Grand Test - 2 - PDF Flipbook
Grand Test - 2
368 Views
15 Downloads
PDF 11,437,271 Bytes
GATE
CIVIL
GrandTests
Test-02Solutions
APTITUDE
One Mark Questions
1. Which of the following options is the closest in meaning to the
word below:
Inexplicable
a) Incomprehensible
b) Indelible
c) Inextricable
d) Infallible
Answer: a
Solution:
Inexplicable means which cannot be explained or cannot be
understood. So clearly the choice is ‘A’ i.e., incomprehensible.
Indelible means which cannot erased/removed.
Inextricable means which cannot be solved infallible means
incapable of error.
2. They were requested not to quarrel with others.
Which one of the following options is the closest in meaning to
the word quarrel?
a) Make out
b) Call out
c) Dig out
d) Fall out
Answer: d
Solution:
1
The meaning of quarrel is fall out which means to cut off
relations over a quarrel.
(a) Make out means to discover the meaning of, to see, to
hear, etc. clearly.
(b) Call out means to summon someone to leave his
house to deal with a situation outside.
(c) Dig out means to find something that has been hidden
or forgotten for a long time or unearth.
(d) Fall out means to quarrel
3. "India is a country of rich heritage and cultural diversity".
Which one of the following facts best supports the claim made in
the above sentence?
a) India is a union of 29 states and 7 union territories.
b) India has a population of over 1.1 billion.
c) India is home to 22 official languages and thousands of
dialects.
d) The Indian cricket team draws players from over ten states.
Answer: c
Solution:
(a) Out of scope
(b) Out of scope
(c) This statement supports the claim made in the above
statement. ‘Rich heritage and cultural diversity has
influence over languages and thousands of dialects’.
(d) Unwarranted (out scope)
2
4. If y: 5x2 + 3, then the tangent at x = 0, y = 3
a) passes through x = 0, y = 0
b) has a slope of +1
c) is parallel to the x - axis
d) has a slope of – 1
Answer: c
Solution:
y = 5x2 + 3
= 10x
x=0
∴ There is no slope, it is parallel to the x- axis.
5. Four cards are randomly selected from a pack of 52 cards. If the
first two cards are kings, what is the probability that the third card
is a king?
a) 4/52
b) 2/50
c) (1/52) × (1/52)
d) (1/52) × (1/51) × (1/50)
Answer: b
Solution:
Probability of an event A, denoted as p(A), is defined as
P(A) = Number of cases favorable to A
Number of possible outcomes
⟶ Number of cases favorable to A = two kings only from a pack
of 52
3
⟶ Number of possible outcomes = 50
∴ The probability that the third card is a king p(A) = 2
50
4
Two Marks Questions
6. Wanted Temporary, Part-time persons for the post of Field
Interviewer to conduct personal interviews to collect and collate
economic data. Requirements: High School-pass, must be
available for Day, Evening and Saturday work. Transportation
paid, expenses reimbursed.
Which one of the following is the best inference from the above
advertisement?
a) Gender-discriminatory
b) Xenophobic
c) Not designed to make the post attractive
d) Not gender-discriminatory
Answer: d
Solution:
The question is about inference which means the reasoning
involved in drawing a conclusion or making a logical judgment
on the basis of circumstantial evidence.
(a) There is no mention of gender in the given advt. so no
question of gender discrimination.
(b) Xenophobic means fear of foreigners.
(c) Day, Evening and Saturday work indicate that it is
designed to make attractive.
(d) Correct: The Advt. is not talking about gender.
5
7. The number of people diagnosed with dengue fever (contracted
from the bite of a mosquito) in north India is twice the number
diagnosed last year. Municipal authorities have concluded that
measures to control the mosquito population have failed in this
region.
Which one of the following statements, if true, does not contradict
this conclusion?
a) A high proportion of the affected population has returned from
neighboring countries where dengue is prevalent
b) More cases of dengue are now reported because of an increase
in the Municipal Office's administrative efficiency
c) Many more cases of dengue are being diagnosed this year since
the introduction of a new and effective diagnostic test
d) The number of people with malarial fever (also contracted
from mosquito bites) has increased this year
Answer: d
Solution:
It is a strengthen the argument question
(a) Unwarranted (out of scope) it doesn’t support the
conclusion. The Passage doesn’t mention ‘neighbouring
countries’
(b) Contradicts the conclusion.
(c) Contradicts the conclusion.
(d) Correct: This is the right answer as it supports the
conclusion. Municipal authorities have concluded that
6
measures to control the mosquito population have failed
in this region.
8. Alexander turned his attention towards India. since he had
conquered Persia. Which one of the statements below is logically
valid and can be inferred from the above sentence?
a) Alexander would not have turned his attention towards India
had he not conquered Persia.
b) Alexander was not ready to rest on his laurels and wanted to
march to India.
c) Alexander was completely in control of his army and could
command it to move towards India.
d) Since Alexander's kingdom extended to Indian borders after
the conquest of Persia, he was keen to move further.
