils with a compacted impervious clay cap so no water can exit, and keep filling the pond
  2. Pump water out of the boils to lower the tailwater, and keep the reservoir level unchanged meanwhile
  3. Add fill to the crest to increase freeboard
  4. Place a graded inverted filter with a weighting berm over the boils, and lower the reservoir
Answer and working

D. Place a graded inverted filter with a weighting berm over the boils, and lower the reservoir

Boils carrying fines show piping has begun. A graded filter weighted with sand and gravel lets water out while holding soil in, raising the effective stress at the exit; lowering the reservoir reduces the head. A clay cap blocks the exit and the water finds a new path; pumping lowers tailwater and raises the gradient; raising the crest does nothing for seepage.

Question 4 · easy

Which roller is best suited for compacting a cohesive clay embankment in layers?

  1. Sheepsfoot roller
  2. Smooth-wheeled (static) roller
  3. Vibratory plate compactor
  4. Grid roller for rock
Answer and working

A. Sheepsfoot roller

Sheepsfoot and padfoot rollers knead cohesive soil and compact each lift from the bottom upward. Vibration is most effective in sands and gravels, where it rearranges grains; smooth-wheeled rollers suit gravels and final finishing. A padfoot roller works the same way and is equally suitable.

Question 5 · easy

A saturated clay specimen fails in an unconfined compression test at an axial stress of 120 kPa. What is its undrained shear strength?

  1. 60 kPa
  2. 120 kPa
  3. 240 kPa
  4. 30 kPa
Answer and working

A. 60 kPa

In unconfined compression σ3 = 0 and σ1 = qu at failure. With φu = 0 the Mohr circle's radius equals cu, so cu = qu/2 = 60 kPa. 120 kPa takes cu as qu itself, and 240 kPa doubles it instead of halving.

Question 6 · medium

In a φ = 0 (Swedish circle) analysis of an embankment on soft clay (cu = 40 kPa), a trial slip circle of radius 12 m has an arc length of 25 m. The sliding mass weighs 1500 kN per metre, acting 4.5 m horizontally from the circle centre. What is the factor of safety for this circle?

  1. 0.67
  2. 0.56
  3. 1.78
  4. 3.56
Answer and working

C. 1.78

F = resisting moment/driving moment = cu·La·R/(W·x) = 40 × 25 × 12/(1500 × 4.5) = 12000/6750 = 1.78. 0.56 is the ratio inverted; 3.56 wrongly uses the unconfined strength (2cu) in place of cu. Many trial circles must be checked; the minimum F governs. 0.67 uses R instead of x in the driving moment (12000/(1500 × 12)).

Question 7 · medium

For the upstream face of a homogeneous earth dam used for a farm reservoir, which loading condition is usually the most critical?

  1. Steady seepage with a full reservoir
  2. End of construction with an empty reservoir
  3. Reservoir held at half depth for a long time
  4. Rapid drawdown of the reservoir
Answer and working

D. Rapid drawdown of the reservoir

After rapid drawdown the stabilising water load on the upstream face is removed while the fill is still saturated and pore pressures remain high, so driving forces rise without a matching increase in effective stress. Steady seepage with a full reservoir usually governs the downstream face.

Question 8 · easy

What is the net ultimate bearing capacity of a shallow footing?

  1. Gross ultimate capacity divided by the factor of safety chosen for the footing design
  2. Gross ultimate capacity minus the overburden pressure γDf at founding level
  3. Gross ultimate capacity plus γDf
  4. The pressure causing 25 mm settlement
Answer and working

B. Gross ultimate capacity minus the overburden pressure γDf at founding level

Net ultimate capacity qnu = qu − γDf is the extra pressure, above what the soil already carried before excavation, that causes shear failure. Dividing it by the factor of safety gives the net safe bearing capacity.

Question 9 · medium

A grain godown is to be built on deep expansive black cotton soil in a region with a long dry season. Cracks in similar buildings nearby are due mainly to seasonal swelling and shrinkage, not bearing failure. Which foundation approach addresses the actual cause?

  1. Widen the footings to reduce bearing pressure so the soil can no longer swell against them
  2. Compact the surface soil dry of optimum
  3. Found below the zone of seasonal moisture change, for example on under-reamed piles
  4. Raise the plinth with compacted black cotton soil so that surface water drains away from the walls
Answer and working

C. Found below the zone of seasonal moisture change, for example on under-reamed piles

Damage in expansive clays comes from volume change as the moisture content of the upper soil changes with the seasons. Taking the foundation below this active zone, with under-reamed piles or deep footings, avoids it. Wider footings reduce bearing pressure but not heave; compacting dry of optimum increases swell potential.

Question 10 · medium

A borrow pit for a small homogeneous earth dam yields a soil classified CL with wL = 38% and Ip = 17%. Which statement about its suitability is most accurate?

  1. It is unsuitable because clays can never be used in dams
  2. It is a free-draining soil suitable for toe filters
  3. Its classification alone proves its permeability is below any required value
  4. It is usually suitable for an impervious section if compacted near optimum, but needs checking for dispersivity and shrinkage cracking
Answer and working

D. It is usually suitable for an impervious section if compacted near optimum, but needs checking for dispersivity and shrinkage cracking

Low-to-medium plasticity inorganic clays are widely used for impervious zones of small dams when compacted near optimum. Classification guides the choice but tests of permeability, compaction and dispersivity (dispersive clays pipe easily) are still needed. A clay is not a filter material.

A study plan for Soil Mechanics in Tamil Nadu PSC

This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.

  1. First, the heavy units (62% of the 26 questions): Seepage and flow nets (7, 27%); Slope stability, including infinite slopes (6, 23%); Bearing capacity basics (3, 12%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Compaction and the Proctor test (3, 12%); Phase relations (3, 12%); Index properties and soil classification (2, 8%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Permeability (1, 4%); Shear strength and the Mohr-Coulomb criterion (1, 4%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Leave time for the hard ones: Compaction and the Proctor test (1 hard of 3), Seepage and flow nets (2 hard of 7).

    Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.

  5. Bank the easy marks: Phase relations (3 easy of 3); Permeability (1 easy of 1).

    Do these early in revision so they are fast on the day.

  6. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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