Practice Test: Civil Engineering (CE)- 7 - Civil Engineering (CE) MCQ

The sentence talks about a past incident and a condition has been mentioned. Thus option 2 fits here correctly. 'Would have' is the correct tense to be used here. Option 1 is illogical. 'Would' must be used here and not 'will' as it is in the past tense. Past perfect tense is incorrect here.

The word that fits here is a noun thus options 3 and 4 are eliminated. The form should be singular as for abstract nouns we do not use the plural form. Option 1 is thus the correct answer.

The word powerful is an adjective and 'empowering' is a verb.

The radius of outer circle = r

∴ Diagonal of square = 2r

Side of square (a)=√2r

Now, the radius of the inner circle = r₁ = 

Average speed = Total distance/Total time

Total distance = a + a + a = 3a

Total time = 

Average speed = 

The sentence mentions a person doing something as per her wishes thus the first word must be a positive reaction. 'Restive' which means 'restless' and 'jinxed' which means 'cursed' are incorrect here. Between options 1 and 3,the word 'elated' which means 'too happy' and the word 'venture' which means 'undertake a risky or daring journey or course of action' fit here correctly. The word 'cease' means 'stop' and does not convey a proper meaning. Thus option 3 is the correct answer.

As, the faster runner is twice as fast as the slowest runner, the faster runner would have complete two rounds by the time the slowest runner completes one round.

Their first meeting takes place after the fastest runner takes 6 min to complete two rounds and the slowest runner completes one round.

Time taken to complete one round by fastest runner = 6/2 = 3 minutes

Rounds needed to complete the race = 1600/400 = 4

Time taken to complete the race = 4 × 3 = 12 min

The young one of a horse is called a foal. Similarly, the young one of a goose is known as gosling. Thus, option 2 contains the pair that best expresses a relationship similar to that expressed in the given pair. In all the other pairs, the first word expresses the male form of the latter.

Since teenagers here have tender age and commit crimes unknown to its consequences, clearly they should not be treated as any criminal. They need to be taught so that they can make difference between right and wrong and do not commit these crimes in future. Hence inference in option 1 is correct.

450 y = n³ implies that 450 y is the cube of an integer. 

When we prime-factorize the cube of an integer, we get 3 (or a multiple of 3) of every prime factor.

8 is the cube of an integer because 8 = 2³ = 2*2*2.

Thus, when we prime-factorize 450 y, we need to get at least 3 of every prime factor. 

Here's the prime factorization of 450 y:

450 y = 2 * 3² * 5² * y 

Since 450 provides only one 2, two 3's, and two 5's, and we need at least 3 of every prime factor, the missing prime factors must be provided by y. Thus, y must provide at at least two more 2's, one more 3, and one more 5. 

Smallest possible case: 

y = 2² * 3 * 5

In this problem, cumulative frequency is given, so converting into frequency

From the table, it is clear that option b, i.e.

60% of the scholars published at least 2 papers is correct.

A matrix is singular if and only if its determinant is zero.

x(3+1)−2(x²+1)=0

−2x²+4x−2 = 0

x²−2x+1 = 0

(x−1)² = 0

x = 1 

According to formula, L(tⁿ f(t)) =

Here, f(t) = sin (2t) ⇒ F(s) = 

∴ L(t²sin (2t)) =  

ii) (4 - x)_x=2 = 2

Hence at x = 2, f(x) is continuous

∴ f(x) is not differentiable.

The given limit is in 0/0 form, So

Applying L hospitals rule

⇒ log b - 1 = -1 - log b

⇒ 2 log b = 0

⇒ b = 1

y log y dx + (x – log y) dy = 0

It is a Leibnitz’s equation in x.

Integrating factor = 

In context of the Real and conjugate Beam:

(i) The fixed end of the real beam will become the free end of the conjugate beam

(ii) The diagram of the Real beam will be loading diagram for the conjugate beam

(iii) Hogging moment diagram (-ve) in the real beam will give downward loading in the conjugate beam

As end A is fixed in the real beam, so it will be free in conjugate beam. Similarly for end B in real beam will become fixed end in conjugate beam. The  diagram of the real beam will become loading diagram for the conjugate beam. 

Concept:

Using the relations/ formula

At surface y = y_max , ε = ε_max

ε_surface = 

Calculation:

ε_surface = 0.00084

L = 250 mm

y_max  = = 0.7mm

ε_surface =

ε_surface =

ε_surface = 0.00084

0.00084 = 

δ = 9.375 mm

As per IS 456 : 2000,

(i) The minimum stripping time for props to slab spanning up to 4.5 m and spanning over 4.5 m is 7 days and 14 days respectively.

