Practice Test: Civil Engineering (CE)- 8 - Civil Engineering (CE) MCQ

The spokes are units which radiate of a wheel and make the wheel a complete entity. Similarly, fingers, tentacles and petals are integral units of hand, octopus and flower respectively. On the other hand, roots and leaves do not share this kind of relationship. Thus option 4 does not expresses a relationship similar to that expressed in the given pair.

Let a = 3x and b = 4x

Similarly b = 2y and c = y

∴ 4x = 2y ⇒ y = 2x

∴ c = 2x

Now a + b + c = 3x + 4x + 2x = 9x

So, the minimum integer value = 9

Friday → 2 days earlier

Therefore, scheduled day = Friday + 2 = Sunday

Sunday + 3 = Wednesday

Therefore, I would have been late by 3 days

The meaning of the given words is:

Impeccable: in accordance with the highest standards

Flawless: without any imperfections or defects

Indelible: (of ink or a pen) making marks that cannot be removed

Intangible: unable to be touched

Collect: bring or gather together (a number of things)

Hence, option 1 is the correct answer.

The sentence implies that it was hoped that that place would become the point from where the African civilization would proceed, but the belief that it would happen was not fulfilled.

Therefore, the correct word to fill in the blank is expectation as it means a strong belief that something will happen or be the case.

Mean = 4

k + m = 8

k ≠ m so k = m = 4 is out.

{k, m} = {1, 7} or {2, 6} or {3, 5}

for median of {3, k, 2, 8, m, 3}

{1, 2, 3, 3, 7, 8} or {2, 2, 3, 3, 6, 8} or {2, 3, 3, 3, 5, 8}

Median: (3 + 3) /2 = 3

A person can take 1^st step in any direction independently

Suppose he moves to A

Now to return to O ( initial point) in 2 steps he can move in 2 directions. Either B or f

Thus probability he will move to either B or F will be =

Suppose he moves to B

Now to return back to O

he has only 1 option

⇒ to move in BO

Probability he will move in BO direction =

Total probability he will return

The nth line drawn will add n more regions to the circle. 

1st line adds 1 region to the circle for a total of 2 regions. 

2nd line adds 2 more regions to the circle, bringing the total number of regions to 2 + 2 = 4. 

3rd line adds 3 more regions to the circle, bringing the total number of regions to 4 + 3 = 7. 

4th line adds 4 more regions to the circle, bringing the total number of regions to 7 + 4 = 11.

5th line adds 5 more regions to the circle, bringing the total number of regions to 11 + 5 = 16. 

6th line adds 6 more regions to the circle, bringing the total number of regions to 16 + 6 = 22. 

7th line adds 7 more regions to the circle, bringing the total number of regions to 22 + 7 = 29. 

8th line adds 8 more regions to the circle, bringing the total number of regions to 29 + 8 = 37. 

9th line adds 9 more regions to the circle, bringing the total number of regions to 37 + 9 = 46. 

10th line adds 10 more regions to the circle, bringing the total number of regions to 46 + 10 = 56. 

11th line adds 11 more regions to the circle, bringing the total number of regions to 56 + 11 = 67. 

12th line adds 12 more regions to the circle, bringing the total number of regions to 67 + 12 = 79. 

The correct answer is option 2 i.e. To implement hardware protocols to minimize risks and reduce electromagnetic radiation production significantly. 

The passage states about the electromagnetic radiations released from the devices and how they affect the individuals. Out of the given options, only option 2 states the possible solution which can be implemented to reduce electromagnetic radiation.

Let the number of students be X who cleared cut off in all sections.

The number of students who cleared the cutoff in section 1 and section 2, only = 20 – X

The number of students who cleared the cutoff in section 2 and section 3, only = 16 – X

The number of students who cleared the cutoff in section 1 and section 3, only = 26 – X

Now, consider section 1:

24 + 20 – x + x + 26 – x = 60

70 – x = 60

x = 10

R₅ → R₁ + R₂ + R₄ + R₅

R₅ → R₅ - R₃, R₃ → R₃ - R₁

R₄ → 2R₄ - R₃

R₃ → R₃ - 4R₂

R₄ → 5R₄ - R₃

Now it is in Echelon form.

Rank of matrix = number of non-zero rows = 4.

f(z) = z̅ = x - iy

u = x, v = -y

u_x = 1, v_y = -1

u_y = 0, v_x = 0

u_x ≠ v_y, as f(z) is not satisfying CR equations, f(x) is not analytic.

g(z) = |z|²

g(z) = x² + y²

u_x = 2x, v_x = 0

u_y = 2y, v_y = 0

at z = 0, u_x = v_y and u_y = -v_x

g(x) is satisfying CR equations at z = 0 but it is not satisfying CR equations at neighborhood of z = 0.

Hence g(z) is not analytic.

The partial differential equation will represent the parabola, if:

B² – 4AC = 0

Now given equation:

16 – 4 x 1 × m = 0

4m = 16

m = 4

Give equation (3xy² + n² x²y) dx + (nx³ + 3x²y) dy = 0, x ≠ 0

Concept:

M dx + N dy = 0, will be exact differential equation if:

Where M and N are functions of x and y.