Answer: a
Solution:
The keyword in the statement ‘since’ which in this context means
‘because’
9. Indian currency notes show the denomination indicated in at least
seventeen languages. If this is not an indication of the nation's
diversity, nothing else is.
Which of the following can be logically inferred from the above
sentences?
a) India is a country of exactly seventeen languages.
b) Linguistic pluralism is the only indicator of a nation's
diversity.
7
c) Indian currency notes have sufficient space for all the Indian
languages.
d) Linguistic pluralism is strong evidence of India's diversity
Answer: d
Solution:
If seventeen languages were not an indication of the nation’s
diversity, nothing else is. If nothing else is so the best inference
is option ‘D’.
10. 5 Skilled workers can build a wall in 20 days, 8 semi-skilled
workers can build a wall in 25 days; 10 unskilled workers can
build a wall in 30 days. If a team has 2 skilled 6 semi-skilled and
5 unskilled workers, how long will it take to build the wall?
a) 20 days
b) 18 days
c) 16 days
d) 15 days
Answer: d
Solution:
5 Skilled workers can build a will in 20 days 1 skilled
workers 5 × 20 = 100 days
1-day work of skilled worker = 1
100
8 semi-skilled workers can build-wall = 25
1 semi-skilled worker = 8 × 25= 200 days
1-day work of semi-skilled worker = 1
200
8
10 unskilled workers can build a wall = 30 days
1 unskilled workers = 10 × 30 = 300 days
1-day work of unskilled worker = 1
300
∴ 2 skilled + 6 semiskilled + 5 unskilled
= 2�1100� + 6 �2100� + 5 �3100�
= 1 + 3 + 1 = 6+9+5 = 1
50 100 60 300 15
∴ 2 skilled + 6 semiskilled + 5 unskilled can build a wall = 15 days
9
TECHNICAL
One Mark Questions
11. l i →m0 sin , where m is an integer, is one of the following:
a) m
b) mπ
c) mθ
d) 1
Answer: a
Solution:
Standard limit formula, l i →m0 sin =
12. Which one of the following is Not associated with vector
calculus?
a) Stoke's theorem
b) Gauss Divergence theorem
c) Green’s theorem
d) Kennedy’s theorem
Answer: d
Solution:
Kennedy’s theorem
13. In a manufacturing plant, the probability of making a defective
bolt is 0.1. The mean and standard deviation of defective bolts in
a total of 900 bolts are respectively
a) 90 and 9
b) 9 and 90
10
c) 81 and 9
d) 9 and 81
Answer: a
Solution:
p = 0.1, n = 900, 1 – p = 0.9
mean = np = 90
S.D = σ = �npq = 9
14. A single die is thrown two times. What is the probability that the
sum is neither 8 nor 9?
a) 1
9
b) 5
36
c) 1
4
d) 3
4
Answer: d
Solution:
Let E be the event of getting the sum 8 or 9
⟹ n(E) = 9 ⟹ p(E) = 9 = 1
36 4
Required probability = 1 – p(E) = 3
4
15. 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
population will increase to 48000 during the next 20 years. What
is the number of years from now when the plant will reach its
11
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
Answer: b
Solution:
Population after n years,
Pn = P0 + n ̅
̅ is arithmetic design growth rate
Given P0 = 28000
For n = 20; P20 = 48000
∴ ̅ = 48000 − 28000 = 1000 per year
20
The population when design capacity will be required,
Pn = 6000 × 1000 = 40,000
150
∴ Number of years to reach the plant at design capacity,
n = − 0 = 4000 × 28000 = 12.0 years
̅ 1000
16. Coefficient of permeability of an underground stratum is 0.001
m/s. Discharge obtained from a well of area 20 m2 dug into this
stratum (with drawdown of 2 m) will be
a) 2400 lpm
b) 2000 lpm
c) 1200 lpm
12
d) 1000 lpm
Answer: c
Solution:
Discharge is given by
q = kiA = 0.001 × 1 × 20
(Assuming L = 1 m)
= 0.02 m3/sec
= 1200 lpm
17. Two reservoirs at different levels are connected by two parallel
pipes of diameter '2d' and 'd'. The ratio if the flows in the two
pipes (larger: smaller) is
a) √2:1
b) 2:1
c) 4:1
d) 4√2:1
Answer: d
Solution:
ℎ = L V2 = 16f L Q2
D 2g π2 D5 2g
Since the pipes are connected in parallel, the head loss will be
same for both pipes. The length of the two pipes is same also.
∴ 1 = � 21�5/2 = �2 �5/2 = 4√2
2
13
18. The product of H+ ions and OH– ions in a strong alkali at 25oC
is
a) 0
b) 1
c) 10 –1
d) 10 –14
Answer: d
Solution:
= [H+][OH–] = 10– 14 at 25oC
19. The ratio of normal stress to volumetric strain in defined as
a) Youngs modulus
b) Bulk modulus
c) Rigidity modulus
d) Tangent modulus
Answer: b
Solution:
Young’s modulus = Tensile Stress
Tensile Strain
Rigidity modulus = Shear Stress
Shear Strain
Tangent modulus = Slope of the stress-strain curve at any
specified stress or strain.
