Practice Test: Civil Engineering (CE)- 11 - Civil Engineering (CE) MCQ

= In (|(e| (sin sin(|x|)|)=|sin sin(|x|)|

Now for drawing the graph follow these steps

1) Draw sinx graph on paper

2) Take reflection about x-axis, you will get graph of sin(|x|)

3) Now, take a reflection again but this time

about y-axis, you will get the graph of sin(|x|)

⇒ Option c is correct.

Option B: Profanity is socially offensive language, which may also be called bad or offensive language. ‘Profanity of a moment’ makes no sense. Therefore, this option can’t be used.

Option C: Penchant is a strong or habitual liking for something or tendency to do something. ‘Penchant of a moment’ makes no sense. Therefore, this option can’t be used.

Option D: Predilection is a preference or special liking for something; a bias in favour of something. ‘Slurry’ is a thin sloppy mud or cement or, in extended use, any fluid mixture of a pulverized solid with a liquid usually water, often used as a convenient way of handling solids in bulk. Thus, this option makes no sense in the context of the sentence.

Therefore, option a is the apt answer.

Analogy = Situation: Appearance. None of the options accept A showcase any situation. Hence A is the right one. The word moment is present in the question, hence any word which suits situation would be the right option.

So 45x is divisible by 8

The number is 45 which is near to 48, the number which gets divided by 6 (8 × 6 = 48)

Thus x = 6 is the correct answer

% compound interest for 2 year = 10+10+(10×10)/100 = 21%

difference in interest = 225 Rs.

So (21% - 18%) = 3% of sum = 225 Rs.

So Sum = 225×100/3 = 7500 Rs.

Sophisticated= Cultured or one who speaks in a respectable manner

Blusterous = loud and aggressive.

Ribald = indecent or bawdy.

Imperative= essential

Let the selling price of the article =Rs.x

If the article is sold in 1/3 price, then selling price =Rs.x/3 and loss =60%

So, the cost price of the article = Rs.( x/3 ) × (100/40) = Rs. (5x/6) Then, profit earned by selling the article at Rs. x = Rs. x – (5x/6) = Rs.x/6

∴ The required profit percentage = [(x/6)/ (5x/6) × 100]% = 20%.

Alternatively:-

Say CP = 100 Rs. say profit X%

SP = 100 + X

New SP = (100 + X)/3 = 100 - 60

X = 20

Alternative Method: (Easier Method)

Let cost price =100 (Because percentages are mentioned here, and the highest number will be 100)

Loss = 60%

Selling price = 40

Actual selling price = 3 × 40 = 120

Profit = 20 %

It is given that Rs. 200 divided among Arun, Bholu and Chetan

Let Bholu’s share = X Rs.

Arun’s Share = X +30 Rs

Chetan’s Share = X+30+20 = X+50

Now, Arun+ Bholu+ Chetan = (X+30) +(X) +(X+50) = 200

3X = 200 – 80

X = 120/3

X = 40

Bholu’s share is 40 Rs.

Hence, (b) is correct option.

The increase in total marks of the class is (64–46)+(74–47)+(34–43) = 36

The average goes up by 36/24 = 1.5 marks

Increase in %age terms = 1.5 x 100/56 = 2.67%.

So, option d is correct.

Present age of Kanika = 16 years

Age of Kanika one year ago = 16 – 1 = 15 years

Age of Monika after four years = 15 × 2 = 30 years

Present age of Monika = 30 – 4 = 26 years

Sum of the present ages of Kanika, Monika and Shweta = 22 × 3 = 66 years

Present age of Shweta = 66 – 16 – 26 = 24 years

Present average age of Kanika and Shweta = x = (24+16)/2=20 years

So, the value of ‘x’ is 20 years

So option (d) is the correct answer.

S.V.I = 200 ml/2000 mg

= 200 ml/2 gm = 100 ml/gm

By putting the given values in the above equation,

ρ_o^e-2t(-2) + ρ.5+ ρ.5 + ρ. λ = 0

Since ρ = ρ_o ^e-2t

Hence

-2ρ + ρ.5+ ρ.5+ ρ. λ = 0

-2+5+5+ λ = 0

Hence λ = -8

JLD of bending moment at C

For the above problem [a = b = 1/2]

M = ab/l = 1/4

Since length of the span is doubled

l' = 2l

Hence the ordinate of M1 will now be = 0.5 × 2 = 1 m

M = 2M₁

∴ Ordinate of M at mid span = 2 × 1 = 2m

• Ca(OH)₂ is not a desirable product in the concrete mass because it is soluble in water and gets leached easily, particularly in hydraulic structure, that’s why cement with the small percentage of C₃S and more C₂S is recommended for use in the hydraulic structure.

• The concrete continues to harden over several months. Hardening is not a drying process and can very well take place in water. Heat speeds up the setting and hardening of cement

• The stiffening of cement without strength development (flash setting of cement) may cause because of C₃A or C4AF.

So option B statement is incorrect.

When unit load at R is acting in the direction of 50kN load, then reaction at R = 2 (downward)

Alternatively:

Applying ΣM_A = 0

–F_BA × 5 + 10 = 0

F_BA = 2 kN

Similarly, ΣM_D = 0

F_CD × 7 – 4 = 0

F_CD = 3 kN

For horizontal equilibrium of BC

P + FBA – FCD = 0

P = 1 kN

Given load, Q = 12000 kN

Depth, z = 3m

Radial distance, r = 2m

vertical load, q = Q/A

Area of tank, A = πr² = 113.04 m²

q = 12000/113.04 = 106.16 kN/m²

⇒ 150 = 0.44q + 2.36s …(i)

Similarly

Q₂ = 50 kN

⇒ 50 = 0.071q + 0.94s ...(ii)

Solving (i) and (ii)

S = 46.13 kN/m

q = 93.48 kN/m²

Now for footing

Q = 279.6 kN

Thus according to Terzaghi’s equation the net safe bearing capacity will be independent of both width and depth of the footing.

We know that

Bending Moment at min-span, M = (45 × 7.3²)/8

= 299.7 kNm

Stress At top fibre fct

= 5.7 N/mm²

Then

Given F(S) = x(S) = log log (S+5) -log log (5+6)

Differentiate both sides

Multiply by negative sign on both

By property (1)

3x₁ + 2x₂ = c₁

4x₁ + x₂ = c₂ Matrix From is (3 2 4 1 )(x₁ x₂) = [c₁ c₂]

AX = B

Characteristic equations of above systems is

|A - λI| = 0

|3 - λ 2 4 1-λ | = 0

By expanding λ₂ - 4λ - 5 = 0

= 14 + 3 - 2 × 8

= 17 - 16

= 1

Now the area to be irrigated = 364500/0.45 = 810000 = 81hectare

Required area = 81 /0.55 = 147.2 hectares

M_A = (−10 × 2 × 1) + 40 × 2 = 60 k Nm

= 90/mm³

τ = 0N/mm² {point is at top}

So principal stress = 90N/mm² = 90MPa