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All questions of Practice Test: Full Length for Civil Engineering (CE) Exam
In an exam, the average was found to be 50 marks. After deducting computational errors the marks of the 100 candidates had to be changed from 90 to 60 each and average came down to 45 marks. Total number of candidates who took the exam was:- a)150
- b)200
- c)300
- d)600
Correct answer is option 'D'. Can you explain this answer?
a)
150
b)
200
c)
300
d)
600
|
Dipika Ahuja answered |
Let number of candidates be n
50 × n – 100 × 30 = 45 × n
n = 600
50 × n – 100 × 30 = 45 × n
n = 600
6 boys and 6 girls sit in a row at random. Find the probability that 1) The six girls sit together 2) The boys and girls sit alternately.- a)1/462
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
6 boys and 6 girls sit in a row at random. Find the probability that 1) The six girls sit together 2) The boys and girls sit alternately.
a)
1/462
b)
c)
d)
|
Vertex Academy answered |
\Total ways of making sit 12 persons 12!
Considering set of 6 girls as a single group. Within the group girls can be arranged in 6! Ways. Now only among 7 persons—> 6 boys + 1 girl group arrangement needs to be done. Thus, number of ways for the same = 7!.
Therefore, probability = Favourable outcomes/ total number of outcomes = (7!) (6!) / (12!) = 1/1322
We can only make sit boys and girls in alternate manner in two possible ways, boys at even and girls at even. But, among boys group and girls group itself, rearrangement can be done in 6! Ways each.
Therefore, probability =(2)(6!)(6!)/ 12! = 1/462
Considering set of 6 girls as a single group. Within the group girls can be arranged in 6! Ways. Now only among 7 persons—> 6 boys + 1 girl group arrangement needs to be done. Thus, number of ways for the same = 7!.
Therefore, probability = Favourable outcomes/ total number of outcomes = (7!) (6!) / (12!) = 1/1322
We can only make sit boys and girls in alternate manner in two possible ways, boys at even and girls at even. But, among boys group and girls group itself, rearrangement can be done in 6! Ways each.
Therefore, probability =(2)(6!)(6!)/ 12! = 1/462
The value of bearing capacity factor for cohesion, Nc, for piles as per Meyerhof, is taken as :- a) 6.2
- b)12.0
- c) 9.0
- d)5.14
Correct answer is option 'D'. Can you explain this answer?
The value of bearing capacity factor for cohesion, Nc, for piles as per Meyerhof, is taken as :
a)
6.2
b)
12.0
c)
9.0
d)
5.14
|
Sanya Agarwal answered |
Meyehoff bearing capacity factors for cohesive clays (phi=0) are
Nc = 5.14; Nq = 1; Ny = 0
Nc = 5.14; Nq = 1; Ny = 0
Following data refers to the activities of a project, where, node 1 refers to the start and node 5 refers to the end of the project.
The critical path (CP) in the network is- a)1-2-3-5
- b)1-4-3-5
- c)1-2-3-4-5
- d)1-4-5
Correct answer is option 'B'. Can you explain this answer?
Following data refers to the activities of a project, where, node 1 refers to the start and node 5 refers to the end of the project.
The critical path (CP) in the network is
a)
1-2-3-5
b)
1-4-3-5
c)
1-2-3-4-5
d)
1-4-5
|
Ravi Singh answered |
The critical path is the sequence of activities with the longest duration. A delay in any of these activities will result in a delay for the whole project.
Critical path = 1‒4‒3‒5 = 3+3+2 = 8
Note: Critical path is the longest path through a network of activities, which in-turn, determines the shortest time possible to complete all tasks in the network (i.e. the shortest time to complete the project).
The length of a new runway to be constructed under the standard condition is 1710 m. The elevation of airport site is 750 m above the mean sea level. The airport reference temperature and the effective gradient of the runway are 25.7oC and 1.6% respectively. If the runway is to be constructed as per the International Civil Aviation Organization (ICAO) recommendations, then the corrected length of the runway after elevation and temperature correction (in km, up to two correct decimal place) will be ______.
Correct answer is '2307-2310'. Can you explain this answer?
The length of a new runway to be constructed under the standard condition is 1710 m. The elevation of airport site is 750 m above the mean sea level. The airport reference temperature and the effective gradient of the runway are 25.7oC and 1.6% respectively. If the runway is to be constructed as per the International Civil Aviation Organization (ICAO) recommendations, then the corrected length of the runway after elevation and temperature correction (in km, up to two correct decimal place) will be ______.
|
Tanvi Shah answered |
Correction for elevation: 7% increase per 300 m
So, correction = 
= 299.25 m
Corrected length = 1710 + 299.25 = 2009.25 m
Correction for temperature: 1% increase per degree rise in temperature
Standard atmospheric temperature = 15 – 0.0065 × 750 = 10.125°C
Rise of temp = 25.7°C – 10.125°C = 15.575°C
15.575 = 311.43375mCorrect length = 2009.25 + 311.43375 = 2320.684 m
Check for total correction for elevation plus temperature
According to ICAO, the total correction for elevation plus temperature should not exceed by 35%.
