Grand Test - 3 - PDF Flipbook
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CIVIL
GrandTests
Test-03Solutions
APTITUDE
One Mark Questions
1. Choose the most appropriate word from the options given below
to complete the following sentence:
His rather casual remarks on politics _____ his lack of
seriousness about the subject.
a) masked
b) believed
c) betrayed
d) suppressed
Answer: c
2. Choose the most appropriate word from the options given below
to complete the following sentence.
It was her view that the country's problems had been _____
by foreign technocrats, so that to invite them to come back
would be counter-productive.
a) identified
b) ascertained
c) exacerbated
d) analyzed
Answer: c
Solution:
The question refers to the country’s problems. To call foreign
technocrats back who had been in the country before, according
to the author would be counterproductive. This implies that the
1
foreign technocrats had not solved the countries problems and
they did just the opposite.
(a) If problems are identified, then problems should be
solved but the word counter productive gives exactly
opposite meaning.
(b) If the foreign technocrats have ascertained what the
countries problems were then they would be expected to
have some idea of countries problems. To state that they
should not be invited back does not make sense.
(c) If the foreign technocrats had exacerbated or worsened
the countries problems then calling those foreign
technocrats would certainly be counterproductive.
(d) To analyze is to separate into parts in order to understand.
If the foreign technocrats had analyzed the countries
problems then that is not necessarily a reason why they
should not be called back.
3. Choose the most appropriate word from the options given below
to complete the following sentence.
If you are trying to make a strong impression on your
audience, you cannot do so by being understated, tentative
or ____.
a) hyperbolic
b) restrained
c) argumentative
d) indifferent
2
Answer: b
Solution:
The given sentence is a contra sentence. If you observe the first
part of the sentence you see an expression strong impression. In
the second part of the sentence you see understated, tentative.
When you compare these two words with the expression ‘strong
impression’ we understated it’s a ‘contrast’ model sentence.
Moreover, the given option should be synonymous to tentative,
understated. Obviously, a linking word should be closer to the
meanings of understated (mute, inexpressive, tentative not
confident). So, it is clear that indifferent is the word that goes with
these words. ‘D’ is the right answer.
4. Choose the most appropriate alternative from the options given
below to complete the following sentence:
If the tired soldier wanted to lie down, he__ the mattress out on
the balcony.
a) should take
b) shall take
c) should have taken
d) will have taken
Answer: a
Solution:
The given sentence is a conditional sentence
(1) If + V1 – will (type -1)
(2) If + V2 – would/should (type – 2)
3
(3) If + had + V3 – would/should have + V3 (type – 3)
The verb in the if clause is in the past tense i.e.,
V2(wanted). So, the main clause should be would/should
(type – 2).
5. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs.
When the smaller wheel has made 21 revolutions, then the
number of revolutions mad by the larger wheel is:
a) 4
b) 9
c) 12
d) 49
Answer: b
Solution:
Let the required number of revolutions made by larger wheel be x.
Then, More cogs, Less revolutions (Indirect Proportion)
∴ 14 : 6 :: 21 : X ⟺ 14 × X = 6 × 21
⟹ X = 6×21 = 9
14
4
Two Marks Questions
6. The part of government which is concerned with making of rules
a) Court
b) Tribunal
c) Bar
d) Legislature
Answer: d
7. In the question below, some part of the sentence has been left
blank. Select the best option which completes the sentence
properly, taking care of semantics, syntax and meaning.
The judgement that this age is too practical for great ideals
may only be a description of the husk that hides __________.
a) the musk
b) the real estate of the society
c) the underlying cornices
d) a very full ear of corn
Answer: d
8. In the sentence given below a part is underlined and for that part
options are given. Choose the most suitable option that can
replace the underlined part.
Days before Congress leader Narayan Rane lost the by-election
in Bandra East assembly constituency, his second consecutive
electoral defeat, Chief Minister Devendra Fadnavis could have
predicted of this.
a) could have predicted of this
5
b) has predicted of the loss.
c) had predicted about the loss.
d) had predicted that this would happen.
Answer: d
Explanation:
Past perfect tense is to be used for prediction.
9. One of the legacies of the Roman legions was discipline. In the
legions, military law prevailed and discipline was brutal.
Discipline on the battlefield kept units obedient, intact and
fighting, even when the odds and conditions were against them.
Which one of the following statements best sums up the meaning
of the above passage?
a) Thorough regimentation was the main reason for the efficiency
of the Roman legions even in adverse circumstances.
b) The legions were treated inhumanly as if the men were
animals.
c) Discipline was the armies' inheritance from their seniors.
d) The harsh discipline to which the legions were subjected to led
to the odds and conditions being against them.
