Practice Test: Civil Engineering (CE)- 15 - Civil Engineering (CE) MCQ

→ (a - b)² = (a + b)²– 4ab ………………….(equation)

Given, (a + b) = 5 and ab = 6

Substituting the values in equation, we get,

(a-b)² = 5² – 4 x 6 = 1

→ (a - b) = 1

We know that, (a² – b²) = (a + b)(a - b)

Substituting the values of (a + b) and (a - b) in above equation, we get,

(a² – b²) = 5 × 1 = 5

Hence, the required answer is 5.

Another Solutions:

let, a= 3 & b=2 a>b

a+b=5

3+2= 5,

ab =6, 2*3=6,

(a² - b²)= 3²- 2² = 9- 4 = 5.

Obligatory (adj.): so customary or fashionable as to be expected of everyone or on every occasion.

Prescribed (adj.): state authoritatively or as a rule that (an action or procedure) should be carried out.

Assured (adj.): made (something) certain to happen.

Avoidable (adj.): able to be avoided or prevented.

So, the correct answer is option D.

= 15% of 60 lacs

= 15 × 60 lacs/ 100

= 9 lacs

Expenditure made by university of psychology

= 10% of 60 lacs

= 10 × 60 lacs/100

= 6 lacs

Difference between the expenditure made by university of publication of journals and psychology

= 9-6

= 3 lacs

Innocuous = not harmful or offensive.

Eg. : The conversation was innocuous and she wasn't identified by name.

Loquacious = tending to talk a great deal; talkative.

Deadly = causing or able to cause death.

Eg. : He was acquitted on charges of assault with a deadly weapon.

Ferocious = frightening and violent.

Harmless = not causing or capable of causing harm.

Hence, option C is the correct answer.

The correct word is “both” as communication and interpersonal skills are two skills that are important.

We ordered/asked the watchman should be used as the sense of the given sentence shows order. Hence, there is an error in (1).

Here the passage is asking, ” what can be inferred ? “, i.e. it’s asking about the hidden conclusion of the passage.

Why not Option A ? – because it is clearly given in the passage : “The exact reasons for the formation of this gap are not clear “.

Why not Option B ? – because the passage nowhere talks about the low-lying regions of Kerala and Tamil Nadu near the Palghat Gap.

Why not option D ? – Because the passage only says : “neighbouring regions of Kerala having higher summer temperatures”, it doesn’t say that this high temperature results in rainfall.

Note: We have to think and conclude only from what is given in the passage. We can’t make our own conclusions.

Why Option C ? – because the passage says : ” It results in the neighbouring regions of Tamil Nadu getting more rainfall from the South West monsoon and the neighbouring regions of Kerala having higher summer temperatures”,

Hence we can conclude here that the weather is getting affected due to the Gap.

Therefore, option C is correct.

Given in the question that; V_half = 30(s)

Let us take drawing rate = s

Total volume = 60 S tank

(s¹)(10) – (s)10 = 30s

s¹(s) – s = 3s

s¹ = 4s

s¹ = 4 times the drawing rate.

Vindicated means freed from any question of guilt, so “substantiated” is the closest.

On solving given equation, we get the intersection points as,

y = x² puty = x

x = x²

x²−x = 0

x(x−1) = x = 0,1

Then from y = x

For x = 0 y = 0

& x = 1 y = 1

We can see that curve y = x² and y = x intersects at point (0,0) and (1,1) So, the area bounded by both the curves is

Area is never Negative.

• Shear parameters ‘c’ and ∅ are dependent of water content of soil

• Vane shear test is a useful method of measuring the shear strength of clay. It is a cheaper and quicker method. The test can also be conducted in the laboratory.

• The laboratory vane shear test for the measurement of shear strength of cohesive soils, is useful for soils of low shear strength (less than 0.3kg/{cm}^2) for which triaxial or unconfined tests cannot be performed.

• The test gives the undrained strength of the soil. The undisturbed and remoulded strength obtained are useful for evaluating the sensitivity of soil.

This test is also useful when the soil is soft and its water content is nearer to the liquid limit.