Hence, the total correction for elevation plus temperature correction = 
1710 = 598.5m
1710 = 598.5mTherefore, the corrected length of the runway = 1710 + 598.5 = 2308.5 m
Two points are chosen randomly on a line 9 cm long. Determine the probability that the distance between them is less than 3 cm
Correct answer is '0.5-0.6'. Can you explain this answer?
Two points are chosen randomly on a line 9 cm long. Determine the probability that the distance between them is less than 3 cm
|
Lavanya Menon answered |
x: The distance of first point from the start of the line
y: distance of second point from the start of the line segment
x,y ϵ[0,9]
So sample space is Area of region bounded by
x ≥ 0, y ≥ 0, x ≤ 9, y ≤ 9
This is square of side 9
Area = 81 cm2
The region of our interest is
|x-y| < 3
0 ≤ x ≤ 9
0 ≤ y ≤ 9
Area of shaded region = 2 (area of triangle) + area of rectangle
= 45
Probability = 45/81 =0.55
In a particular bank 60% of the customers applied for a loan are rejected. If 4 new loan applications are received, find the probability that exactly 3 of them are accepted.- a)0.1536
- b)0.3456
- c)0.45
- d)0.75
Correct answer is option 'A'. Can you explain this answer?
In a particular bank 60% of the customers applied for a loan are rejected. If 4 new loan applications are received, find the probability that exactly 3 of them are accepted.
a)
0.1536
b)
0.3456
c)
0.45
d)
0.75
|
|
Lavanya Menon answered |
n = 4, p = 0.4, q = 0.6
Applying Binomial theorem:
Applying Binomial theorem:
p(X = 3) = 
p(X = 3) = 
= 0.1536
A river 5 m deep consists of a sand bed with saturated unit weight of 20 kN/m3 . vw = 9.81 kN/m3. The effective vertical stress at 5 m from the top of sand bed is- a)41 kN/m2
- b)51 kN/m2
- c)55 kN/m2
- d)53 kN/m2
Correct answer is option 'B'. Can you explain this answer?
A river 5 m deep consists of a sand bed with saturated unit weight of 20 kN/m3 . vw = 9.81 kN/m3. The effective vertical stress at 5 m from the top of sand bed is
a)
41 kN/m2
b)
51 kN/m2
c)
55 kN/m2
d)
53 kN/m2
|
Bijoy Kapoor answered |
= 5x9.81+5x20=149.05 kn/m3
pore water pressure at 10mm
=9.81x10=98.1 kn/m3
∴ effective verticle stress = 50.95 kn/m3
A transmission tower carries a vertical load of 10000 kg, the vertical stress increment at a depth 6 m directly below the tower as per Boussinesq’s theory is____Pa.
Correct answer is between '1300,1350'. Can you explain this answer?
A transmission tower carries a vertical load of 10000 kg, the vertical stress increment at a depth 6 m directly below the tower as per Boussinesq’s theory is____Pa.
|
|
Tanvi Shah answered |
Concept:
Vertical stress increment at a any depth (z) at a radial distance (r) as per Boussines’q theory is:
Where,
KB = Boussinesq’s influence factor
Calculation:
At r = 0 m,
Q = 10000 × 9.81 = 98100 N
z = 6 m
= log(a+b), then:
Standard 5-day BOD of a waste water sample is nearly x% of the ultimate BOD, where x is- a)48
- b) 58
- c)68
- d)78
Correct answer is option 'C'. Can you explain this answer?
Standard 5-day BOD of a waste water sample is nearly x% of the ultimate BOD, where x is
a)
48
b)
58
c)
68
d)
78
|
|
Ravi Singh answered |
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%
Substituting, the value of k BOD5 = BODu [0.6837]
Therefore, value of x will be 68.37%
Consider the following statements
1) Cambium layer is between sapwood and heartwood
2) Heartwood is otherwise termed as deadwood
3) Timber used for construction is obtained from heartwood
Which of the following is/are correct statements?- a)1, 2 and 3
- b)2 and 3 only
- c) 1 and 3 only
- d) 2 only
Correct answer is option 'B'. Can you explain this answer?
