Quiz: Civil Engineering
Topic: Soil Mechanics

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 10 Minutes

Q1. An odd shaped body weighing 7.5 kg and occupying 0.01 cubic meter volume will be completely submerged in a fluid having specific gravity of ………………..
(a) 1
(b) 1.2
(c) 0.8
(d) 0.75

Q2. For perfectly dry soil, degree of saturation is:
(a) Zero
(b) 0-1
(c) 1
(d) Infinite

Q3. The relationship between air content of soil (a_c) and its degree of saturation (S_r) is expressed as
(a) a_c=1+S_r
(b) a_c=S_r-1
(c) a_c=1-S_r
(d) None of these

Q4. The meniscus and dispersing agent corrections, in the hydrometer analysis, respectively are
(a) Positive and negative
(b) Negative and positive
(c) Positive and positive
(d) Negative and negative

Q5. A moist sample of soil weight 24 grams in a tin lid. The tin lid above weight 14 grams. The over dry weight of tin and sample is 22 grams. What is the water content of the soil?
(a) 15.5%
(b) 39.5%
(c) 25%
(d) 60%

Q6. Toughness index is defined as the ratio of
(a) Plasticity index to consistency index
(b) Liquidity index to flow index
(c) Consistency index to liquidity index
(d) Plasticity index to flow index

Q7. If a soil sample is dried beyond its limit, this sample will show –
(a) No volume change
(b) Moderate volume change
(c) Low volume change
(d) large volume change

Q8. When the consistency index of a soil is equal to zero, then it is side to be at its
(a) Liquid limit
(b) Plastic limit
(c) Shrinkage limit
(d) Any one of the above limits

Q9. A negative value of the group index of a soil is reported as:
(a) A positive value of the same magnitude dropping the negative sign.
(b) Zero
(c) negative value, as GI may be negative.
(d) Gl is reported as non-existent.

Q10. Nuclear density gauge can be used for all the following purposes, except
(a) Moisture content
(b) Wet density
(c) Dry density
(d) Standard penetration reading


S1. Ans.(d)
Sol. weight = 7.5 kg
Volume = 0.01 m³
Weight density (r) = 7.5/0.01
= 750 kg/m³
Specific gravity (G) = r/r_w =750/1000
= 0.75
S2. Ans.(a)
Sol. for perfectly dry soil degree of saturation equal to zero.
S3. Ans.(c)
Sol. ▭(S+a_c=1)
S → degree of saturation
a_c → air content
S4. Ans.(a)
Sol. the observed hydrometer reading is further being corrected by following –
Meniscus correction (Cm)
Temperature correction (C_T )
Dispersing agent correction (C_D )
→ Meniscus correction is always positive & dispersing agent correction is always negative
→ if temperature greater then 27°c then temperature correction is Positive and when temperature is less than 27°C then temperature correction is negative.
S5. Ans.(c)
Sol. weight of moist (soil +tin lid) = 24 gm
Weight of tin lid = 14gm.
Oven dry weight of (soil +tin lid) = 22 gm.
Weight of water = 24-22 = 2gm.
Weight of solid = 22 – 14 = 8 gm.
Water content = 2/8×100
= 25%
S6. Ans.(d)
Sol. ▭(Toughness index=(Plasticity index)/(flow index))
S7. Ans.(a)
Sol. if soil sample dried beyond its limit then no volume change in soil because water is replaced by air & volume of soil same
S8. Ans.(a)
Sol. Consistency index (I_C ) = (W_L-W)/(W_L-W_P )
If I_C=0
▭(W=W_L )
S9. Ans.(b)
Sol. when value of group index is negative than we will assumed as zero.
S10. Ans.(d)
Sol. nuclear density gauge used in standard penetration reading.

Leave a comment

Your email address will not be published. Required fields are marked *