20. The delta for a crop having base period 120 days is 70 cm. What
is the duty?
a) 243 hectare/cumec
b) 1481 hectare/cumec
14
c) 148 hectare/cumec
d) 1.481 hectare/cumec
Answer: b
Solution:
Duty = 864 ×120 = 1481 ha/cumec
70
21. During a particular stage of the growth of a crop, consumptive
use of water is 2.8 mm/day. If the amount of water available in
the soil is 25% of 80 mm depth of water, what is the frequency of
irrigation?
a) 9 days
b) 13 days
c) 21 days
d) 25 days
Answer: c
Solution:
Water available to the soil = 25 × 80 = 20 mm
100
Additional water required = 80 – 20 = 60 mm
∴ Frequency of irrigation = 60 = 21.43 days
2.8
Thus, the crop should be irrigated after every 21 days.
22. The material in which large deformation is possible before the
absolute failure or rupture is termed as
a) Brittle
b) Elastic
c) Ductile
15
d) Plastic
Answer: c
23. Shrinkage deflection in case of rectangular beams and slabs can
be eliminated by putting
a) Compression steel equal to tensile steel
b) Compression steel more than tensile steel
c) Compression steel less than tensile steel
d) Compression steel 25% greater than tensile steel
Answer: a
Solution:
The bond between concrete and steel restrains shrinkage.
Therefore, the shrinkage curvature is influenced by the area of
tension and compression reinforcement equally.
The shrinkage curvature
ϕsh = βesh/D
esh ⟶ free shrinkage = 0.0003
D ⟶ Overall depth of beam
Β = 0.72Pt�−PtPc ≤ 1.0
for 0.25 ≤ Pt − Pc ≤ 1
= 0.65Pt�−PtPc ≤ 1.0
for Pt − Pc ≥ 1
Pc, Pt ⟶ percentage of compression and tension steel respectively
Thus, shrinkage deflection can be eliminated by making
compression steel equal to tension steel.
16
24. The number of observations required of an operation to produce
results having a specified accuracy
a) Varies inversely with the square of the confidence interval
b) Varies directly with the square of the confidence interval
c) Increases with the possible magnitude of the average expected
value
d) Increases with the range of observed values, namely, the
difference between the maximum and the minimum values of
the observations.
Answer: a
Solution:
Probable error of single observation of unit weight
ES = ± 0.6745�(∑ − 12)
Where
‘e’ is residual error
‘n’ is number of observations
So, number of observations (n)vary inversely with
square of confidence interval (Es)
n = �0.6 7 45�2 ∑ 2 + 1
25. A scale of 1 inch = 50 ft. is mentioned on an old map. What is
the corresponding equivalent scale?
a) 1 cm = 5m
b) 1cm = 6m
c) 1cm = 10m
17
d) 1cm = 12m
Answer: b
Solution:
1 inch = 2.54 cm
1 feet = 0.3048 m
Given that 1 inch = 50 ft.
⟹ 2.54 cm = 50 × 0.3048 m
⟹ 1 cm = 15.24 m
2.54
⟹ 1 cm = 6 m
26. The side of a square land was measured as 150m and is in error
by 0.05 m. What is the corresponding error in the computed area
of the land?
a) 5 m2
b) 10 m2
c) 15 m2
d) 20 m2
Answer: c
Solution:
A = 2
A = 2a. A
= 2 × 150 × 0.05
= 15 m2
18
27. The dry density of a soil is 1.5 g/cc. If the saturation water
content was 50% then its saturated density and submerged density
would, respectively by
a) 1.5 g/cc and 1.0 g/cc
b) 2.0 g/cc and 1.0 g/cc
c) 2.25 g/cc and 1.25 g/cc
d) 2.50 g/cc and 1.50 g/cc
Answer: c
Solution:
= (1 + w) = 1.5 × 1.5
= 2.25 g/cc
= − = 2.25 – 1.0 = 1.25 g/cc
28. The length of National Highways as per 3rd 20-year (Lucknow)
road plan is given by
a) area of the country/75
b) area of the country/50
c) area of the country/40
d) area of the country/25
Answer: b
Solution:
1) Length of National Highways (NH)
= Area of the country/5O
2) Length of State Highways (SH)
= Area of the state/25
= 62.5 × number of towns in the state – Area of the state/50
19
3) Length of the Major District Roads (MDR)
= Area of the State/12.5
= 90 × number of towns in the state
4) Total length of all categories of roads in a
NH + SH + MDR + ODR + VR
= Area of District × 0.82
29. In which one of the following grades of a highway is an
emergency escape ramp provided?