Answer: a
Solution:
The question is about sum up. In sum up questions we have to
make sure that all the points mentioned in the paragraph should
be reflected in the answer choice.
(a) This is the right choice.
6
(b) This is irrelevant as there is no mention of inhuman
treatment.
(c) This is a general statement, not reflecting the given
points in the paragraph.
(d) The interpretation of the sentence is completely opposite
to the given paragraph.
10. Budhan covers a distance of 19 km in 2 hours by cycling one
fourth of the time and walking the rest. The next day he cycles (at
the same speed as before) for half the time and walks the rest (at
the same speed as before) and covers 26 km in 2 hours. The speed
in km/h at which Budhan walks is
a) 1
b) 4
c) 5
d) 6
Answer: d
Solution:
Cycling ⟹ 14th = 2 × 1 = 1
4 2
19 km → 2 hrs
Walking ⟹ 14th = 2 × 3 = 3
4 2
Let cycling speed = C
Walking speed = W
C + 3W = 19 ……... (i)
2 2
7
Cycling ⟹ 21th = 2 × 1 = 1
2
26 km → 2 hrs
Walking ⟹ 12th = 2 × 1 = 1
2
C + W = 26 ………. (ii)
By solving equation no (i) and (ii)
W = 6 km/hr
∴ The speed at which Budhan walks = 6 kmph
8
TECHNICAL
One Mark Questions
11. The smallest and largest Eigen values of the following matrix
3 −2 2
are:�4 −4 6�
2 −3 5
a) 1.5 and 2.5
b) 0.5 and 2.5
c) 1.0 and 3.0
d) 1.0 and 2.0
Answer: d
Solution:
The characteristic equation of the given matrix is
− 3 + 4 2 − 5 + 2 = 0
⇒ ( − 2)(− 2 + 2 − 1) = 0
∴ = 2,1
12. Number of inflection points for the curve y = x + 2x4 is _______
Answer: 1
Solution:
Equate 2 = 0 ⇒ = 0, = 0
2
∴ (0, 0) is the only point of inflection
13. l i →m0 2 = ____
Answer: 0
Solution:
l i →m0 � × � = 1 × 0 = 0
9
14. Match List-I with List-II and select the correct answer using the
codes given below the lists:
List – I
A. Soil pipe
B. Intercepting trap
C. P-trap
D. Cowl
List – II
1. Ventilating pipe
2. Wash basin
3. Water closet waste
4. House drainage
Codes:
ABCD
a) 3 4 1 2
b) 3 4 2 1
c) 4 3 2 1
d) 4 3 1 2
Answer: b
15. The least expensive and most suitable excreta disposal unit for
rural areas would be the
a) soak pit
b) pit privy
c) leaching cesspool
d) septic tank
10
Answer: b
16. Modulus of rigidity is expressed as
a) compressive stress/compressive strain
b) tensile stress/tensile strain
c) shear stress/shear strain
d) stress/volumetric strain
Answer: c
Solution:
The shear modulus is concerned with the deformation of a solid
when it experiences a force parallel lo one of its surfaces while
its opposite face experiences an opposing force (such as friction.)
17. Acidic soils are reclaimed by
a) leaching of the soil
b) using limestone as a soil amendment
c) using gypsum as a soil amendment
d) provision of drainage
Answer: b
Solution:
The salts injurious to the health of plants are alkali salts e.g.
Na2CO3, Na2SO4 and NaCl. Such a salt affected soil is
unproductive and is known as saline soil. If the soil is saline and
the efflorescence continues for a long period, then soils become
illaerated and highly unproductive and these are called alkaline
soils, these soils can be reclaimed by
i. Leaching
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ii. Providing surface and subsurface drainage of the area.
iii. Crop rotation
iv. Lining of canals and water courses
v. Reducing the intensity of irrigation.
vi. Optimum use of water
vii. Gypsum addition
Acidic soils, on the other hand, can be reclaimed by basic salts
such as limestone.
18. The conditions to be satisfied for a channel in 'Regime' as per-
Lacey are
1. Constant discharge
2. Silt grade and silt concentration are constant
3. The channel is flowing in unlimited in coherent alluvium of
the same alluvial character as that transported.
Which of the above statements are correct?
a) l and 2 only
b) 1, 2 and 3
c) 1 and 3 only
d) 2 and 3 only
Answer: b
Solution:
According to Lacey for a channel to be in initial regime.
i. Silt grade and silt concentration are constant.
ii. The channel is made up of the same material as that of being
transported.
