1. According to the unified soil classification system, what property do the symbols L, M and H indicate?
a) Viscosity
b) Grain size
c) Density
d) Plasticity
Explanation: The symbol L represents soil with low plasticity i.e. it has liquid limit less than 35%. The symbol M represents soil with medium plasticity and has liquid limit in between 35% and 50%. The symbol H represents soil with high plasticity and has liquid limit greater than 50%.
2. The tests to evaluate the strength of subgrade soil can be broadly classified as ______
a) Shear and penetration tests
b) Shear and bearing tests
c) Bearing and penetration tests
d) Shear, bearing and penetration tests
Explanation: There are three broad classifications of test types to evaluate subgrade soil strength. They are shear, bearing and penetration tests. Each contains several other tests under them to evaluate the strength.
3. The ______ test can be conducted in-situ as well as in the laboratory.?
a) CBR
b) Unconfined compression
c) Triaxial
d) Direct shear
Explanation: CBR is a type of penetration test and there have been procedures developed to allow conducting the test both in-situ and in the laboratory. Unconfined compression test, triaxial test and direct shear test are all types of shear tests. These are usually conducted in the laboratory only.
4. What does CBR stand for?
a) California Binding Ratio
b) Combined Binding Ratio
c) California Bearing Ratio
d) Combined Bearing Ratio
Explanation: CBR stands for California Bearing Ratio and it is a test used to determine the strength of subgrade soil. It was developed by the California State Highway Department and hence the name.
5. What is the angle of internal friction for a saturated soil sample that fails under axial stress of 100 kN/m2 in an unconfined compression test and the failure plane makes an angle of 50° with the horizontal??
a) 10°
b) 20°
c) 15°
d) 25°
Explanation: The angle of internal friction, Φ is related to the failure angle, θ by the equation:
Φ=2θ-90
Substituting θ=50° from the question,
Φ=2×50°-90=10°
6. Modulus of subgrade reaction is defined as pressure sustained per unit volume of subgrade.
a) true
b) false
Explanation: Modulus of subgrade reaction is represented by the letter k and it is defined as the pressure sustained per unit deformation of the subgrade. It is obtained after conducting the plate load test.
7. Determine the CBR value of the specimen at 2.5 mm penetration if the dial reading obtained from the graph is 35 divisions and 100 division of load dial represents 190 kg load in the calibration chart of proving ring.
a) 4.8%
b) 4.9%
c) 4.8
d) 4.9
Explanation: The equation for computing CBR value of the soil is as below:
CBR at 2.5mm penetration in %
= \(\frac{(Load \,sustained \,by \,specimen \,at \,2.5mm \,penetration)}{(Load \,sustained \,by \,standard \,aggregate \,at \,2.5mm \,penetration)}×100%\)
Load sustained by specimen=\(35×\frac{190}{100}=66.5 kg\)
CBR at 2.5mm penetration=\(\frac{66.5}{1370}×100=4.85\approx \)4.9%
8. The test to obtain values of consistency limits are carried out as per which IS code?
a) IS 2720 Part V
b) IS 2720 Part VII
c) IS 2720 Part IV
d) IS 2720 Part III
Explanation: Consistency limits include liquid limit, plastic limit and shrinkage limit. These are tested using the Casagrande apparatus and the guidelines to be followed for performing the test are outlined in IS 2720 Part V. IS 2720 Part VII gives guidelines on tests for water content determination, IS 2720 Part IV gives guidelines on the grain size distribution and IS 2720 Part III gives guidelines on specific gravity tests.
9. Which of the below indices is a measure of shearing strength of soil at the plastic limit?
a) Plasticity index
b) Consistency index
c) Flow index
d) Toughness index
Explanation: Toughness index is obtained as the ratio between the plasticity index and flow index. It represents the shearing strength of soil at the plastic limit. Plasticity index gives a measure of clay content, consistency index gives a measure of firmness of soil and flow index gives a measure of shear strength.
10. As per the Mohr-Coulomb theory, the point inside the Mohr’s circle is considered to be ______
a) Not possible
b) Failure
c) Stable
d) Unstable
Explanation: If the point is inside the Mohr’s circle, then the soil is considered to be stable as the point lies below the failure envelope. If the failure envelope just touches the circle, then the soil is considered to have failed. It is not possible to have the point above the failure envelope.