a) 1 in 200
b) Zero grade
c) Down grade
d) Up grade
Answer: c
Solution:
The provision of emergency escape ramp on long, descending
grades is appropriate for slowing or stopping out-of-control
vehicles away from the main stream of traffic. Loss of breaking
ability through overheating or mechanical failure results in the
driver loosing the control of vehicles. Four basic types of
emergency escape ramps are commonly used. They are:
1. Sandpile
2. Descending grade
3. Horizontal grade
4. Ascending grade
20
30. It a vehicle travelling at 40 kmph was stopped within 1.8 sec
after the application of the brakes, then the average skid resistance
coefficient is ______
a) 0.63
b) 0.73
c) 0.83
d) 0.93
Answer: a
Solution:
Data given, u = 40 kmph
V = 0, t = 1.8 sec
∴ V = u – at
u = at
a = = 40 × 185 = 6.1728 m/sec2
1.8
Average skid resistance,
= 6.1728 = 0.6292
9.81
≃ 0.63
31. What is the significant purpose of monitoring a project through
its implementation phase?
a) To fix responsibility for delays
b) To rerail the project with control over cost overrun
c) To rerail the project with minimum time overrun
d) To rerail the project with optimal time and cost over-run
Answer: c
21
Solution:
Controlling and monitoring is done to complete the project by
the stipulated data.
32. The flow net of the activities of a project are shown in the
network given below indicating the duration of the activities
along their arrows.
The critical path of the activities is along
a) 1 → 2 → 4 → 5 → 7 → 8
b) 1 → 2 → 3 → 6 → 7 → 8
c) 1 → 2 → 3 → 5 → 7 → 8
d) 1 → 2 → 4 → 5 → 3 → 6 → 7 → 8
Answer: c
Solution:
Path Time taken
1 - 2 - 4 - 5 - 7 - 8 7 + 3 + 4 + 8 + 7 = 29
1 - 2 - 3 - 5 - 7 - 8 7 + 8 + 0 + 8 + 7 = 30
The path 1 - 2 - 4 - 5 - 3 - 6 - 7 - 8 is not possible. Hence, critical
path is 1 - 2 - 3 - 5 - 7 - 8.
22
33. Three weights are hung from a level surface and a force parallel
with the surface is applied as shown in the figure below. What is
the magnitude of the resultant force?
a) 3
b) √18
c) 6
d) √45
Answer: d
Solution:
Resultant force is the result of adding the forces, which can be
done using their vector representations as shown.
∴ 2 = 62 + 32
⟹ = √45
f4
f1 3
6 f2 fr
f3
23
34. For two plates of equal thickness, full strength of fillet weld can
be ensured if its maximum size, for square edge, is limited to
a) 1.5 mm less than the thickness
b) 75% of the thickness
c) 80% of the thickness
d) Thickness of the plate
Answer: a
Solution:
The maximum size of a fillet weld is obtained by subtracting 1.5
mm from the thickness of the thinner member to be jointed. This
specification limits the size of the fillet weld so that total strength
may be developed without overstressing the adjacent metal.
35. The type of stress induced in the foundation bolts fixing a
column to its footing is
a) Pure compression
b) Bearing
c) Pure tension
d) Bending
Answer: c
24
Two Marks Questions
0 0
36. The Eigen vector (s) of the matrix �0 0 0� , ≠ 0 is (are)
000
a) (0, 0, α)
b) (α, 1, 0)
c) (0, 0, 1)
d) (0, α, 0)
Answer: d
Solution:
0 0
Given matrix is = �0 0 0�
000
Eigen values:
Given matrix is an upper triangular matrix. Therefore,
diagonal elements of A are eigen values of A.
i.e. = 0, 0, 0.
Eigen vectors:
Consider (A- )X=0 1 0
0 � � 2� = �0� − − − −(1)
0 − 0 0 − 3 0
⇒ � 0 0 −
00
Put = 0 (1) 1 0
0� � 2� = �0� − − − −(2)
00 0 3 0
. . �0 0
00
From (2), we have
3 = 0
25
3 = 0
Therefore, any non-zero vector with z-component as zero is an
eigen vector. Hence in the given options (b) and (d) are correct.
2 1 and = �03 21� then × is
37. Given matrix = �3 2�
5
4
81
a) �13 2�
22 5
65
b) � 9 8 �
12 13
18
c) �2 13�
5 22
62
d) �9 4�
05
Answer: b
Solution:
2 1 �30 21� = 6 5
= �3 2� �9 8�
5 12 13
4
38. Changing the order of integration in the double integral =
∫08 ∫ 2 �4 ( , ) leads to = ∫ ∫ ( , ) . What is
q?