12
19. The best design of an arch dam is when
a) all horizontal water loads are transferred horizontally to the
abutments
b) the dam is safe against sliding at various levels
c) the load is divided between the arches and cantilevers and the
deflections at the conjugal points being equal
d) the deflections of the cantilevers are equal at different points
Answer: c
Solution:
Arch dams are solid walls curved in plan. They behave as a
cantilever retaining wall standing up from its base, and partly the
load will be transferred to the two ends of the arch span by
horizontal action. In the trial load method of design, the load is
assumed to be distributed between the arch elements and the
cantilever elements. The deflection calculated at conjugal points
should be the same.
20. A bar of circular cross-section varies uniformly from a cross-
section 2D to D. If extension of the bar is calculated treating it as
a bar of average diameter, then the percentage error will be
a) 10
b) 25
c) 33.33
d) 50
Answer: a
Solution:
13
∆0 = 4PL
πED1D2
Actual extension,
∆0 = 4PL = 2PL
π(2D2)E πD2E
The average diameter of bar
2D+D = 1.5 D
2
Approximate extension,
∆ = 4PL = 4PL
π×(1.5D)2E 2.25πD2E
∴ Error in calculation = �1 − ∆∆0� × 100
= �1 − 2.225� × 100 = 11.11% = 10
21. In an experiment it found that the bulk modulus of a material is
equal to its shear modulus. The Poisson's ratio is
a) 0.125
b) 0.250
c) 0.375
d) 0.500
Answer: a
Solution:
We know that, μ = 3K−2G
6K+2G
But here, K = G
μ = 3−2 = 0.125
6+2
14
22. In limit state approach, spacing of main reinforcement controls
primarily
a) Collapse
b) Cracking
c) Deflection
d) Durability
Answer: b
Solution:
The code specifies minimum and maximum limits for the spacing
between parallel reinforcing bars in a layer. The minimum limits
are necessary to ensure that the concrete can be placed easily in
between and around the bars during the placement of fresh
concrete. The maximum limits are specified for bars in tension
for the purpose of controlling crack-widths and improving bond.
23. The reinforcement for tension, when required in members, shall
consist of
a) Only longitudinal reinforcement in the tension face
b) Only longitudinal reinforcement in the compression face
c) Only two legged closed loops enclosing the corner
reinforcement
d) Both longitudinal and transverse reinforcement
Answer: d
Solution:
Transverse reinforcement is required to hold the longitudinal bars
also.
15
24. The characteristic strength of concrete is
a) higher than the average cube strength
b) lower than the average cube strength
c) the same as the average cube strength
d) higher than 90% of the average cube strength
Answer: b
Solution:
Characteristic strength,
= − 1.65σ
∴ <
25. A simply supported isotropically reinforced square slab of side
4 m is subject to a service load of 6 kPa. The thickness of the slab
is 120 mm. The moment of resistance required as per yield line
theory is _____
Answer: 9 kN-m/m
Solution:
Moment of resistance = wuL2
24
Self-weight of slab = 0.12 × 25 = 3 kN/m2
Service load = 6 kN/m2
Ultimate load = 1.5 × (6 + 3) = 13.5 kN/m2
Ultimate moment capacity = 13.5 × 104 = 9 kN-m/m
24
16
26. Match List-I (Corrections)with List-II (Name) and select the
correct answer using the codes given below the lists:
List – I
A. ± �1 − ℎ �
B. ± � �2
24
C. ± � − �
D. ± �� − � �
List-II
1. Sag correction
2. Pull correction
3. Temperature correction
4. Mean sea level correction
Codes:
ABCD
a) 4 1 3 2
b) 1 4 3 2
c) 4 1 2 3
d) 1 4 2 3
Answer: a
27. The clogging of chain rings with mud introduces (with 'error'
defined in the standard way)
1. Negative cumulative error.
2. Positive cumulative error.
3. Compensating error.
17
a) 1 only
b) 2 only
c) 3 only
d) 1, 2 and 3
Answer: b
Solution:
Length of chain is reduced
Measured length = Less
Noted down value = More
Error = (+) ve
Correction = (–) ve
Positive Cumulative Error
28. Consistency as applied to cohesive soils is an indicator of its
a) Density
b) Moisture content
c) Shear strength
d) Porosity
Answer: c
Solution:
Consistency of soil refers to the resistance offered by it against
forces that tend to deform or rupture the soil aggregate. It is
related to strength.
18
29. Consider the following statements:
1. Mica is a clay mineral.
2. Rock dust particles, even of clay size are non-plastic.
3. A particle of kaolinite is electrically charged.
Which of these statements are correct?
a) 1, 2 and 3
b) Only 1 and 2
c) Only 2 and 3
d) Only 1 and 3
Answer: c
Solution:
Feldspars are the most common rock minerals which account for
the abundance of clays derived from the feldspars on the earth's
surface.
Quartz comes next in order of frequency. Most sands are
composed of quartz.
