Grand Test - 1 - PDF Flipbook
Grand Test - 1
419 Views
93 Downloads
PDF 10,778,668 Bytes
GATE
CIVIL
GrandTests
Test-01Solutions
APTITUDE
One Mark Questions
1. Which of the following options is the closest in meaning to the
word below:
Circuitous
a) cyclic
b) indirect
c) confusing
d) crooked
Answer: b
Solution:
Circuitous means of a route or journey, long and not direct which
means indirect.
2. Which of the following options is the closest in meaning to the
sentence below?
“As a woman, I have no country"
a) Women have no country.
b) Women are not citizens of any country.
c) Women's solidarity knows no national boundaries
d) Women of all countries have equal legal rights.
Answer: c
Solution:
‘As a woman I have no country’
A. It doesn’t mean that woman have no country. The
expression here, it is quit literal in its sense.
1
B. How woman are not citizens of any country? It is not
taking about citizenship.
C. It is the right answer. As it is illustrated above the
sentence is quite metaphysical in its sense.
D. One nation of woman rights are different from the other
nations. Sentence has nothing to do woman’s rights.
3. Choose the statement where underlined word is used correctly.
a) The minister insured the victims that everything would be all
right.
b) He ensured that the company will not have to bear any loss
c) The actor got himself ensured against any accident
d) The teacher insured students of good results
Answer: b
Solution:
‘ensured’ means ‘make sure’ or ‘assure’.
4. There are two candidates P and Q in an election. During the
campaign, 40% of the voters promised to vote for P, and rest for
Q' However, on the day of election 15% of the voters went back
on their promise to vote for P and instead voted for Q 25% of the
voters went back on their promise to vote for Q and instead voted
for P. Suppose, P lost by 2votes, then what was the total number
of voters?
a) 100
b) 10
c) 90
2
d) 95
Answer: a
Solution:
P:Q
40 : 60 15% of 40 = 8 × 40 = 6
100
−6 ∶ +6 25% of 60 = 25 × 60 = 15
34 66 100
15% of 40 = 25 × 60 = 15
100
The difference between P and Q
∴ The total No. of voters are 100
5. The sum of the digits of a two-digit number is 12. If the new
number formed by reversing the digits is greater than the original
number by 54, find the original number.
a) 39
b) 57
c) 66
d) 93
Answer: a
Solution:
The new number formed by reversing the digits is greater than the
original number is possible in options A and B only.
Option ‘a’:
Sum of two digits in the number = 3 + 9 = 12
After reversing the two digits number = 93
3
The difference between the new number formed and original
number = 93 – 39 = 54
∴ option ‘A’ is correct.
Option ‘b’:
Original number = 57
After reversing, the number is formed = 75
The difference between these two numbers = 75 – 57 = 18
∴ It is not.
4
Two Marks Questions
6. Statement: There were different streams of freedom movements
in colonial India carried out by the moderates, liberals, radicals
socialists, and so on. Which one of the following is the best
inference from the above statement?
a) The emergence of nationalism in colonial India led to our
independence
b) Nationalism in India emerged in the context of colonialism
c) Nationalism in India is homogeneous
d) Nationalism in India is heterogeneous
Answer: d
Solution:
“Inference” is an act of drawing a conclusion from the given
information. We can draw the conclusion clearly from the words
‘different streams- moderates, liberals, radials, socialists, and so
on’.
A. It doesn’t support the above statement
B. It doesn’t support the above statement
C. It is definitely false as it is just opposite to the argument.
D. It can be inferred from the given information. i.e., different
streams is equal to heterogeneous.
5
7. Humpty Dumpty sits on a everyday while having lunch. The wall
sometimes breaks. A person sitting on the wall falls if the wall
breaks.
Which one of the statements below is logically valid and can be
inferred from the above sentences?
a) Humpty Dumpty always falls while having lunch.
b) Humpty Dumpty does not fall sometimes while having lunch.
c) Humpty Dumpty never falls during dinner.
d) When Humpty Dumpty does not sit on the wall, the wall does
not break.
Answer: b
Solution:
The key word to solve this question is ‘sometimes’
8. Today, we consider Ashoka as a great ruler because of the
copious evidence he left behind in the form of stone carved edicts'
Historians tend to correlate greatness of a king at his time with
the availability of evidence today. Which of the following can be
logically inferred from the above sentences?
a) Emperors who do not leave significant sculpted evidence are
completely forgotten.
b) Ashoka produced stone carved edicts to ensure that later
historians will respect him.
c) Statues of kings are reminder of their greatness.
d) A king's greatness, as we know him today, is interpreted by
historians.
6
Answer: d
Solution:
‘Today, historians correlate greatness of a king at his time with
the availability of evidence.’ This statement leads to the best
inference option ‘d’.