a) 4y
b) 16y2
c) x
d) 8
26
Answer: a
Solution:
Given limits are = 0 = 28�
= =
4
Variable limits for y Changing the limits
x = 0 to x = 4y
y = 0 to y = 2
∴ = ∫02 ∫04 ( , )
39. The line integral ∫ . of the vector function V(r) = 2 ̅ +
2 ̅ + 2 � from the origin to the point P (1, 1, 1)
a) is 1
b) is zero
c) is – 1
d) cannot be determined without specifying the path
Answer: a
Solution:
V(r) = 2xyz ̅ + x2z ̅ + x2y � ……. (1)
Curl V� = 0� ⟹ V� is irrotational
There exists a scalar function ϕ (x, y, z) such that
V� = ∇ϕ = ϕ + ϕ + ϕ …….. (2)
27
⟹ ϕ = x2yz (from 1 & 2)
∫((01,0,1,0,1)) V. dr̅ = ∫((01,0,1,0,1)) ∇ϕ. dr̅ = ∫((01,0,1,0,1)) dϕ
= ϕ(1, 1, 1) – ϕ(0, 0, 0) = 1
40. If f(x, y, z) = ( 2 + 2 + 2)−1�2, 2 + 2 + 2 is equal to
2 2 2
___
a) 0
b) 1
c) 2
d) −3( 2 + 2 + 2)−5�2
Answer: a
Solution:
2 = 2 + 2 + 2, = , = , =
= 1 ⇒ = − 1 . = −
2 3
⇒ = − � 1 3 + . −3 . � = −1 + 3 2
4 3 5
∴ + + = −3 + 3( 2+ 2+ 2) = 0
3 5
41. If tomato juice is having a pH of 4.1, the hydrogen ion
concentration will be
a) 10.94 x 10-5 mol/l
b) 9.94 x 10-5 mol/l
c) 8.94 x 10-5 mol/l
d) 7.94 x 10-5 mol/l
Answer: d
28
Solution:
PH = log10 �H1+�
4.1 = log10 �H1+�
H+ = 7.94 × 10– 5 mol/l
42. A portion of waste water sample was subjected to standard BOD
test (5 days, 20oC), yielding a value of l80 mg/l. The reaction rate
constant (to the base ‘e’) at 20oC was taken as 0.18 per day. The
reaction rate constant at other temperature may be estimated by
kT = k20 (1.047) T-20. The temperature at which the other portion
of the sample should be tested, to exert the same BOD in 2.5 days,
is
a) 4.9oC
b) 24.9oC
c) 31.7oC
d) 35.0oC
Answer: d
Solution:
520° = 180 mg/l K20 = 0.18d– 1
y2T.°5C = 180 ⟹ ToC =?
520° = y2T.5 = 180
520° = Lo[1 – e– kt]
Lo = [1 Y520° = [1 YT2.°5
– e– kt] – e– kt]
180 = Lo�1 − − 20 ×5� = Lo�1 − − T ×2.5�
29
K20 × 5 = KT × 2.5
0.18 × 5 = K20(1.047)T – 20 × 2.5
0.18 × 5 = 0.18[1.047]T – 20 × 2.5
[1.047]T – 20 = 5 = 2
2.5
(T – 20)ln[1.047] = ln(2)
⟹ T – 20 = 15.09
T = 15.09 + 20 = 35.09oC = 35oC
43. Consider a primary sedimentation tank (PST) in a water
treatment plant with Surface Overflow Rate (SOR) of 40 m3/m2/d.
The diameter of the spherical particle which will have 90 percent
theoretical removal efficiency in this tank is _______µm. Assume
that settling velocity of the particles in water is described by
Stokes’s Law.
Given: Density of water = 1000 kg/m3; density of particle = 2650
kg/m3; g = 9.81 m/s2; Kinematic viscosity of water (v) = 1.10 x
10-6 m2/s
Answer: 22.576
Solution:
η = Vs × 100
Vo
Vo = 40 m3/m2/day = 40 m/day
= 4.629 × 10– 4 m/sec
η = Vs × 100
Vo
90 = Vs 10−4 × 100
4.629 ×
30
Vs = 4.166 × 10– 4 m/sec
Vs = [ −1] 2
18
→ 4.166 × 10– 4 = 9.81[2.65−1] 2
18 ×1.1 × 10−6
d = 22.576 × 10– 6 m
= 22.576 m
44. A water treatment plant of capacity, 1 m3/s has filter boxes of
dimensions 6 m x 10 m. Loading rate to the filters is 120
m3/day/m2. When two of the filters are out of service for back
washing, the loading rate (in rn3/day/m2) is _______
Answer: 144
Solution:
Q = 1 m3/sec
Size of the each filter = 6 × 10
Area of the each filter = 60 m2
Loading rate = 120 m3/day/m2
Total area of filter = Q rate
loading
Total area of filter = 1 ×24 ×60 ×60 = 720 m2
120
No. of filters required = Total area
Area of each filter
= 720 = 12
60
No. of filters in operation = 12 – 2 = 10
Loading rate for 10 filter = Q
Total area
= 1 ×24 ×60 ×60 = 144 m3/day/m2
10 ×60
31
45. For the plane frame with an overhang as shown below, assuming
negligible axial deformation, the degree of static indeterminacy,
d, and the degree of kinematic indeterminacy, k, are
a) d = 3 and k = 10
b) d = 3 and k = 13
c) d = 9 and k = 10
d) d = 9 and k = 13
Answer: d
Solution:
Dk = 0 + 3 × 7 + 2 + 1 = 24 and after neglecting axial
deformation,
Dk = 24 – 11 = 13
Dse = (3+2+1) – 3 = 3
Dsi = 3 × = 3 × 2 = 6
∴ Ds = Dse + Dsi = 3 + 6 = 9
46. Consider the structural system shown in the figure under the
action of weight W. All the joints are hinged. The properties of
the members in terms of length (L), area (A) and the modulus of
elasticity (E) are also given in the figure. Let L, A and E be l m,
0.05 m2 and 30 x 106 N/m2, respectively, and W be 100kN.