Mica constitute very small fraction of original igneous rocks and
thus it cannot be said to be clay mineral.
Rock flour (rock dust particles) generally consists of more or less
uniform size grains of quartz and they are least plastic varieties.
Electrical charge on kaolinite is responsible for plastic behaviour
of soils and cohesion in clays.
19
30. Consider the following properties for clays X and Y:
S. No Properties Clay X (%) Clay Y (%)
1 Liquid limit 42 56
2 Plastic limit 20 34
3 Natural water content 30 50
Which of the clays, X or Y, experiences larger settlement under
identical loads; is more plastic; and is softer in consistency?
a) X, Y and X
b) Y, X and X
c) Y, X and Y
d) X, X and Y
Answer: c
Solution:
Plasticity index (Ip) is the numerical difference between liquid
limit and plastic limit. The soil with higher value of 1. will be
more plastic.
(IP)X = 42 – 20 = 22
(IP)Y = 56 – 34 = 22
But if plasticity index is same for both soils, then the soil with
larger difference in liquid limit and natural water content is more
plastic.
Consistency index,
IC = Liquid limit − Natural water content
IP
(IC)X = 42 − 30 = 0.545
22
20
(IC)Y = 56 − 50 = 0.273
22
Soil which has low value of consistency index is softer. Thus, the
softer soil will undergo large settlements under identical loads.
31. A specimen of clayey silt contains 20% silt size particles. Its
liquid limit = 40 and plastic limit = 20. In liquid limit test, at
moisture content of 30%, required number of blows was 50. Its
plasticity index, activity and consistency index will respectively
be
a) 20, 0.67 and 0.5
b) 20, 1.5 and 2.0
c) 30, 1 .5 and 0.72
d) 20, 0.286 and 0.38
Answer: a
Solution:
Plasticity Index, IP = WL – WP
= 40 – 20 = 20
Activity of clay,
AC = Plasticity index 2μ
per cent finer than
= per cent Plasticity index in the soil
clay size particles
= 20 70 = 0.67
100 −
Consistency Index,
IC = WL− WN = 40 − 30
IP 20
= 0.50
21
32. Match List-I (Deposit) with List-II (Soil structure) and select the
correct answer using the codes given below the lists:
List – I
A. Coarse grained soil
B. Silt deposit
C. Clay deposit
D. Composite soil
List – II
1. Flocculated
2. Cohesive matrix
3. Honeycomb
4. Single-grained
Codes:
ABCD
a) 2 3 1 4
b) 4 3 1 2
c) 2 1 3 4
d) 4 1 3 2
Answer: b
Solution:
Single grained structure is formed by the settlement of coarse
grained soils in suspension in water. A flocculent structure is
formed by the deposition of fine soil fraction (like clay particles)
in water. The particles oriented in flocculent structure will have
edge to face contact. Honey comb structure is formed by the
22
disintegration of flocculent structure under superimposed load.
The particles oriented in honeycomb structure will have face to
face contact.
33. Four-lane divided highway, with each carriage way being 7.0 m
wide, is to constructed in a zone of high rainfall. In this stretch,
the highway has a longitudinal slope of 3% and is provided a
camber of 2%. What is the hydraulic gradient on this highway in
this stretch?
a) 4.0%
b) 3.6%
c) 4.5%
d) 3.0%
Answer: a
Solution:
Hydraulic gradient is twice of camber
Given, camber = 2%
∴ Gradient = 2 × 2% = 4%
34. The maximum super-elevation to be provided on a road curve is
1 in 15. If the rate of change of super elevation is specified as 1
in 120 and the road width is 10 m. then the minimum length of
the transition curve on either end will be ______ m
Answer: 80 m
Solution:
Length of transition curve,
L = eWN
23
Given, e = 1 ; 1 = 1120;
15 N
W = 10 m
∴ L = 1 × 10 × 120 = 80 m
15
35. Assertion (A): In structural bearing type joints each connection
is assumed to transmit its proportional share of the applied load.
Reason(R): Applied load passes through the centroid of the
connector group.
a) both A and R are true and R is the correct explanation of A
b) both A and R are true but R is not a correct explanation of A
c) A is true but R is false
d) A is false but R is true
Answer: c
Solution:
The transmission of load in a riveted (or bolted) joint occurs ether
by friction between the connected plates due to large gripping
forces produces by tension in the rivet (or bolt), or by shearing
action on the cross-section of the rivet and bearing stresses on the
faces of rivet and plates in contact with each other' ln the bearing
of joint friction is overcome and the joint slips The contribution
of friction is neglected and the load is fully transmitted through
the rivet' All the rivets are assumed to resist the load equally. The
applied load may or may not pass through the centroid of the
connector group.