1) Cambium layer is between sapwood and heartwood
2) Heartwood is otherwise termed as deadwood
3) Timber used for construction is obtained from heartwood
Which of the following is/are correct statements?
a)
1, 2 and 3
b)
2 and 3 only
c)
1 and 3 only
d)
2 only
|
Yash Patel answered |
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.
A clayey soil sample in a consolidometer test showed decrease in void ratio from 1.15 to 1.05 when the pressure was increased from 0.30 kgf/cm2 to 0.60 kgf/cm2. If the same soil shows 10% more decrease in void ratio when the pressure was increased from 0.30 kgf/cm2 to 0.80 kgf/cm2, then change in coefficient of consolidation observed during the test will be?Take initial coefficient of consolidation observed from test when the pressure was increased from 0.30 kgf/cm2 to 0.60 kgf/cm2 to be 12 m2/year.- a)4.5 m2/year (decrease)
- b)3.22 m2/year (increase)
- c)12 m2/year (increase)
- d)3.22 m2/year (decrease)
Correct answer is option 'B'. Can you explain this answer?
A clayey soil sample in a consolidometer test showed decrease in void ratio from 1.15 to 1.05 when the pressure was increased from 0.30 kgf/cm2 to 0.60 kgf/cm2. If the same soil shows 10% more decrease in void ratio when the pressure was increased from 0.30 kgf/cm2 to 0.80 kgf/cm2, then change in coefficient of consolidation observed during the test will be?
Take initial coefficient of consolidation observed from test when the pressure was increased from 0.30 kgf/cm2 to 0.60 kgf/cm2 to be 12 m2/year.
a)
4.5 m2/year (decrease)
b)
3.22 m2/year (increase)
c)
12 m2/year (increase)
d)
3.22 m2/year (decrease)
|
|
Sanvi Kapoor answered |
Coefficient of compressibility (av)=
Δe = change in void ratio
increase in effective stressCoefficient of volume compressibity
e0 = initial void ratio
Also,
k = cvmvγw
k = permeability of soil Sample
cv = coefficient of consolidation
mv = coefficient of volume compressibility
Calculation:
Δe1 = 1.15 – 1.05
Δe1 = 0.10
Δσ1 = 0.60 – 0.30
Δσ1 = 0.30 kgf/cm2
= 3.80×10−3cm2/sec k = 3.80 × 10–3 × 0.155 × γw
Now,
Δe2 = 1.10 × 0.10 = 0.11
Δσ2 = 0.80 – 0.30
Δσ2 = 0.50
As permeability of both the sample is Sames,
So,
3.80×10−3×0.155=cv2×0.122
cv2=4.83×10−3cm2/sec
cv2=15.22m2/ year
Change in coefficient of consolidation:
Δcv=cv2−cv1
Δcv = 15.22 – 12
Δcv = 3.22 m2/year
to overtake a vehicle moving at a speed of 30 kmph is : 
Portion BC is semi-circulate. If after several oscillation ball of mass 10 Kg comes to rest. The bending moment at A when ball is at rest will be ? Take g = 9.81 m/s2.- a)1226 N-m
- b)1345 N-m
- c)1468 N-m
- d)1534 N-m
Correct answer is option 'A'. Can you explain this answer?
Portion BC is semi-circulate. If after several oscillation ball of mass 10 Kg comes to rest. The bending moment at A when ball is at rest will be ? Take g = 9.81 m/s2.
a)
1226 N-m
b)
1345 N-m
c)
1468 N-m
d)
1534 N-m
|
|
Lavanya Menon answered |
At rest, ball is at lower most potion of semi-circle.
BM at A = 10g ´ (10 2.5)
= 10 ´ 9.81 ´ 12.5
= 1226.25 N - m
An undisturbed soil sample of thickness 50 mm undergoes 60% consolidation in 28 minutes when double drainage is allowed in the laboratory. If the same clay layer from which the sample was obtained is 3.5 m thick in field conditions, estimate the degree of consolidation occurs when the clay is allowed to consolidate to 600 days with single drainage allowed. Assume the consolidation pressure to be same throughout.- a)47.6%
- b)60%
- c)82.16%
- d)23.4%
Correct answer is option 'D'. Can you explain this answer?
An undisturbed soil sample of thickness 50 mm undergoes 60% consolidation in 28 minutes when double drainage is allowed in the laboratory. If the same clay layer from which the sample was obtained is 3.5 m thick in field conditions, estimate the degree of consolidation occurs when the clay is allowed to consolidate to 600 days with single drainage allowed. Assume the consolidation pressure to be same throughout.
a)
47.6%
b)
60%
c)
82.16%
d)
23.4%
|
Uday Kumar answered |
U1 = 60%
In lab:
H1 = 50 mm
Double drainage
t = 28 minutes
CV1 = 6.310mm2/min
In field:
t2 = 600 days = 600 × 24 × 60
t2 = 864000 minutes
d2 = H2 = 3.5 m = 3500 mm [single drainage]
As consolidation pressure is same
CV1 = CV2 = CV = 6.310mm2/min
TV2 = 0.04450
For U ≤ 60%, TV = 0.2827
So, it is obvious that U < 60%
U2 = 23.4%
“Wanted a two bedroom flat in the court area for immediate possession” - An Advertisement.