32
Which one of the following sets gives the correct values of the
force, stress and change in length of the horizontal member QR?
a) Compressive force = 25 kN; stress = 250 kN/m2, Shortening
= 0.0118m
b) Compressive force= 14.14 kN; stress = 141.4 kN/m2,
Extension = 0.0118m
c) Compressive force = 100 kN; stress = 1000 kN/m2
Shortening = 0.0417m
d) Compressive force = 100 kN; stress = 1000 kN/m2 Extension
= 0.0417m
Answer: c
Solution:
The free body diagram removing the top support is shown below.
Due to symmetry, the reaction is W upwards.
33
= =
sin 90 sin 45 sin 45
= ( )
√2
, = ( )
√2
Σ = 0
∴ = 45 + 45
= W (compressive)
= 100 ( )
Given, A = 0.05m2
Area of bar QR = AQR = 2A = 0.1m2
Stress in the bar QR = = 100
0.1
= 1000 / 2
Length of bar QR = LQR = L√2 = √2
Shortening of member QR
= . = 100×103×1×√2
. 0.1×30×106
= 0.0417
47. The culturable command area for a distributary channel is
20,000 hectares. Wheat grown in the entire area and the intensity
of irrigation is 50%. The kor period for wheat is 30 days and the
kor water depth is 120 mm. The outlet discharge for the
distributary should be
34
a) 2.85 m3/s
b) 3.21 m3/s
c) 4.63 m3/s
d) 5.23 m3/s
Answer: c
Solution:
Area to be irrigated
= C.C.A × intensity of irrigation
= 20,000 × 0.50 = 10,000 ha.
Period, B = 30 days
Water depth, ∆ = 120 mm = 0.12 m
∴ Duty, D = 8.64 ×B = 8.64 ×30 = 2160 ha/cumec
∆ 0.120
Discharge, Q = A = 10000 = 4.63 m3/sec.
D 2160
48. Uplift pressures at points E and D (Figure A) of a straight
horizontal floor of negligible thickness with a sheet pile at
downstream end are 28% and 20%, respectively. If the sheet pile
is at upstream end of the floor (Figure B), the uplift pressures at
points D1 and C1 are
a) 68% and 60% respectively
b) 80% and 72% respectively
c) 88% and 70% respectively
35
d) 100% and zero respectively
Answer: b
Solution:
Uplift pressure at D1 = 100 – 20% = 80%
Uplift pressure at C1 = 100 – 28 = 72%
49. A field was supplied water from an irrigation tank at a rate of
120 lit/sec to irrigate an area of 2.5 hectares. The duration of
irrigation is 8 hours. It was found that the actual delivery at the
field, which is about 4 km from the tank, was 100 lit/sec. The
runoff loss in the field was estimated as 800 m3. The application
efficiency in this situation is______%
a) 62%
b) 72%
c) 76%
d) 80%
Answer: b
Solution:
Water conveyance efficiency,
ηc = water Water delivered to the field × 100
supplied into the canal at the reservior
= 100 × 100 = 83.33%
120
Water application efficiency,
ηc = Water strored in root zone × 100
water delivered to the field
Water supplied to the field in 8 hrs @ 100 lit/sec
= 100 × 8 × 60 × 60 = 2880 × 103 Lit
36
= 2880 m3
Run off losses in the field = 800 m3
∴ Water application efficiency,
= 2880 – 800 = 2080 m3
∴ Water application efficiency,
ηa = 2080 × 100 = 72.22%
2880
50. The transplantation of rice requires l0 days and total depth of
water required during transplantation is 48 cm. During
transplantation, there is an effective rainfall (useful for irrigation)
of 8 cm. The duty of irrigation water (in hectares/cumec) is: ____
a) 612
b) 216
c) 30
d) 108
Answer: b
Solution:
Irrigation water depth = Total water depth required –
effective rainfall
∆ = 48 – 8 = 40 cm = 0.4 m
Irrigation period, B = 10 days
Duty, D = 8.64B∆
= 8.64 × 10 = 216 ha/cumec
0.4
37
51. A field channel has cultivable commanded area of 2000
hectares. The intensities of irrigation for gram and wheat are 30%
and, 50% respectively. Gram has a kor period of 18 days, kor
depth of 12 cm, while wheat has a kor period of 18 days and a kor
depth of 15 cm. The discharge (in m3/s) required in the field
channel to supply water to the commanded area during the kor
period is ______
Answer: 1.428
Solution:
CCA = 2000 ha
Area under gram = 30% of CCA
= 600 ha
Area under wheat = 50% of CCA
= 1000 ha
D = 8.64B∆
For gram: Duty, D = 8.64 × 18 = 1296 ha/cumec
0.12
For wheat, D = 8.64 × 18 = 1036.8 ha/cumec
0.15
Discharge required, Q = A
D
For gram, Q1 = 600 = 0.463 m3/s
1296
For wheat, Q2 = 1000 = 0.965 m3/s
1036.8
Total discharge = Q1 + Q2 = 1.428 m3/sec
38
52. The culturable command area of a canal is 10,000 ha. The area
grows only two crops rice in the Kharif season and wheat in the
Rabi season. The design discharge of the canal is based on the
rice requirements, which has an irrigated area of 2500 ha, base
period of 150 days and delta of 130 cm. The maximum
permissible irrigated area (in ha) for wheat, with a base period of
120 days and delta of 50 cm is _______
Answer: 5200
Solution:
Rice Wheat
A = 2500 ha A =?