24
Two Marks Questions
36. The value of ( , l )i→m(0,0) 2− is
√ −√
a) 0
b) 1
2
c) 1
d) ∞
Answer: a
Solution:
Along the path y = mx
( , l )i→m(0,0) 2− = l i →m0 2− 2
√ −√ √ −√
2(1− ) 3
√ (1−√ ) 2(1− )
= l i →m0 = l i →m0 (1−√ ) =0
37. Divergence of the vector field v (x, y, z) = = – (x cosxy + y) +̅
(y cosxy) ̅ + [(sinz2) + x2+ y2] � is
a) 2z cosz2
b) sin xy + 2z cosz2
c) x sinxy - cosz
d) none of these
Answer: a
Solution:
V(x, y, z) = – (x cos xy + y) ̅ + (y cos xy) ̅ + {sin z2 + x2 + y2] �
divV� = ∂ [−(x cos xy + y)] + ∂ [y cos xy] + [sin 2 + 2 + 2]
∂x ∂y
= 2 z cos z2
25
38. The maximum value of f(x) = x3 – 9x2 + 24x + 5 in the interval
[1, 6] is _______
a) 21
b) 25
c) 41
d) 46
Answer: c
Solution:
f(x) = x3 - 9x2 + 24 x + 5 in [1, 6]
f(l) = 21, f(6) = 41
f '(x) = 3x2 - 18x + 24 = 0
⇒x = 2, 4 are the stationary points
f "(x) = 6x -18
f "(2) = - 6 < 0, f"(4) = 6 > 0
and f(2) = 25 and f(4) = 21
∴ Maximum value of f(x) in [1, 6] is f(6) = 41
39. The velocity field on an incompressible flow is given by
V = (a1x + a2y + a3z)I + (b1x + b2y + b3z)j + (c1x + c2y + c3z)k,
where a1 = 2 and c3 = – 4. The value of b2 is_____
Answer: 2
Solution:
For an incompressible (solenoidal) flow of a fluid, divergence is
zero.
i.e., div V� = 0
⟹ a1 + b2 + c3 = 0
26
⟹ 2 + b2 – 4 = 0
∴ b2 = 2
40. The 2-day and 4-day BOD values of a sewage sample are 100
mg/L and 155 mg/L, respectively. The value of BOD rate
constant (expressed in per day) is ______
Answer: 0.3
Solution:
y2 = 100 mg/l
y4 = 155 mg/l
y2 = L0�1 − e−k×2�
100 = L0�1 − e−k×2� ………... I
y4 = L0�1 − e−k×4�
155 = L0�1 − e−k×4� ………… II
By solving I and II
K = 0.3 d– 1 (base e) (or) 0.13 d– 1 (base 10)
41. Mathematical idealization of a crane has three bars with their
vertices arranged as shown in the figure with a load of 80 kN
hanging vertically. The coordinates of the vertices are given in
parentheses. The force in the member QR, FQR will be
27
a) 30 kN Compressive
b) 30 kN Tensile
c) 50 kN Compressive
d) 50 kN Tensile
Answer: a
Solution:
Σ = 0
(2) = 80(3)
= 120
Consider FBD of joint Q
Σ = 0
+ cos 14.03 = 0
= − = −120
14.03 14.03
Σ = 0
28
sin 14.03 − = 0
−120 sin 14.03 − = 0
14.03
= −30.58
Negative sign indicates that the assumed direction of force for
QR (tensile) is wrong.
Hence, FQR = 30kN (Compressive)
42. Determine the degrees of freedom of the following frame. _____
Answer: 13
Solution:
Degrees of freedom of various supports (or) joints are shown in
fig.
Dk = 0 + 3 × 7 + (1 + 2)
= 24 (with axial deformation)
= 24 – 11 = 13 (neglecting axial deformation)
29
43. Water flows in a rectangular channel at a depth of 1.20 m and a
velocity of 2.4 m/s. A local rise in the bed of 0.60 m will be cause
a) the surface to rise
b) the surface to fall
c) a stationary jump to form
d) a surge to travel upstream
Answer: d
Solution:
Rectangular Channel
y1 = 1.2 m
V1 = 2.4 m/s
∆Z = 0.6 m
E1 = y1 + V12 = 1.2 + (2.4)2 = 1.49 m
2g 2 × 9.81
Q = 2.4 × 1.2 = 2.88 m3/s/m
Assuming channel width as constant, the critical depth
1
= �gQb22�3 = 0.94 m
Critical specific energy for rectangular channel Ec = 3 yc
2
Ec = 23(0.94) = 1.41
We know for critical flow in the hump portion E1 = E2 +
(∆Z) = EC + (∆Z)C
⟹ 1.49 = 1.14 + (∆Z)C
30
∴ (∆Z)C = 0.08 m
If the hump provided is more than the critical hump height the u/s
flow gats affected.
