Practice Test: Civil Engineering (CE)- 6 - Civil Engineering (CE) MCQ

The correct spelling of the word is 'definitive' which means '(of a conclusion or agreement) done or reached decisively and with authority.' Thus option 2 is the correct answer.

The preposition 'about' is mandatory here thus option 1 and 4 are eliminated. Option 2 is correct as the tense present perfect continuous fits here. It conveys the meaning that the person usually talked about the place. 

Options 1 and 2 are incorrect as they change the meaning of what is mentioned. Option 3 is incorrect as with 'less' we use the adjective in positive form and not comparative form. The correct word here should be 'brave.' Option 4 is thus the correct answer.

Let number is N

N = 65 k +29

Now 

∴ Remainder = 3

The word 'here' and 'there' both can be used here but should be followed with 'waiting' as no other word can fit here. The word 'caught' is correct here as 'catch a sight' means 'to see something.' Option 2 thus has the correct combination of words. The word 'cot' means 'a small bed with high barred sides for a baby or very young child.'

For a number to be greater than 5000, d₁ should be filled with either 5 or 7

∴ Total numbers formed when the digits are repeated = 2 × 5 × 5 × 5 = 250

total cases = 250 -1 = 249 ( case of 5000 is not included)

Now, For the number to be divisible by 5, unit digit d₄ should be either 0 or 5.

∴ Total no. of ways = 2 × 5 × 5 × 2 = 100

favorable cases = 100 - 1=9 ( 5000 is not included))

∴Required Probability = =0.397

The question asks which of the statements given in the options can weaken the argument put by the author that all bacteria from Venus must have died out.

The passage states that since there is a single strain of bacteria which exists on the Earth, they all must be belonging to the Earth or let's say a single planet. But here the author does not take into consideration (as can be argued from his theory) the fact that may be all the bacteria came from Venus and there are none which originally belong to the Earth. So this criticism as mentioned in option 3 makes the argument of the author vulnerable. 

Options 1, 2 and 4 are completely irrelevant criticisms as they do not address the main argument. The argument claims that if there were Venusian bacteria on Earth, then they must have died out by now. Whether there are bacteria originally from Earth that have also disappeared from Earth is irrelevant to the question and has no effect on the given argument.

Let a, b and c be the cost prices of the three articles A, B and C.

SP = CP + Profit (or) SP = CP – Loss

⇒ SP of A = 1.1a; SP of B = 1.2b; SP of C = 0.9c

By question,

1.1a + 0.9c = a + c ⇒ 0.1a = 0.1c ⇒ a = c

1.2b + 0.9c = 1.05(b + c) ⇒ 0.15b = 0.15c ⇒ b = c = a

Gain% = {(SP – CP)/CP} × 100

⇒ Net gain on the sale of all the articles =   =

∴ Net gain on the sale of all the articles = 6.66%

Option 1 is false as graph says there is decrease in students qualifying in Physics in 2015 compared to 2014.

Option 2

Let no. of students qualifying in Biology in 2013 be 100

⇒ No. of students qualifying in Biology in 2014 = 100 – 10% of 100 = 90

⇒ No. of students qualifying in Biology in 2015 = 90 + 10% of 100 = 99

∴ The number of students qualifying in Biology in 2015 is less than that in 2013

Option 3 and option 4 are incorrect since no detail is given regarding how many students qualified the subject in 2013.

Area of triangle PAB = Area of triangle RCB (By symmetry)

∴ Area of Δ PAB = ½ × PA × AB

= ½ × 2A/3 × A = A²/3

Area of ΔRCB = A²/3

Now, Area of ABCD = Area of DRQP + Area of PAB + Area of RCB + Area of PBRQ

A² = a² + A²/3 + A²/3 + Area of PBRQ

As, A/a = 3 ⇒ a = A/3

⇒ Area of PBRQ = 

Y(s) = L(y(t)) =

Apply inverse Laplace transform,

⇒ y(t) = 
Differentiate with respect to ‘t’.

⇒ y(0) + y'(0) = 1

= 1

A matrix is said to be singular, if determinant of that matrix is zero.

= 1 (18 – 12) - 1 (9 – 4) + 1 (3 – 2)

= 6 – 5 + 1 = 2 ≠ 0

M is non singular

(Q) A matrix can be diagonalizable when it has distinct eigen values

S is a diagonalizable matrix. Hence, has distinct eigen values.

Let S be a 3 × 3 matrix and the eigen values of s are λ₁, λ₂, λ₃

Given that, S + 5T = I

From the properties of Eigen values,

(a) If λ₁ is an eigen value of matrix A, then -λ_{1 ­}will be on eigen value of matrix -A.

(b) If λ₁ is an eigen value of matrix A, then (λ₁ + 1) will be an eigen value of matrix (A + I)

(c) If λ_1­ is an eigen value of matrix A, then  will be an eigen value of matrix  where K is a scalar.

From the above properties, eigen values of T are,

As λ₁, λ₂, λ₃ are distinct values, λ′₁,λ′₂,λ′₃ will be distinct.

Hence, matrix T is diagonalizable

So, only Q is true.

A.E is, (D² – 3D + 2) = 0

⇒ (D – 1) (D – 2) = 0

⇒ D = 1, 2

x(t) = C₁e^t + C₂e^2t

x = 0 at t = 0

⇒ 0 = C₁ + C₂ ⇒ C₁ – C₂

x = 1 at t = 1

⇒ 1 = C₁e¹ + C₂e²

⇒ 1 = C₁e¹ + (-C₁) e²

⇒ C₁(e¹ – e²) = 1

x(2) = C₁e² + C₂e⁴

= e + e²

Static indeterminacy of the Planer structure:

D_S = R – r

R → no. of unknowns = 3

r → no. of equilibrium equations = 3

D_S = 3 – 3 = 0

As all the three Reactions (1, 2, 3) are non-parallel, non-concurrent and non-trival, So structure is externally stable.