B = 150 d B = 120 d
∆ = 130 m ∆ = 50 cm
Q = Constant
Q = A & D = 8.64B∆
C
Q = A∆
8.64B
A1∆ = A2∆2
B1 B2
2500 ×130 = A2 ×50
150 120
A2 = 130 × 40 = 5200 ha
53. A concrete gravity dam section is shown in the figure. Assuming
unit weight of water as 10 kN/m3 and unit weight of concrete as
24 kN/m3, the uplift force per unit length of the dam (expressed
in kN/m) at PQ is _______
39
Answer: 10500
Solution:
Area of uplift pressure distribution diagram gives uplift force
Total Area = (1) + (2) + (3) + (4)
(1) 10 × 250 = 2500 kN/m
(2) 1 × 10 × 400 = 2000 kN/m
2
(3) 40 × 50 = 2000 kN/m
(4) 1 × 40 × 200 = 4000 kN/m
2
Total uplift force on base = 10,500 kN/m
40
54. A metal bar of length 100 mm is inserted between two rigid
supports and its temperature is increased by 10°C. If the
coefficient of thermal expansion is 12 x 10-6 per °C and the
Young's modulus is 2 x 105 MPa, the stress in the bar is
a) Zero
b) 12 MPa
c) 24 MPa
d) 2400 MPa
Answer: c
Solution:
σ = αtE
= 12 × 10−6 × 10 × 2 × 105
= 24
55. A frame ABCD is supported by a roller at A and is on a hinge
at C as shown below:
The reaction at the roller end A is given by
a) P
b) 2P
c)
2
d) Zero
41
Answer: d
Solution:
∑ = 0
The load acting on the frame is balanced with respect to the
central support at C. Therefore, horizontal hinge reaction at A is
zero. (HA = 0). Just by observation answer can be picked up.
56. A reinforced concrete (RC) beam with width of 250 mm and
effective depth of 400 mm is reinforced with Fe 415 steel. As per
the provisions of IS 456-2000, the minimum and maximum
amount of tensile reinforcement (expressed in mm2) for the
section is, respectively
a) 250 and 3500
b) 205 and 4000
c) 270 and 2000
d) 300 and 2500
Answer: b
Solution:
Min % steel
=( ) 0.85
∙
( ) = 0.85 × 250 × 400
415
= 205 mm2
Maximum reinforcement cannot be calculated as total depth is not
given
∴ = 4% = 4% ( × )
42
In fact, there is no need to calculate ( ) as per options
given.
57. Consider a bar of diameter ‘D’ embedded in a large concrete
block as shown in the adjoining figure, with a pull-out force P
being applied. Let and , be the bond strength (between the
bar and concrete) and the tensile strength of the bar, respectively.
If the block is held in position and it is assumed that the material
of the block does not fail, which of the following options
represents the maximum value of P?
a) Maximum of � 4 2 � and ( )
b) Maximum of � 4 2 � and ( )
c) Minimum of � 4 2 � and ( )
d) Minimum of � 4 2 � and ( )
Answer: c
Solution:
Strength of the bar in direct tension = � 4 2 �
43
Strength of the bar in based on bond strength of the bar =
( )
The minimum of the above two should be considered for the
safety of the ba, so that is should not fail in any of the above cases
58. A tall tower was photographed from an elevation of 700 m
above the datum. The radial distance of the top and bottom of the
tower from the principal points are 112.50 mm and 82.40 mm,
respectively. If the bottom of the tower is at an elevation 250 m
above the datum, then the height (expressed in mm) of the tower
is _________
Answer: 120.4 mm
Given:
H = 700 m, ℎ = 250
Relief distance,
d = 1 −
= 112.5 – 82.40 = 30.1 mm
∴ d = hr1 (where h is height if tower)
H−havg
⟹ 30.1 = ℎ ×112.5
700−250
∴ h = 120.4 mm
44
59. The water content of a saturated soil and the specific gravity of
soil solids were found to be 30% and 2.70, respectively.