(Or)
Fr1 = V1 = 2.4 = 0.69 < 1
�gy1 √9.81 ×1.2
⟹ Hence sub-critical.
If the approaching flow is sub-critical the level of water will fall
in the hump portion.
Optical (b) is correct if the hump height provided is less than
critical hump height.
As the hump height provided is more than critical, the u/s flow
gets affected with the increase of the specific energy from E1 to
Eˈ1.
In the sub-critical region as the specific energy increases, the level
of water rises from y1 to yˈ1 in the from of a surge.
Eˈ1 = yˈ1 + vˈ1
2g
Eˈ1 = yˈ1 + 2 ……… (1)
2g ˈ21
Also, Eˈ1 = EC + (∆Z) provided
= 1.41 + 0.6 = 2.01 m
31
∴ 2.01 = yˈ1 + 2 × 2.882 21
9.81 ×
Solve by trial & error for yˈ1 > 1.2 m
44. Oil flows through a 200mm diameter horizontal cast iron pipe
(friction factor, f = 0.0225) of length 500m. The volumetric flow
rate is 0.2 m3/s. The head loss (in m) due to friction is______ m
(assume g: 9.81 m/s2)
Answer: 116.18
Solution:
ℎ = . .v2 = . .�QA�2
2gd 2gd
= . .Q2 = 0.0225 ×500 × (0.2)2 = 116.18 m
12.1 × 5 12.1 × (0.2)2
45. Water is pumped through a pipe line to a height of 10 m at the
rate of 0.1 m3/sec. Frictional and other minor losses are 5 m. Then
the power of pump in kw required is _____.
Answer: 15
Solution:
H = 10 m
Q = 0.1 m3/sec
ℎ = 5 m
P =?
P = ρgQ(H + hf)kW
= γ. Q(H + hf)
= 9.81 × 0.1 × (10 + 5) = 9.81 × 0.1 × 15
= 14.715 kW ≃ 15 kW
32
46. Critical depth at a section of a rectangular channel is 1.5 m. The
specific energy at the section is ____ m
a) 0.75 m
b) 1.0 m
c) 1.5 m
d) 2.25 m
Answer: d
Solution:
= 1.5 m
Ec = 3 yc = 3 × 1.5 = 2.25 m
2 2
47. The channel width is to be contracted. The minimum width to
which the channel can be contracted without affecting the
upstream flow condition is ______ m
a) 3.0 m
b) 3.8 m
c) 4.1 m
d) 5.4 m
Answer: c
Solution:
In this case assume channel bottom is horizontal.
33
If the width of contraction is less than critical width the u/s flow
is not affected. As the channel bottom is assumed to be horizontal.
E1 = E2 = EC
EC = 3 ⟹ 1.74 = 3
2 2
∴ = 1.16
1
= �gQb2c2�3,
Because flow is critical in the throat portion
1.16 = �9.8116×2 1
b2c �3
bc = 4.1 m
48. Water is to be lifted by a net head of 180 m. pumps each with
specific speed of 30 and rotational speed of 1450 rpm with design
discharge of 0.2 m3/sec are available. The minimum number of
pumps required is _____
Answer: 3 pumps
Solution:
Total head H = 180 m
Specific speed NS = 30
Rotational speed of each N = 1450 rpm
Discharge through a pump
Q = 0.2 m3/sec
No. of pumps n =?
NS = N�Q ⟹ 1450√0.2
(H)3/4 (H)3/4
Head lift capacity of each pump = H = 60.21 m
34
∴ No of pumps required = 180 = 2.99 = 3 pumps
60.21
49. Glycerine (specific weight 1260 kg/m3, dynamic viscosity 8.00
x 10-2 kg-s/m2) is spread freely to a thickness of 1 mm between a
bottom stationary plate and a top movable plate of 10 cm2 area.
The top plate is to be moved at a uniform speed of 1 m/s. The
force to be exerted on the top plate is ______ kg.
a) 1.6 kg
b) 0.8 kg
c) 0.16 kg
d) 0.08 kg
Answer: d
Solution:
Newton’s viscosity law
= μ du
dy
Therefore force required for top plate
F = . A
= 8 × 10– 2 × 1 × 10 × 10−4
1×10−3
= 0.08 kg.
50. The number of independent elastic constants for a linear elastic
isotropic and homogeneous material is _____
Answer: 2
Solution:
For homogeneous isotropic material independent elastic
constants are two. ( & )
35
51. In a material under a state of plane strain, a 10 × 10 mm square
centered at a point gets deformed as shown in the figure.