Assumptions:
I. Flats are available in the court area
II. Some people will respond to the advertisement
III. It is a practice to give such an advertisement- a)All assumptions are implicit
- b)Only assumption II is implicit
- c) Only I and II are implicit
- d)None is implicit
Correct answer is option 'B'. Can you explain this answer?
Assumptions:
I. Flats are available in the court area
II. Some people will respond to the advertisement
III. It is a practice to give such an advertisement
a)
All assumptions are implicit
b)
Only assumption II is implicit
c)
Only I and II are implicit
d)
None is implicit
|
Gate Funda answered |
The person needs a flat in court area. It may be or may not be possible to have a court area.
The person hopes that he will get the room because if available people will respond to the advertisement.
It may or may not be practice to give such an advertisement and nothing can be said because of insufficient information in the question.
The person hopes that he will get the room because if available people will respond to the advertisement.
It may or may not be practice to give such an advertisement and nothing can be said because of insufficient information in the question.
Plate bearing test with 20 cm diameter plate on soil subgrade yielded a pressure of 1.25 x 105 N/m2 at 0.5 cm deflection. What is the elastic modulus of subgrade ?- a)56.18 x 105 N/m2
- b)22.10 x 105 N/m2
- c)44.25 x 105 N/m2
- d)None.
Correct answer is option 'D'. Can you explain this answer?
Plate bearing test with 20 cm diameter plate on soil subgrade yielded a pressure of 1.25 x 105 N/m2 at 0.5 cm deflection. What is the elastic modulus of subgrade ?
a)
56.18 x 105 N/m2
b)
22.10 x 105 N/m2
c)
44.25 x 105 N/m2
d)
None.
|
Rahul Chatterjee answered |
The maximum permissible deflection for a gantry girder, spanning over 6 m, on which an EOT (electric overhead travelling) crane of capacity 200 kN is operating, is- a)8 mm
- b)10 mm
- c)12 mm
- d)18 mm
Correct answer is option 'A'. Can you explain this answer?
The maximum permissible deflection for a gantry girder, spanning over 6 m, on which an EOT (electric overhead travelling) crane of capacity 200 kN is operating, is
a)
8 mm
b)
10 mm
c)
12 mm
d)
18 mm
|
Rishabh Choudhury answered |
Maximum vertical deflection of Gantry girder for EOT of capacity less than 500KN is ( L/750), where L in mm
For EOT having capacity more then 500KN deflection is (L/1000) and for manually operated crane Deflection is (L/500)
For EOT having capacity more then 500KN deflection is (L/1000) and for manually operated crane Deflection is (L/500)
A trapezoidal section of an open channel has side slope 2H : 1V. If bottom width is 'b’ and depth d’, the relation between b & d for most economical trapezoidal section of the channel is:- a)b = 0.47 d
- b)b/d = 0.5
- c)b = 2d
- d)d2 b = 2
Correct answer is option 'A'. Can you explain this answer?
A trapezoidal section of an open channel has side slope 2H : 1V. If bottom width is 'b’ and depth d’, the relation between b & d for most economical trapezoidal section of the channel is:
a)
b = 0.47 d
b)
b/d = 0.5
c)
b = 2d
d)
d2 b = 2
|
Tanishq Yadav answered |
For most of the economical open channel Trapezoidal section,
Half of top width should be equal to length of the side slope (n:1=H:V)
A square plate 5m × 5m hangs in water from one of its corners and its centroid lies at 8m from the free water surface. Find the total pressure force on the top half of the plate. 
- a)2000 kN
- b)1000 kN
- c) 850 kN
- d) None of these
Correct answer is option 'C'. Can you explain this answer?
a)
2000 kN
b)
1000 kN
c)
850 kN
d)
None of these
|
Rounak Choudhary answered |
Pressure force = Pressure at CG × Area of plate
h is the height of CG from top of water surface.
A reinforced Concrete column of size 450mm × 550mm (overall depth) is provided with area of steel in compression as 4 bars of 25mm diameter. If the effective length of the column is 3.5m, the ultimate load carrying capacity of the column will be _______kN.Use M-20 concrete and Fe-415 steel.
Correct answer is '2500-2520'. Can you explain this answer?