The specific gravity of the manometer fluid is given to be 0.86

ρ = SG(ρH₂O) = (0.86) (1000 kg/m³) =860 kg/m³

Then from Eq. ,

(5)-(7) = Total float = t_e-tL= 0

(11)-(12) = Total Float = t_e-t_L = 0

simplifying and neglecting the products and squares of small quantities,

i.e .δdδL , hence = 2d.δd.L+δL.d²/d²L

= δL/L+ 2.δd/d Bt defintion δL/L

= Longitudinal strain

δdd= hoop strain

Thus Volumetric strain = longitudinal strain + 2 × hoop strain

ϵV = 2ϵh + ϵL

25000/π/4(150)² × 1000

= pD/4tE(5-4μ) ⇒ 1.4147×10^-3

i.e. if Yc < Yn

Then, the flow profile will be M₂.

σ = 3 = (t_p-t₀)/6

t_p-t₀ = 6×3, t_p-t₀ = 18

We know that, t₀ = 8

So,

t_p- 8 =18 And, therefore, t_p = 26

We have, t_e = t₀+4tm+t_p/6

106 m³ of air has 30g SO₂

= 30/64molSO²

Concentration of SO₂ in ppm = 0.0113 ppm

Probability getting head = 1/2 = 0.5.

⇒ f'(x) = 0

⇒ 6x²−6x−36=0

⇒ x²−x−6=0

⇒ x²−3x+2x−6=0

⇒ x(x-3)+2(x-3)=0

⇒ x = 3, -2

∴ f''(x) = 12x-6

⇒ f''(x)<0 @x=-2

So, f(x) has a maximum value at -2 only.

Tension in each bolt

The ratio of tension and shear in each bolt

= Tension in each bolt / Shear in each bolt

= 34.62/14.42 = 2.40

N = 3, ∑x = 6, ∑y = 21, ∑x2 = 14 and ∑xy = 46

There are two normal equations to evaluate a and b of the equation

y = a + bx

∑y = Na + b∑x …(i)

and ∑xy = a∑x + b∑x2 …(ii)

From Eq. (i), 21 = 3a + 6b …(iii)

From Eq. (ii) 46 = 6a + 14b

⇒ 23 = 3a + 7b …(iv)

Now, Eq. (iv) – Eq. (iii),

3a + 7b = 23

3a + 6b = 21

- - -

b = 2

From Eq. (iii), 3a = 21 – 12

3a = 9 ⇒ a = 3

We have from the interaction diagram ∴Mu 0.3 x 20 x 400 x 400² x 10^-3 ⇒ M_u = 384000 kN -mm

e = Mu/Pu = 384000/640 = 600 mm

Mathematically, we can say,

Cost slope = [Crash cost – Normal cost]/[Normal time – Crash time]

In simple words, we can say that the cost slope of an activity is the increase in cost of an activity by reducing the duration by one unit.

So, Cost Slope = (Crash Cost-Normal Cost)/(Normal time - Crash time) = 800-400/8-6 = 200.

Q_u = C_ubN_cA_b + aC_{u av}A_s

Q_u = 500 × FOS = 500×3 = 1500

1500 = 110/2 × 9 × 0.5² + 0.6 × 110/2 × 4 × 0.5 × L

L = 20.8522 m.

Shear resistance of concrete,

V_us = τ_cbd = 0.6 × 250 × 450 = 67.5kN

∴ Total shear capacity = Vus + Vus = 169.58kn

Depth of water = 60cm

Effective rainfall = 15 cm

Depth of irrigating water = 60 – 15 = 45 cm

⇒ Delta = 45 cm = 0.45 m

B = 15 days

From relation, ∆= 8.64×B/D

⇒ Duty(D) = 8.64×15/0.45 = 288ha/cumec

Due to loss of water, duty at head of channel is reduced

Here losses are 15%

So, duty at head of channel = 288 × 85/100

= 244.80 ha/cumecs

Velocity of water (υ) =10 m/s

Cross section of jet, A = 10 mm²

Force on Plate due to jet of water F = ρυ2A

= 1000 kg/m³×(10 m/sec)² × 10 × 10^-6 m²

= 1 N