If the shear strain at this point is expressed as 0.001 k (in
radians), the value of k is ______
a) 0.5
b) 0.25
c) -0.25
d) -0.5
Answer: d
The shear strain = –0.0005
= 0.001 (k)
= 0.001 x (-0.5)
∴k = (–) 0.5
52. Un-factored maximum bending moments at a section of a
reinforced concrete beam resulting from a frame analysis are 50,
80, 120 and 180 kN-m under dead, live, wind and earthquake
loads respectively. The design moment (kNm) as per IS: 456-
2000 for the limit state of collapse (flexure) is ______
a) 195
b) 250
36
c) 345
d) 372
Answer: d
Solution:
Load combinations
1. (DL + LL) =1.5 (50 + 80)
= 195 kN- m
2. 1.5(DL + EQ) = 1.5 (50 + 180)
= 345 kN- m
3. 1.2(DL + LL + EQ)
= 1.2 (50 + 80 + 180)
= 372 kN- m
Use maximum of above combinations. i.e., 372 kN-m
53. A rectangular concrete beam of width 120 mm and depth 200
mm is prestressed by pretensioning to a force of 150 kN at an
eccentricity of 20 mm. The cross-sectional area of the
prestressing steel is 187.5 mm2. Take modulus of elasticity of
steel and concrete as 2.1 x 105 MPa and 3.0 x 104 MPa
respectively. The percentage loss of stress in the prestressing steel
due to elastic deformation of concrete is ______
a) 8.75
b) 6.125
c) 4.81
d) 2.19
Answer: b
37
Solution:
Stress in concrete at the level of steel is
= + ( )
= 150 × 103 + 150 × 103 × (20)2
120 × 200 �120 ×122002�
Loss of prestress due to elastic shortening = m
= ( )
= 2.1 × 105 (7) = 49 Mpa
3 × 104
Initial prestress in steel, =
= 150 × 103 = 800 Mpa
187.5
% loss = 49 × 100 = 6.125%
800
54. The observations from a closed loop traverse around an obstacle
are
Segment Observation Length Azimuth (clockwise
from station (m) from magnetic north)
PQ P Missing 33.7500o
QR Q 300.00 86.3847o
RS R 354.524 169.3819o
ST S 450.000 243.9003o
TP T 268.00 317.5000o
What is the value of the missing measurement (rounded off to the
nearest 10 mm)?
a) 396.86 m
38
b) 396.79 m
c) 396.05 m
d) 396.94 m
Answer: b
Solution:
In a closed traverse
∑ = 0 & also
∑ = 0
Using, ∑ = 0
l cos 33.75 + 300 cos86.3847
+ 354.524 cos 169.3819
+ 450 cos 243.9003
+ 268 cos 317.5 = 0
∴ l = 396.79 m
55. The chainage of the intersection point of two straights is
1585.60 m and the angle of intersection is 140o. If the radius of a
circular curve is 600.00 m, the tangent distance (in m) and length
of the curve (in m), respectively are
a) 418.88 and 1466.08
b) 218.38 and 1648.49
c) 218.38 and 418.88
d) 418.88 and 218.38
Answer: c
Solution:
Angle of intersection = 140o
39
∴ Deflection angle, ∆ = 180 – 140 = 40o
Tangent length = R tan�∆2�
= 600 tan�420� = 218.38 m
Length of curve, L = R.∆ �1 8 0�
= 600 × 40 × �1 8 0�
L = 418.8 m
56. The void ratio of a soil sample is 1. The corresponding property
of the sample is _____
Answer: 0.5
Solution:
Porosity, n = = 1 = 0.5
1+ 1+1
57. The specific gravity and in-situ void ratio of a soil deposit are
2.71 and 0.85 respectively. The value of the critical hydraulic
gradient is ______
a) 0.82
b) 0.85
c) 0.92
d) 0.95
Answer: c
Solution:
G = 2.71, e = 0.85
= −1 = 2.71−1 = 0.924
1+ 1+0.85
40
58. A uniformly distributed line load of 500 kN/m is acting on the
ground surface. Based on Boussinesq's theory, the ratio of vertical
stress at a depth 2 m to that at 4 m, right below the line of loading,
is _____
a) 0.25
b) 0.5
c) 2.0
d) 4.0
Answer: c
Solution:
Extending Boussinesq's equation for point load, the vertical stress
directly below the line load,
= 2 ˈ
∴ 1
2 ℎ = �45� = 2
4 ℎ
59. A 3 m thick clay layer is subjected to an initial uniform pore
pressure of 145 kPa as shown in the figure.