A reinforced Concrete column of size 450mm × 550mm (overall depth) is provided with area of steel in compression as 4 bars of 25mm diameter. If the effective length of the column is 3.5m, the ultimate load carrying capacity of the column will be _______kN.
Use M-20 concrete and Fe-415 steel.
|
Saikat Gupta answered |
Follow step by step method:
(i) check for short or long column
λ = 7.77 < 12
it is a short column
(ii) check for Minimum eccentricity
Minimum eccentricity (ex-x) is greater of
(ii) 20 mm
So, ex-x = 25.33 mm
ex-x < 0.05 × 550
25.33 < 27.5 mm
(O.K)
For ey−y = 
= 22mm
= 22mmey-y < 0.05 × 450
22 mm < 22.5 mm
(O.K)
So, Load carrying capacity of the column will be
Pu = 0.4 fck Ac + 0.67fy Asc
fck = 20 N/mm2
= 1963.50mm2
fy = 415 N/mm2
Pu = 0.4 × 20 × [450 × 550 – 1963.50] + 0.67 × 415 × 1963.50
Pu = 2510.24 kN
A man sells three articles A, B, C and gains 10% on A, 20% on B and loses 10% on C. He breaks even when combined selling prices of A and C are considered, whereas he gains 5% when combined selling prices of B and C are considered. What is his net loss or gain on the sale of all the articles?- a)10% gain
- b)20% gain
- c)10.66% gain
- d)6.66% gain
Correct answer is option 'D'. Can you explain this answer?
A man sells three articles A, B, C and gains 10% on A, 20% on B and loses 10% on C. He breaks even when combined selling prices of A and C are considered, whereas he gains 5% when combined selling prices of B and C are considered. What is his net loss or gain on the sale of all the articles?
a)
10% gain
b)
20% gain
c)
10.66% gain
d)
6.66% gain
|
Parth Patel answered |
Let a, b and c be the cost prices of the three articles A, B and C.
SP = CP + Profit (or) SP = CP – Loss
⇒ SP of A = 1.1a; SP of B = 1.2b; SP of C = 0.9c
By question,
1.1a + 0.9c = a + c ⇒ 0.1a = 0.1c ⇒ a = c
1.2b + 0.9c = 1.05(b + c) ⇒ 0.15b = 0.15c ⇒ b = c = a
Gain% = {(SP – CP)/CP} × 100
⇒ Net gain on the sale of all the articles = 
= 
= ∴ Net gain on the sale of all the articles = 6.66%
The present population of a community is 28000 with an average water demand of 150 Lpcd. The existing water treatment plant has a design capacity of 6000 m3/d. It is expected that the next 20 years. What is the number of years from now when the plant will reach its design capacity assuming an arithmetic rate of population growth??- a) 8.6 years
- b)12.0 years
- c)15.0 years
- d)16.5 years
Correct answer is option 'B'. Can you explain this answer?
The present population of a community is 28000 with an average water demand of 150 Lpcd. The existing water treatment plant has a design capacity of 6000 m3/d. It is expected that the next 20 years. What is the number of years from now when the plant will reach its design capacity assuming an arithmetic rate of population growth??
a)
8.6 years
b)
12.0 years
c)
15.0 years
d)
16.5 years
|
Vandana Mukherjee answered |
A car starts from rest and accelerates at a speed of 35 m/s in time 18 seconds. Find the average acceleration of the car?- a) 2.9 m/s2
- b)3.75 m/s2
- c)7.25 m/s2
- d)1.94 m/s2
Correct answer is option 'D'. Can you explain this answer?
A car starts from rest and accelerates at a speed of 35 m/s in time 18 seconds. Find the average acceleration of the car?
a)
2.9 m/s2
b)
3.75 m/s2
c)
7.25 m/s2
d)
1.94 m/s2
|
Kabir Verma answered |
From the question:
Acceleration of car can be obtained using:
Now:
=(35 m / s)/18s
= 1.944 m/s2
= 1.944 m/s2
As per IS 10500: 2012, the permissible limits in absence of alternate source of water for calcium and magnesium in the drinking water respectively will be (in ppm)- a)75 and 30
- b)45 and 30
- c)200 and 100
- d)250 and 125
Correct answer is option 'C'. Can you explain this answer?
As per IS 10500: 2012, the permissible limits in absence of alternate source of water for calcium and magnesium in the drinking water respectively will be (in ppm)
a)
75 and 30
b)
45 and 30
c)
200 and 100
d)
250 and 125
|
|
Shail Rane answered |
As per IS 10500: 2012,
As per IS 1055: 2012, the permissible limits in absence of alternate source of water for calcium and magnesium in the drinking water respectively will be 200 ppm and 100 ppm.