(ii) The minimum stripping time for props to aches spanning up to 6 m and spanning over 6 m is 14 days and 21 days respectively.

(iii) When bar of two different diameter are to be spliced the lap length shall be calculated on the basis of bar of larger diameter.

(iv) A random sampling procedure shall be adopted to ensure that each concrete batch shall have a reasonable chance of being tested that is, the sampling should be spread over the entire period of concreting and cover all mixing units.

Member AB and AG are co-linear,

So F_AB = F_AG = 0 kN

Now removing the member AB and AG. Now member BC and BG are also collinear. So, F_BC = F_BG = 0 kN

Consider Joint G

∑F_x = 0

F_GC Sin 63.43° = 0

F_GC = 0 kN

∑F_V = 0

20 + F_GF + F_GC Cos 63.43° = 0

F_GF = - 20 kN 

Concept:

Tensile strength of a section, (T_dn)

where A_n = net section area, mm²

f_u = ultimate strength of the material

γ_m = pantial safety factor.

Calculation:

For Fe 410 grade of steel

f_u = 410 N/mm², f_y = 250 N/mm²

γ_m = 1.25

T_dn = 826.56 kN.

Concept:

Vertical stress increment is given by:

r = radius of circular area

z = depth of soil stata

q = uniformly distributed circular load

Calculation:

z = 6 m

q = 200 kN/m²

σ_z = 29.2 kN/m²

Concept:

Using Peck, Hanson equation:

The net allowable bearing pressure (kN/m²) is given by

q_a(net) = 0.41 N S C_W

N = Final corrected SPT Number

C_w = Water table correction factor

S = Permissible Settlement of foundation (mm)

Calculation:

N = 28

S = 40 mm

C_w = 0.8

q_a(net) = 0.41 × 28 × 40 × 0.8

q_a(net) = 367.36 kN/m²

Consistency Index (I_C) of the soil is:

I_P = W_L – W_P

W_n = 30%

W_L = 55%

W_P = 25%

Liquidity Index (I_L) will be:

W_n = 30%

W_P = 25%

W_L = 55%

As the soil is in Plastic stage of consistency because both liquidity index and consistency index lies in between 0 to 1.

Concept:

Height of capillary rise is given by:

Where,

σ = Surface tension of fluid at 20° C

ρ = Density of fluid (kg/m³)

D = Diameter of the glass tube

h = height of capillary rise/fall

Calculation:

σ = 0.0825 N/m

θ = 0°

ρ = 1000 × 9.6 = 960 kg/m³

Diameter (D) = 5.60 mm

For 30% increase in “h”

h_I = 8.125 mm

D_I = 4.31 mm

% Decrease 

As per IS10500:2012, the total concentration of manganese (as Mn) and iron (as Fe) for drinking water purpose shall not exceed 0.3 mg/L.

Concept:

Carbonate hardness = minimum of {Alkalinity, Total Hardness}

Total hardness = Carbonate Hardness + Non-carbonate Hardness.

Calculation:

Carbonate Hardness= minimum of {200 mg/L , 250 mg/L}

Carbonate Hardness = 200 mg/L

Non – Carbonate Hardness = Total Hardness – Carbonate Hardness

= 250 – 200

= 50 mg/L

Concept:

Ultimate BOD = L_o

BOD_5­ = 5day BOD, of water/waste water

BOD at any time ‘t’ (L_t)

L_t = L_o(1−e^−kDt) 

Where L_o = ultimate BOD

K_D = Rate constant

Calculation:

L_t = 300 × (1 – e^{-0.3585 × t})

L₅ = BOD₅ = 300 × (1 – e^{-0.3585 × 5})

= 250 mg/L

Capacity of different types of roads in rural areas.

Concept:

We know that by Martin’s formula for a Broad Gauge track in a transition curve. For speed greater than 100 kmph.

Speed. 

Where V is speed in kmph

R is the radius of curvature in m.

Calculation:

Bearing of PQ = 60°

Bearing of RQ = N50° W (WCB)

Bearing of RQ = 360° - 50° = 310° (QBS)

Bearing of line QR = Bearing (RQ) – 180°

Bearing of line QR = 310° - 180° = 130°

Bearing of line QP = 180° + Bearing of PQ = 180° + 60° = 240°

∠PQR = (Bearing)_QP = - (Bearing)_QR

∠PQR = 240° - 130°

∠PQR = 110°