For the given ground conditions, the time (in days, rounded to the
nearest integer) required for 90% consolidation would be ____
Answer: 1771
41
Solution:
For the given clay, drainage path, d = H
(Single drainage)
= 3m = 3000 mm
Cv = 3 mm2/min
TV(90) = 0.85
Tv = Cv.t
d2
0.80 = 3×t
(3000)2
∴ t = 25 × 106 min = 177.83 days
Say 1771 days
60. List-I below gives a list of physical properties of aggregates
which should be determined to judge their suitability in road
construction. List-II gives a list of laboratory tests which are
conducted to determine these properties. Match List-I with List-
II and select the correct answer from the codes given below the
lists:
List – I
A. Hardness
B. Porosity
C. Toughness
D. Durability
List – II
1. Water adsorption
2. Impact test
42
3. Soundness test
4. Abrasion test
Codes:
ABCD
a) 1 2 3 4
b) 4 1 2 3
c) 3 4 1 2
d) 2 3 4 1
Answer: b
61. While aligning a hill road with a ruling gradient of 6%, a
horizontal curve of radius 50 m is encountered. The grade
compensation (in percentage, up to two decimal places) to be
provided for this case would be _____
Answer: 1.5
Solution:
Ruling gradient, G = 6% (> 4% grade compensation is allowed as
per IRC)
Radius of curve, R = 50 m
Grade compensation, GC = 30+R
R
= 30+50 = 1.6%
50
= 75 = 75 = 1.5
R 50
∴ Use min value of GC = 1.5%
43
62. The penetration value of a bitumen sample tested at 25oC is 80.
When this sample is heated to 60oC and tested again, the needle
of the penetration test apparatus penetrates the bitumen sample
by d mm. The value of d CANNOT be less than ___ mm.
Answer: 8
Solution:
At 25oC, penetration value is 110(80) = 8 mm
At 60oC, penetration value should be more than (or should not be
less than) 8 mm
63. Two activities A and B are segmented into four identically
executable parts each, as shown in the ladder diagram. The
expected duration of each of total A and total B is 16 days; the
standard deviation of the expected duration of total A is1.6 days
and that of total B is 2.4 days. What is the standard deviation of
the laddered network through its critical path?
a)2.20 days
b)2.53 days
c)2.66 days
d)2.80 days
Answer: d
44
Solution:
2.80 days
Laddered network through critical path
Standard deviation for A1 = 1.6 = 0.4 days
4
Standard deviation for B1, B2, B3, B4
= �24.4� × 4 = 2.4 days
∴ Total standard deviation = 2.4 + 0.4 = 2.8 days
64. Two plates of 8 mm thickness ea.ch are connected by a fillet
weld of 6 mm thickness as shown in the figure.
The permissible stresses in the plate and the weld are 150 Mpa
and 110 Mpa, respectively. Assuming the length of the weld
shown in the figure to be the effective length the permissible load
P (in KN) is______
Answer: 60
Solution:
The permissible stress in the plate
= 150 Mpa
The permissible stress in the weld
= 110 Mpa
45
Strength of small width plate
= × = 50 × 8 × 150
= 60 × 103 N = 60 KN
( = gross sectional area of smaller width plate = 50 × 8 = 400 mm2)
(since smaller width place has least sectional area and least strength)
Effective length of fillet weld
= 2 × 100 + 50 = 250 mm
Size of fillet weld S = 6 mm
Effective throat thickness
= 0.707 × S = 0.707 × 6 = 4.242 mm
Strength of fillet weld
PS = × ×
= 250 × 4.242 × 110 = 116.65 × 103 N
= 116.65 KN
The permissible load P = smaller of and PS = 60 KN
The permissible load P (in KN) is 60
65. A Bracket plate connected to a column flange transmits a load
of 100 KN as shown in the following figure. The maximum total
force for which the bolt should be designed is ____ KN
Answer: 156.19
46
Solution:
Critical bolt or maximum forced bolt is one which is subjected to
maximum resultant shear force, which are farthest from c.g of
bolt group and which may be close to the applied load line. Here
Bolt No:1 and Bolt No:2 are critical and designed for maximum
total force.
An eccentric load may be replaced by set of one direct concentric
load (P) and twisting moment (M)
P = 100 KN;
M = P. e = 100 (600) = 6 × 104 KN – mm
Radial distance 1 = 2 = 3 = 4 = 106.07 mm;
cos = 75 = 0.707
106.07
Force in each bolt due to direct concentric load ( )
= = 100 = 20 KN
5
Force in critical bolt due to twisting moment ( )
= . . = 6 × 104 ×106.07 = 141.416 KN
∑ 2 4 ×106.072+0
Maximum total resultant force in critical bolt 1 & 2 is
FRmax = �F2a + F2m + 2FaFm cos θ
= √202 + 141.4162 + 2 × 20 × 141.416 × 0.707
FRmax = 156.19 KN
47
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