When a four digit number is divided by 65, it leaves a remainder of 29. If the same number is divided by 13, the remainder would be______
Correct answer is '3'. Can you explain this answer?
When a four digit number is divided by 65, it leaves a remainder of 29. If the same number is divided by 13, the remainder would be______
|
Simran Dasgupta answered |
Let number is N
N = 65 k +29
Now 
∴ Remainder = 3
MPN index is a measure of one of the following :- a)Coliform bacteria
- b)BOD5
- c)Dissolved Oxygen Content
- d)Hardness
Correct answer is option 'A'. Can you explain this answer?
MPN index is a measure of one of the following :
a)
Coliform bacteria
b)
BOD5
c)
Dissolved Oxygen Content
d)
Hardness
|
|
Ravi Singh answered |
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.
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.
On sag (or valley) curves the available sight distance is determined based on :- a) design speed
- b)height of obstacle
- c)height of driver eye
- d)night-time driving conditions
Correct answer is option 'C'. Can you explain this answer?
On sag (or valley) curves the available sight distance is determined based on :
a)
design speed
b)
height of obstacle
c)
height of driver eye
d)
night-time driving conditions
|
Anirban Khanna answered |
Option (C) is Correct Because driver can see the obstacles or vehicle which moves at either side of curvature & reduce the accident.
Pervious sand having air content 15% is used in foundation of a masonry dam. If the maximum permissible upward gradient for a factor of safety of 3 against boiling is 0.315. The percentage air voids in the pervious sand will be ______ %.
Correct answer is '6.2-6.6'. Can you explain this answer?
Pervious sand having air content 15% is used in foundation of a masonry dam. If the maximum permissible upward gradient for a factor of safety of 3 against boiling is 0.315. The percentage air voids in the pervious sand will be ______ %.
|
Jyoti Choudhury answered |
Concept:
F.O.S against boiling is giving by
ier = critical hydraulic gradient
iper = permissible hydraulic gradient
η = percentage air voids
ac = air content
η = porosity
calculation:
icr = iper x FOS
icr = 0.315 × 3
icr = 0.945
e = 0.746
η = 0.427 = 42.7%
ac = 15%
ηa = 0.427 × 0.15
ηa = 0.06405
ηa = 6.405%
The Thiessen weights of 4 raingauges A, B, C and D covering a river basin are 0.15, 0.25, 0.30 and 0.30 respectively. If the average depth of rainfall for the basin is 5 cm, the rainfall recorded at B, C and D are 5 cm, 4 cm and 5 cm respectively. What is the rainfall at A?- a)5 cm
- b)6 cm
- c)7 cm
- d)8 cm
Correct answer is option 'C'. Can you explain this answer?
a)
5 cm
b)
6 cm
c)
7 cm
d)
8 cm
|
|
Sanya Agarwal answered |
The problem of lateral buckling can arise only in those steel beams which have- a)Moment of inertia about the bending axis larger than the other
- b)Moment of inertia about the bending axis smaller than the other
- c) Fully supported compression flange
- d)None of the above.
Correct answer is option 'B'. Can you explain this answer?
The problem of lateral buckling can arise only in those steel beams which have
a)
Moment of inertia about the bending axis larger than the other
b)
Moment of inertia about the bending axis smaller than the other
c)
Fully supported compression flange
d)
None of the above.
|
|
Ravi Singh answered |
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.
A geometrically similar model is made to the scale ratio of 1:14 to study the characterstics of spillway. If the energy dissipated per second in prototype is 15000 N-m, what will be the Energy dissipated per sec in model ________ N – m.
Correct answer is '20-21'. Can you explain this answer?
A geometrically similar model is made to the scale ratio of 1:14 to study the characterstics of spillway. If the energy dissipated per second in prototype is 15000 N-m, what will be the Energy dissipated per sec in model ________ N – m.
|
|
Rashi Shah answered |
Concept: As it is the case of spillway, So Fraude model law is applicable
Pm = Power in model = ρQmgHm
Pp = Power in Prototype = ρQpgHp
Calculation:
Pm = 20.45 N – m per second
Hence power in model will be 20.45 N – m per second
Can you explain the answer of this question below:Man does not live by __________ alone.- A:Food
- B:Bread
- C: Meals
- D:Diet
The answer is B.
|
Shivam Sharma answered |
The answer is B.Physical nourishment is not sufficient for a healthy life; man also has spiritual needs.
Let A be a 3 × 3 matrix such that 
and 
suppose 
Then AQ is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Let A be a 3 × 3 matrix such that and suppose Then AQ is
and suppose Then AQ isa)
b)
c)
d)
|
Gate Gurus answered |
These are in the form of AX = λX
Where λ is an Eigen value of A and X is the Eigen vector of A corresponding to the Eigen value λ.