For Internal stability of truss:

m ≥ 2J -3

m → no. of member

J → no. of Joints

m = 6

J = 5

6 ≥ 2 × 5 – 3

6 ≥ 10 – 3 = 7

As no. of members is less than (2J – 3), therefore the truss is internally unstable.

Factor affecting concrete strength

a) concrete porosity

b) water/cement ratio

c) Soundness of aggregate

d) Aggregate paste bond

e) Cement related parameters.

Hence, The water cement ratio (w/c) is not only factor influencing the strength of concrete.

At water/cement ratio more than 0.6, the increase in volume of hydrated product will not be able to occupy the space already filled with water. Hence, porosity increase and strength decreases.

The addition of the surfactant in the concrete mix results in the decrease in the cement ratio.

As per clause 5.4 of IS 456 : 2000

Potable water is considered satisfactory for mixing Concrete and the permissible limits for solids is shown in table below:

Concept:

Using the relation,

E = 3k(1 – 2μ)       ……(i)

E = 2G(1 + μ)         ……(ii)

From equation (i) and (ii)

3k(1 – 2μ) = 2G (1 + μ)

where k = Bulk modulus of the material

G = shear modulus of the material

μ = poisson’s ratio

Calculation:

μ = 0.21

3k(1 – 2μ) = 2G (1 + μ)

Soil sample when water content is 26.5%:

Using the relation Se = wG

a_c = 20%

a_c + S = 100%

S = 100% - a_c

S = 100% - 20%

S = 80%

w = 26.5%

0.80 × e = 0.265 × 2.65

e = 0.878

e = 87.8% (ans)

Soil sample when water content is 30%:

w_I = 30%

G = 2.65

e = 0.878

S_I e = W_i G

S_I × 0.878 = × 2.65

S_I = 0.905

S_I = 90.5%

a_c1+S_I = 100%

a_c1 = (100 − 90.5)%

a_c1+S_I = 100%

a_c1 = (100 − 90.5)%

change in the air content = a_c − a_cI

Δa_c = (20 – 9.5)%

Δa_c = 10.5% (Ans)

Concept:

Using the relation

Where

S_f = settlement of foundation

S_p = settlement of plate

B_f = width of footing/foundation

B_p = width of plate

Calculation:

S_f = 35 mm

S_p = 10 mm

B_p = 450 mm

B_f = 1575 mm

Therefore width of foundation required will be 1575 mm.

From the test results:

(i) % of particles retained over 0.075mm IS sieve = (100 – 4)% = 96%

So, soil is coarse Grained soil

(ii) % fineness = 4% < 5%

(iii) As 62% (more than 50%) coarse fraction is retained over 4.75 mm IS sieve, the soil is termed as Gravel

Also, C_u = 5 > 4

C_c = 1.5 which lies between 1 and 3

So, soil is termed as well graded Gravel

Hence, soil is GW

Base Exchange Capacity: The ability of the clay particles to absorb ions on its surface or edges is called Base or cation exchange capacity. Base exchange capacity depends upon the size of particles and and mineral structure.

Free float (F_F) = Total float (F_T) – Head event slack (S_j)

Total float is the maximum available time in excess to activity completion time.

FF for the activity [(3) – (5)] = F_T [(3) – (5)] – S₍₅₎

= (14 – 8) – 4 – 2

= 0 days

Concept:

Critical depth of flow:

q = discharge per unit width

Calculation:

Specific Energy at critical depth =
= 1.1122m

Specific Energy at depth “y₁” = 1.20 E_c

E₁ = 1.20 × 1.1122

E₁ = 1.3347m

y₁ = 1.191m

As per IS 10500: 2012,

As per IS 1055: 2012, the permissible limits in absence of alternate source of water for calcium and magnesium in the drinking water respectively will be 200 ppm and 100 ppm.

Discharge form well in confined aquifer (Q)

where, k = hydraulic conductivity

d = depth of the aquifer

S₁ = drawdown in the observation well 1

S₂ = drawdown in the observation well 2

(S₁ - S₂) = 0.5m given

r₁ = distance of observation well 1 form test well

r₂ = distance of observation well 2 from test well.

⇒ k = 0.00637 m/s = 6.37 mm/s

The Vehicle Damage Factor (VDF) is a multiplier to convert the number of commercial vehicles of different axle loads and axle configuration into the number of repetitions of standard axle load of magnitude 80 kN.

It may also define as equivalent number of axles per commercial vehicles. The VDF varies with the vehicle axle configuration and axle loading.

Theoretical maximum capacity of single lane can be expressed as

where, S = Capacity of single lane in vehicle per hour.

V = Speed in kmph

S = Space headway

Theoretical capacity can also be written as

Where, H_t = time headway

Calculation:

Space Headway (S) =

Time Headway (H_t) = 3600/900 sec

= 4 sec 

Cutback bitumen is a bitumen with less viscosity which, is achieved by addition of volatile diluent. Hence, to increase fluidity of the bitumen binder at low temperature the binder is blended with a volatile solvent.

The viscosity of the cut book and the rate of which hardness on the road depends on the characteristics and quantity of both bitumen and volatile oil used as diluent.

Cutback bitumen are available in there types:

1) Rapid curing → Naptha, gasoline

2) Medium curing → Kerosene or high diesel oil

3) Slow Curing → High boiling point gas oil.