Assuming the unit weight of water to be 10 kN/m3, the saturated
unit weight (kN/m3), and the void ratio of the soil are
a) 19.4, 0.81
b) 18.5, 0.30
c) 19.4, 0.45
d) 18.5, 0.45
Answer: a
Solution:
Given w = 30%, GS = 2.70
γw = 10 kN/m3, Sr = 1 or 100%
e = = 0.30 ×2.70 = 0.81
1
γsat = γw � 1 + + �
= 10�2.17+0+0.08.181� = 19.40 kN/m3
60. A non-homogeneous soil deposit consists of a silt layer
sandwiched between a fine-sand layer at top and a clay layer
below. Permeability of the silt layer is 10 times the permeability
of the clay layer and one-tenth of the permeability of the sand
layer. Thickness of the silt layer is 2 Times the thickness of the
sand layer and two-third of the thickness of the clay layer. The
ratio of equivalent horizontal and equivalent vertical permeability
of the deposit is ______
Answer: 11
45
Solution:
k2 = 10k3;
k2 = 110k1
∴ k1 = 10 k2
k3 = 0.1 k2
z2 = 2z1
z2 = 23z3
∴ z1 = 0.5 z2
z3 = 1.5 z2
kH = z1k1+ z2k2+ z3k3
z1+ z2+ z3
= 0.5z2 ×10k2+ z2k2+ 1.5z2 ×0.1k2 = 2.05 k2
0.5z2+ z2+ 1.5z2
z1+ z2+ z3 0.5z2+ z2+ 1.5z2 3z2
K = = =v
kz11+ kz22+ z3 01.05kz22+ kz22+ 1.5z2 16.05 z2
k3 0.1k2 k2
= 0.1869 k2
kH = 2.05k2 = 10.97 say 11
kv 0.1869k2
61. A two lane, one-way road with radius of 50 m is predominantly
carrying lorries with wheelbase of 5 m. The speed of lorries is
restricted to be between 60 kmph and 80 kmph. The mechanical
widening and psychological widening required at 60 kmph are
designated as Wme,6o and Wps,6o, respectively. The mechanical
widening and psychological widening required at 80 kmph are
46
designated as Wme,80 and Wps,80, respectively. The correct values
of Wps,60, Wme,80, Wps,80, respectively are
a) 0.89 m, 0.50 m, 1.19 m, and 0.50 m
b) 0.50 m, 0.89 m, 0.50 m, and 1.19 m
c) 0.50 m, 1.19 m, 0.50 m, and 0.89 m
d) 1.19 m, 0.50 m, 0.89 m, and 0.50 m
Answer: b
Solution:
No. of lanes, n = 2
R = 50 m
l=5m
for V = 60 kmph for V = 80 kmph
Wm = 2
2
Wm = 2 = 2(5)2 = 0.5 = (2)(5)2 = 0.5
2 2×50 2×50
Wp = V
9.5√R
Wp = V = 60 = 0.89
9.5√R 9.5√50
m = 80 = 1.19 m
9.5√50
62. If the design speed is 80 kmph, perception reaction time is 3sec
and coefficient of friction is 0.5, then the SSD is _______ m
a) 61.3 m
b) 117 m
c) 85.5 m
d) 171 m
47
Answer: b
Solution:
SSD = vt + v2
2gf
= 0.278 × 80 × 3 + (0.278 ×80)2
2×9.81 ×0.5
= 117 m
63. At the five different feasible levels of investment (expressed as
present worth), the summated benefits (also expressed as present
worth) are tabulated for a possible project. (All values are in 104
₹)
Levels of investment, as P.W Benefits as P.W
800 1200
900 1400
1000 1520
1100 1740
1200 1895
What will be the most profitable investment level; and the next
best?
a)900 and 1200
b)1100 and 1200
c)1200 and 1100
d)1100 and 900
Answer: b
Solution:
1100 and 1200
48
Level of instrument 800 900 1000 1100 1200
B/C ratio 1.525 1.556 1.52 1.58 1.579
Most profitable – 1100
Next Best – 1200
64. ISA 100 x 100 x 10 mm (cross sectional area = 1903 mm2) is
welded along A and B (refer to figure in the below question) such
that the lengths of weld along A and B are l1 and l2 respectively.
Which of the following is a possibly acceptable combination of l1
and l2?
a) l1 = 60 mm and l2 = 150 mm
b) l1 = 150 mm and l2 = 60 mm
c) l1 = 150 mm and l2 = 150 mm
d) Any of the above, depending on the size of the weld
Answer: a
Solution:
The C.G of the angle is close to B compared to A. Therefore, the
strength of the weld shall be more close to B (to have more
strength of weld for given size of weld, the length of the weld
(Close to the B), 2 shall be more than 1 so that the algebraic sum
of moments of the two welds 1 & 2 about C.G of the angle will
be zero and there is no eccentricity.
49
Data Loading...