Now, λ1 = 1, λ2 = 1, λ3 = 1
From the properties of matrix, we can write the matrix A as
A = PDP-1
Where P is the matrix formed by Eigen vectors, D is the diagonal matrix and Eigen values as diagonal elements.
⇒ AP = PD
Here,
⇒ AP = PD = 
From the given question, Q = P
⇒ AP = AQ
For the state of stress (MPa) shown in the figure below. The ratio of Major principal stress to minor principal stress will be _____ (up to two decimal places)
Correct answer is '2.5-2.7'. Can you explain this answer?
For the state of stress (MPa) shown in the figure below. The ratio of Major principal stress to minor principal stress will be _____ (up to two decimal places)
|
|
Lavanya Menon answered |
Concept:
Major (σ1) and Minor (σ2) principal stresses are given by:
Calculation:
σx = 15 MPa
σy = 10 MPa
τxy = 5 MPa
σ1 = 18.10 MPa
σ2 = 12.5 – 5.59
σ2 = 6.91 MPa
In an irrigation project, in a certain year, 70% and 46% of the culturable command area in Kharif and Rabi, respectively, remained without water and the rest of the area got irrigation water. The intensity of irrigation in that year for the project was- a)116%
- b)84%
- c)42%
- d)58%
Correct answer is option 'B'. Can you explain this answer?
In an irrigation project, in a certain year, 70% and 46% of the culturable command area in Kharif and Rabi, respectively, remained without water and the rest of the area got irrigation water. The intensity of irrigation in that year for the project was
a)
116%
b)
84%
c)
42%
d)
58%
|
Akash Rane answered |
Given:
- In a certain year, 70% of the culturable command area remained without water in Kharif season.
- In the same year, 46% of the culturable command area remained without water in Rabi season.
- The rest of the area received irrigation water.
To Find:
The intensity of irrigation in that year for the project.
Solution:
To find the intensity of irrigation, we need to determine the percentage of the culturable command area that received irrigation water.
Step 1: Find the percentage of the culturable command area without water in each season:
- In Kharif season, 70% of the culturable command area remained without water.
- In Rabi season, 46% of the culturable command area remained without water.
Step 2: Find the percentage of the culturable command area with water in each season:
- In Kharif season, the percentage of the culturable command area with water is 100% - 70% = 30%.
- In Rabi season, the percentage of the culturable command area with water is 100% - 46% = 54%.
Step 3: Calculate the overall percentage of the culturable command area with water:
- Since the rest of the area (100% - 30% = 70%) received irrigation water in Kharif season,
the percentage of the culturable command area with water after Kharif season = 30% + 70% = 100%.
- Since the rest of the area (100% - 54% = 46%) received irrigation water in Rabi season,
the percentage of the culturable command area with water after Rabi season = 54% + 46% = 100%.
Step 4: Calculate the intensity of irrigation:
- The intensity of irrigation is the percentage of the culturable command area with water divided by the total area.
- Since 100% of the culturable command area received irrigation water after both Kharif and Rabi seasons,
the intensity of irrigation = 100% / 100% = 1.
- Multiplying by 100 to convert it into a percentage, we get the intensity of irrigation as 100%.
Therefore, the correct answer is option 'B' - 84%.
- In a certain year, 70% of the culturable command area remained without water in Kharif season.
- In the same year, 46% of the culturable command area remained without water in Rabi season.
- The rest of the area received irrigation water.
To Find:
The intensity of irrigation in that year for the project.
Solution:
To find the intensity of irrigation, we need to determine the percentage of the culturable command area that received irrigation water.
Step 1: Find the percentage of the culturable command area without water in each season:
- In Kharif season, 70% of the culturable command area remained without water.
- In Rabi season, 46% of the culturable command area remained without water.
Step 2: Find the percentage of the culturable command area with water in each season:
- In Kharif season, the percentage of the culturable command area with water is 100% - 70% = 30%.
- In Rabi season, the percentage of the culturable command area with water is 100% - 46% = 54%.
Step 3: Calculate the overall percentage of the culturable command area with water:
- Since the rest of the area (100% - 30% = 70%) received irrigation water in Kharif season,
the percentage of the culturable command area with water after Kharif season = 30% + 70% = 100%.
- Since the rest of the area (100% - 54% = 46%) received irrigation water in Rabi season,
the percentage of the culturable command area with water after Rabi season = 54% + 46% = 100%.
Step 4: Calculate the intensity of irrigation:
- The intensity of irrigation is the percentage of the culturable command area with water divided by the total area.
- Since 100% of the culturable command area received irrigation water after both Kharif and Rabi seasons,
the intensity of irrigation = 100% / 100% = 1.
