Practice Test: Civil Engineering (CE)- 1 - Civil Engineering (CE) MCQ

Since the middle term 'cakes' is not distributed even once in the premises, no definite conclusion follows. However, I and II involve only the extreme terms and form a complementary pair. So, either I or II follows.

Man does not live by bread alone.

Extraneous – irrelevant or unrelated to the subject being dealt with

Light cannot be seen by blind.
On the similar logic, Speech cannot be spoken by dumb. Sound cannot be heard by deaf. 

Let the total number of candidates who took the exam be $N$.

Step 1: Compute the total marks before the corrections

The average marks before corrections were $50$. Hence, the total marks before corrections can be written as:

$$\text{Total Marks Before Correction} = 50N$$

Step 2: Compute the adjustment due to the corrections

100 candidates originally scored $90$ marks each, contributing:

$$100 \times 90 = 9000 \, \text{marks}.$$

After the correction, these candidates scored $60$ marks each, contributing:

$$100 \times 60 = 6000 \, \text{marks}.$$

Thus, the total reduction in marks due to the correction is:

$$9000 - 6000 = 3000 \, \text{marks}.$$

Step 3: Compute the total marks after the corrections

The new total marks after corrections are:

$$\text{Total Marks After Correction} = 50N - 3000$$

The new average marks are $45$, so:

$$\text{Total Marks After Correction} = 45N$$

Step 4: Set up the equation and solve for $N$

Equating the two expressions for total marks after corrections:

$$50N - 3000 = 45N$$

Simplify:

$$50N - 45N = 3000$$ $$5N = 3000$$ $$N = 600$$

Final Answer:

The total number of candidates who took the exam was 600.

Number of cubes with 3 faces red = 8
Number of cubes with 2 faces red = 24
Number of cubes with 1 face red = 24
Number of cubes with no face red = 64-(24+24+8) = 8

Continuity equation and momentum equation are used for the hydraulic jump in terms of the conjugate depth and Froude number.
Energy equation is further used to determine the loss of energy in the jump.

Let Amit’s age be n years
Age of his father = (4n+n) = 5n years
After 8 years:
Amit’s age = n+8
Father’s age =
3(n+8) = 5n+8
3n+24 = 5n+8
16 = 2n
n = 8 years
After further 8 years:
Amit’s age = 24 years
Father’s age = 5n+16 = 56 years
So father is 2.33 times older than Amit

Let pipe A, B, C fills the tank in a, b, c minutes respectively.
Pipe A is kept open for t minutes
Pipe B is kept open for 10 minutes
Pipe C is kept open for t-10-2 minutes

t/a+10/b+(t-12)/c=1
t/a=10/b=(t-12)/c since each pipe has equal share of tank
Also, 
t/a+t/b=1
Both pipes fill the tank in time t.
Solving 1st two equations we get 30/b =1  b= 30 minutes
Solving last 2 equations we get 10+t = b   t =20 minutes
Therefore, time taken by C to fill it will be  24 minutes from equation 2.   

Moment Capacity of the Section = Plastic Modulus of Section x Yield Stress of Material
120 x 106 Nmm = 4.8 x 10-4+9 mm³ x Yield stress of Material
Yield stress of Material = 250 N/mm²

MPN- Most Probable Number used to measure Coliform Bacterial
Wrinkler’s Method is used to measure dissolved oxygen and further, which is used to determine BOD5
Hardness is measured in mg/L as CaCO3 via various methods like lime treatment, boiling.

Lower the water cement ratio higher the concrete strength.
A rich concrete due to greater cement content will possess higher strength in comparison to lean cement.
Higher the water content in mix higher will be the shrinkage in concrete.

BOD5 = BODu [ 1-10-k x 5] and the usual value of deoxygenation constant is 0.1
Substituting, the value of k BOD5 = BODu [0.6837] 
Therefore, value of x will be 68.37%

In a beam, the flexural rigidity (EI) varies along the length as a function of x as shown in equation:

Where E is the young's modulus (in Pasual, Pa), I is the second moment of area (in m⁴). Y is the traverse displacement of the beam x and M(x) is the bending moment at x. The SI unit of flexural rigidity is thus Pa. m⁴ or Nm⁴
∴ Dimension is ML³T^–2

A Correct the options by inserting a zero after the decimal in each option.

e- superelevation; f- coefficient of friction = 0.15; v= speed of vehicle in m/s = 60 kmph = 60x0.278 m/s = 16.68 m/s; R = radius of curvature = 128m
ⅇ+f=v²/gR  ⅇ+0.15=〖16.68〗²/(9.81 x 128)  e= 0.0716 

Weighing bucket type

Effective Vertical Stress = Total Vertical Stress – Pore Water Pressure
= γ_(sat ) z - γ_(w ) z = 20x 5 – 9.81 x 5 = 50.95 kN/m²

Q = A×V
Spillway model obeys Froude’s Law

For a droplet;

Minimum Grade of Concrete on Sea Cost:
Plain Cement Concrete – M20
Reinforced Cement Concrete – M30

Order = 2
Degree = 3

The distribution curve for the time taken in to complete each activity of a project resembles a beta-distribution curve and the distribution curve for the time taken to complete entire project ( consisting of several activities) in general resembles a normal distribution curve.

Cambium layer is between inner bark and sapwood. Heartwood is dead wood which does not take part in the growth of tree and is used for construction.

Number of restraints required = 6
Number of cuts required = 3
= 3×3-6 = 3

Lateral buckling occurs when an applied load causes lateral displacement of a member which generally occurs when moment of inertia about the bending axis smaller than the other.

Available moisture holding capacity = 15×0.8 = 12cm
Readily available moisture content = 0.4×12 = 4.8cm
Frequency of irrigation = 48/6 = 8 days