Practice Test: Civil Engineering (CE)- 9 - Civil Engineering (CE) MCQ

In sentences where the structure is "one of + noun + who" is used, the verb after "who" agrees with the noun used after "one of". In the sentence, the noun is "those writers" which is in plural form. Therefore, it must take a plural verb. This makes the use of "has" incorrect in the sentence. The correct plural verb would be "have". The correct sentence would be:
He is one of those writers who have won acclaim the world over.

 Irrespective of the value of ‘x’ (odd or even), x³ -x² - 2x will always be even which means the complete expression x³ - x² - 2x - 19 can NEVER be zero. 

The pair of such numbers are

11/12; 22/23; 33/34; 44/45; 55/56; 66/67; 77/78; 88/89;

Middle number is the average of the numbers on both sides.

Average of 6 and 4 is 5

Average of (7+4) and (2+1) is 7

Average of (1+9+2) and (1+2+1) is 8

Average of (4+1) and (2+3) is 5

Therefore, Average of (3) and (3) is 3

x²-y²= 702 i.e.(x-y)(x+y)= 702

Factories 702 as 702= 2×3×3×3×13 Since x and y are both natural numbers, (x-y)  and (x+y) should be both even or both should be odd. In order to get possible values of x and y we should write 702 as a product of 2 natural numbers which should be a like i.e. both even or both odd. This is not possible since 702 has only 1 even factor leading to (d) as the answer.

Expenditure made by university on research work
= 8% of 60 lacs = 8 × 60 lacs/100 = 4.8 lacs
Expenditure made by university on purchase of books for library
= 6% of 60 lacs = 6 × 60 lacs/100 = 3.6 lacs
Hence ratio between the expenditure made by university on research work and purchase of books for library
= 4.8 : 3.6 = 4:3

Option C is the correct answer.

'Since' is used when a specific time is mentioned.

'For' is used to signify the total time period.

In the question. We have to count 4 digit numbers which are multiples of 11, 13 and 143 (product of 11 and 13).

The first 4 digit multiple of 11, 13 as well as 143 is 1001 = 11×91= 13×77= 143×7

The last 4 digit multiple of 11 is 9999 =11×909 and that of 13 is 9997 =13×769 and for 143 is 9867 (=143×69)

Number of 4 digit multiple of 11, 13 and 143 are (909-91+1), (769-77+1) and (69-7+1)

i.e. 819,693 and 63 respectively.

The number of 4 digit roll numbers which are divisible by 11 and 13 are 819+693-63=1449 Number of 4 digit numbers are from 1000 to 9999 i.e. 9000.

Probability of selection the required roll number = 

5x + y + 2Z = 34
4y - 3z = 12
10x - 2y + z = -4
5, 0, 10
1st Pivoting
10x - 2y + z = -4.......eq1
4y - 3z = 12..............eq2
5x + y + 2z = 34.......eq3
eq3 - (eq1)/2,
10x - 2y + z = -4
4y - 3z = 12
2y + 3Z/2 = 36
4, 2
Hence, Pivot elements are x=10 & y=4

Substituting the coordinates of the straight line
we get the correct option as C which is  

For bacterial concentration in water MPN test is used in which lactose broth is used, In DO test alkali iodide azide is used to get accurate value of dissolved oxygen , in chlorine determination silver nitrate is used as titrating agent and in COD sodium thiosulphate is used as titrating agent.

Let V_w = Velocity of surge wave

Superimpose a velocity V_w to the left to simulate steady flow as in figure

Given, y₂ = 4.0 m, V₂ = 5 m/s and V_w = 10 m/s

By the continuity equation,

⇒ y₁ (V_w - V₁ )= y₂ (V_w - V₂ )

⇒ y₁ (10 - V₁ )= 4(10 - 5)

Strength of one rivet in double shear

Strength of the riveted joint in double shear

= 16 x 86.75 = 1388 kN

Strength of one rivet in bearing = d'×t×σb

=(22+1.5)×16×300×10^-3

=112.8 kN

Strength of rivet joint in bearing

=16 x 112.8 = 1804.8 kN

Strength of plate in tearing = (250-4 x 23.5) x 16 x 150 x10^-3

= 374.4 kN

Thus, the strength of joint will be governed by tearing and it will be equal to 374.4 kN.

Equilibrium of joint C,

Let the compressive force in each of the members OB and OC be P.

Let the tension in OA be R.

Resolving the forces at O vertically, we have

R + 2P cos 60⁰  = W

∴ P = W -R

∴  stain energy stored by the frame

R is determined from the condition of minimum strain energy i.e. by the condition,

Let W be the total weight of the piston and the slab.

Since pressure below the piston = Pressure at the same level in the vertical tube

In pretension, tension is induced in the tendons before concreting.Once the concrete sets and hardens, tendons are cut and prestress is transferred to concrete. Hence after application of prestressing, concrete is in compression.

R_A  = 40+60+80=180 kN
Force P₁ on portion AB = 180 kN (tensile)
Force P₂ on portion BC = 180-40 = 140 kN (tensile)
Force P₃ on portion CD = 80 kN ( tensile)

The three soil parameters are the Effective size of the particle, Uniformity Coefficient and Coefficient of Curvature.

The Particle size corresponding to 10% finer is Effective size. i.e. D₁₀ = 0.16 mm

Uniformity Coefficient, 

Coefficient of Curvature 

Flexibility Matrix, 

Unit force in direction 1,

Using Mohr’s circle

Angularity number= 67-% Solid Volume

Solid Volume= 51%= 0.51 ×3000= 1530 cc

For critical flow condition,

For a circular conduit flowing half full,

Area A = 4 ×2 = 8, Perimeter P = 4 + 4 = 8

R = (A/P) = 1 m

Discharge per unit width q = 8/4 = 2

We know that the minimum specific energy corresponds to the point of critical depth and is given by 1.5 times the critical depth

Critical depth = (q²/g)^(1/3) = 0.74 m

Minimum specific energy is 1.5 × 0.74 = 1.11 m

Specific energy should be same at all points (neglecting hydraulic losses)

So, E = E_minimum + h (this energy is at position of hump)

E = y + v²/(2*g) = 2.05

h = 2.05 – 1.11 = 0.94 m

We know that
C_d = C_c × C_v
C_d = coefficient of discharge
C_c = coefficient of contraction
C_v = coefficient of velocity
Now, C_c = area of venacontracta /area of orifice