- Multiplying by 100 to convert it into a percentage, we get the intensity of irrigation as 100%.
Therefore, the correct answer is option 'B' - 84%.
, 0
, 0
A capillary rise in a glass tube when immersed in oil having surface tension 0.0825 N/m at 20° C and specific gravity of 0.96 is 6.25 mm. For 30% increase in height of capillarity, the diameter of glass tube should be decreased by ______ (percentage)Take angle of contact between glass tube and oil will be zero
Correct answer is '22.5-23.5'. Can you explain this answer?
A capillary rise in a glass tube when immersed in oil having surface tension 0.0825 N/m at 20° C and specific gravity of 0.96 is 6.25 mm. For 30% increase in height of capillarity, the diameter of glass tube should be decreased by ______ (percentage)
Take angle of contact between glass tube and oil will be zero
|
Avik Chaudhary answered |
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”
hI = 8.125 mm
DI = 4.31 mm
% Decrease 
The dimensions for the flexural rigidity of a beam element in mass (M), length (L) and time (T) is given by:- a)MT-2
- b)M-1T2
- c)ML-1T-2
- d)ML3T-2
Correct answer is option 'D'. Can you explain this answer?
The dimensions for the flexural rigidity of a beam element in mass (M), length (L) and time (T) is given by:
a)
MT-2
b)
M-1T2
c)
ML-1T-2
d)
ML3T-2
|
|
Sanvi Kapoor answered |
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
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
The probability distribution taken to represent the completion time in PERT analysis is:- a)Gamma distribution
- b)Normal distribution
- c)Beta distribution
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
a)
Gamma distribution
b)
Normal distribution
c)
Beta distribution
d)
none of these
|
|
Shivam Sharma answered |
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.
The nominal bearing strength of the bolt is given by 2.5 Kb dt fu, where Kb depends upon:
1) Edge distance
2) Bolt diameter
3) Diameter of bolt hole
4) Ultimate tensile stress of bolt
5) Ultimate tensile stress of plate Of the above- a)1, 3, 4 and 5 are correct
- b)1, 2 and 4 are correct
- c)3, 4 and 5 are correct
- d)2, 3, 4 and 5 are correct
Correct answer is option 'A'. Can you explain this answer?
The nominal bearing strength of the bolt is given by 2.5 Kb dt fu, where Kb depends upon:
1) Edge distance
2) Bolt diameter
3) Diameter of bolt hole
4) Ultimate tensile stress of bolt
5) Ultimate tensile stress of plate Of the above
1) Edge distance
2) Bolt diameter
3) Diameter of bolt hole
4) Ultimate tensile stress of bolt
5) Ultimate tensile stress of plate Of the above
a)
1, 3, 4 and 5 are correct
b)
1, 2 and 4 are correct
c)
3, 4 and 5 are correct
d)
2, 3, 4 and 5 are correct
|
|
Simran Saha answered |
Kb does not depend on bolt diameter.
Consider the following activities:
1. Pouring of concrete
2. Erection of formwork
3. Curing of concrete
4. Removal of formwork
What is the correct sequence on a network of these activities?- a)1-2-3-4
- b)2-1-4-3
- c)2-1-3-4
- d)1-3-2-4
Correct answer is option 'B'. Can you explain this answer?
1. Pouring of concrete
2. Erection of formwork
3. Curing of concrete
4. Removal of formwork
What is the correct sequence on a network of these activities?
a)
1-2-3-4
b)
2-1-4-3
c)
2-1-3-4
d)
1-3-2-4
|
|
Lavanya Menon answered |
Correct sequence of activities is:
a) Erection of formwork
b) Placing of reinforcement
c) Pouring of concrete
d) Compaction of concrete
e) Removal of formwork
f) Curing
a) Erection of formwork
b) Placing of reinforcement
c) Pouring of concrete
d) Compaction of concrete
e) Removal of formwork
f) Curing
at a point (1, -2, -1) in the direction of 
Extreme focus on syllabus and studying for tests has become such a dominant concern of Indian students that they close their minds to anything ___________ to the requirements of the exam.- a)Related
- b)Extraneous
- c)Outside
- d)Useful
Correct answer is option 'B'. Can you explain this answer?
a)
Related
b)
Extraneous
c)
Outside
d)
Useful
|
Ashutosh Kulkarni answered |
Extraneous – irrelevant or unrelated to the subject being dealt with
Chapter doubts & questions for Practice Test: Full Length - GATE Civil Engineering (CE) 2026 Mock Test Series 2025 is part of Civil Engineering (CE) exam preparation. The chapters have been prepared according to the Civil Engineering (CE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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