Here M = 3xy² + n² x²y

N = nx³ + 3x²y

Given equation is exact, thus:

6xy + n² x² = 3nx² + 6xy

n² x² = 3nx²

n = 3

The increase in direct cost for 3 days crashing

= 1500×3 

= 4500 rupees

The decrease in indirect cost for 2 days

= 1000×2 

= 2000 rupees

The total cost of the activities after crashing

=  50000 + 4500 - 2000

= 52500 rupees

Seasoning of timber results in:

1. Increased strength

2. Increased durability

3. Increased resilience

Through the application of preservation of timber,

the life span of the timber can be increased and prevent the growth of fungi in the timber.

The specific gravity of wood is less than 1.

Abel's process treatment is used to make timber fire-resistance.

As the maximum permissible stress in the outermost fibers in both steel and concrete reaches at the same time, therefore it is the case of balanced section:

So,

X_u = X_ulim

For Fe 500

X_ulim = 0.46 d

d = effective depth of the beam = 550 mm

X_ulim = 0.46 x 550 = 253 mm

d' = effective cover to compression reinforcement = 50 mm

Maximum strain in concrete at outermost compression reinforcement as per IS 456: 2000 = 0.0035

Now from similar triangle,

ϵ_sc = 0.00281

Stiffness of member AB at 

Stiffness of member AD at

Distribution factor for AB  = 

Stiffness of member BC at B = 

Stiffness of member BA at B = 

Distribution factor (BC) = 

For simply supported and continuous beam: 

clear distance between lateral restrains shall not be greater than:

  whichever is lesser 

Where,

B = width of the beam = 250 mm

d = effective depth of the beam = 500 mm

i) 60 B = 60 × 250 = 15000 mm = 15 m

= 31250mm = 31.25m

lesser will be 15 m.

Number of independent degrees of freedom is nothing but kinematic indeterminacy of the beam/structure

D_k = 3j – r_e + r_r

J → No. of Joints

r_r = No. of released reactions

r_e = External support Reactions

J = 3

r_e = 5

r_r = m^’ - 1

m^’ = no. of members meeting at a Joint (hinge)

m^’ = 2

r_r = 2 – 1 = 1

D_k = 3 × 3 – 5 + 1

D_k = 5

Concept: At x – x

σ′_x = τ_xysin2θ 

σ′_y = -τ_xysin2θ 

τ_x′_y′ = τ_xycos2θ

Calculation:

τ_xy = 50 MPa

θ = 30°

σ′_x = 50sin60^∘ 

σ′_x = 43.30MPa 

σ′_y = −50sin60^∘ 

σ′_y = −43.30MPa 

τ_x′_y′ = 50cos60^∘ =25MPa 

Centre of moh’r circle = (a, 0)

So, centre = (0, 0)

Radius of moh’r circle

r = 25 MPa

Concept:

Vertical stress increment at a any depth (z) at a radial distance (r) as per Boussines’q theory is:

Where,

K_B = Boussinesq’s influence factor

Calculation:

At r = 0 m,

Q = 10000 × 9.81 = 98100 N

z = 6 m

Velocity at point A, V_A = 6 cm × ω

⇒ ω = 0.0556 rad/sec

∴ Velocity of point B, V_B

= 0.08 × ω

= 0.08 × 0.0556 rad/sec

= 0.444 m/sec

Activity of the soil is defined as the ratio of Plasticity Index of the soil to the % by weight of particles finer than 2μ.

I_P = w_L - w_P

w_L = 60%

w_P = 45%

I_P = 60% - 45% = 15%

As Activity is less than 0.75, Soil is classify as Inactive.

Toughness index is defined as the ratio of Plasticity index to the flow index.

Toughness Index=

I_P = Plasticity Index

I_f = Flow index, which represent the rate of loss of shear strength in soils.

More will be the toughness index, more will be the soil firm.

Magnetic bearing of line PQ = 9° 30’

Magnetic Declination = 3° W

True bearing of line PQ = M.B – Declination

= 9° 30’ – 3°

Present Magnetic Bearing = TB + Present Declination = 6° 30’ + 5° 30’ = 12°

Concept:

Froude number is used in model analysis of spillways

(F)_m = (F)_P

Calculation:

= 35000 

Concept:

Velocity field for a flow is represented as:

Where,

u = velocity component in x – direction

v = velocity component in y – direction

Rate of shear strain in x – y plane is given by:

Calculation:

u = 2xy – 4y³

Check for the flow

   (OK, flow is possible)

For rate of shear stain in x – y plane

 = 

The ratio of oxygen available in the wastewater (as dissolved oxygen) to the total oxygen required to satisfy its first state BOD is called relative stability.

Relative stability,

Where, t₂₀ and t₃₇ represent the time in days for a sewage sample to decolorize a standard volume of methylene blue solution, when incubation is done at 20° or 37°C respectively.

S = 100 × 0.842

S = 84.2%