9. There friends, R, S and T shared toffees from a bowl. R took 1/3rd
of the toffees but returned four to the bowl. S took 1/4th of what
was left but returned three toffees to the bowl. T took half of the
remainder but returned two back into the bowl. If the bowl had
17 toffees left, how many toffees were originally there in the
bowl?
a) 38
b) 31
c) 48
d) 41
Answer: c
Solution:
The total No. of toffees in a bowl = x
= 2 +12 − �2 1 −224�
3
For R ⇒ = 8 +48−2 +24 � 3 − 4� = 6 +72
12 12
The remaining toffees in a bowl
− � 3 − 4�
= − + 4 = 3 − + 24 = (2 +12)
3 3 3
7
2 +12 2 +12 2 +12−36
12 12
For S ⇒ 3 −2 ⇒ − 3 =
4
= 2 +24
12
The remaining toffees in a bowl
2 +12 − �2 1 −224�
12
= 8 +48−2 +24
12
= 6 −72
12
6 +72 6 +72
24
For T ⇒ 12 −2 ⇒ − 2
2
= 6 +72−48 = 6 +24
24 24
The remaining toffees in a bowl
6 +72 − �6 2 −424�
12
= 12 +144−6 −24
24
= 6 +120
24
∴ 6 +120 = 17
24
6x + 120 = 17 × 24 = 408
6x = 408 – 120 = 288
∴ x = 280 = 48
6
Short cut Method:
Among all given alternatives, divisible by 3 is 48 only
For R ⇒ 48 = 16 ⇒ 48 – 16 = 32 + 4 = 36
3
For S ⇒ 36 = 9 ⇒ 36 – 9 = 27 + 3 = 30
4
8
For T ⇒ 30 = 15 ⇒ 30 – 15 = 15 + 2 = 17
2
∴ Answer 48
10. Few school curriculum include a unit on how to deal with
bereavement and grief, and yet all students at some point in their
lives suffer from losses through death and parting. Based on the
above passage which topic would not be included in a unit on
bereavement?
a) How to write a letter of condolence
b) What emotional stages are passed through in the healing
process
c) What the leading causes of death are
d) How to give support to a grieving friend
Answer: c
Solution:
a) Letter of condolence is obviously related to bereavement
and grief because it is written to the family which is going
through pain and grief.
b) This option even related to the situation and mental agony
a family goes through after a death in the family.
c) The leading causes of death are clearly not related to
bereavement and parting of deceased person. This is the
right choice.
d) Giving support to a grieving friend means dealing with
situation after death.
9
Explanation:
It is clear that we are supposed to find out the choice that is not
related to death or situation before the death. Its about healing
process a person goes through after losing a relative.
10
TECHNICAL
One Mark Questions 11� are
11. The eigen values of the matrix �
a) (a+1), 0
b) a, 0
c) (a-1), 0
d) 0, 0.
Answer: a
Solution:
= � 11�
⇒ | − | = 0
⇒ � − 1 � = 0
−
1
⇒ 2 − ( + 1) + 0 = 0
∴ = 0, + 1 ℎ
12. The integration of ∫ has the value
a) (x logx- 1)
b) log x – x
c) x (logx – 1)
d) None of the above
Answer: c
Solution:
Using integration by parts
∫ = − ∫ 1 . = −
11
13. l i →m0 1 1− − 5 = ____
10 1− −
Answer: 0.5
Solution:
Using L’ Hospital rule
l i →m0 1 × 5 − 5 = 5 = 0.5
10 − 10
14. Suppose Xi for i = 1, 2, 3 are independent and identically
distributed random variables whose probability mass functions
are Pr[Xi = 0] = Pr[Xi = 1] = Pr[Xi = 1] =1/2 for i = 1, 2, 3.
Define another random variable Y = X1 X2 (+) X3, where (+)
denotes XOR. Then Pr[Y = 0/X3 = 0] = _____
Answer: 0.75
Solution:
When X3 = 0, Y = X1X2
P[Y = 0 | X3 = 0]
= (X1=1,X2=0,X3=0)+ (X1=0,X2=1,X3=0)+ (X1= X2= X3= 0)
( 3=0)
= 21∙21∙21+ 12∙21∙21+ 12∙12∙12 = 3 = 0.75
4
1
2
15. Which one of the following Acts/Rules has a provision for "No
right to appeal"?
a) Environment (Protection) Act, 1986
b) The Hazardous Waste (Management and Handling)Rules,
1989
c) Manufacture, Storage and import of Hazardous Chemicals
Rules, 1989
12
d) Environment (Protection) Rules 1992
Answer: a
Solution:
Section-22 of Environment (Protection) Act, 1986has a provision
for 'No right to appeal'.
16. The entry of foul-smelling gases into the house coming from the
sewers can be prevented by
a) providing water seals for all the fixtures
b) providing water seals for all the fixtures and a vent pipe in the
plumbing system
c) providing sufficient vent pipes in the plumbing system
d) exhaust fans
Answer: b
17. Match List-I (Unit) with List-II (Purpose) and select the correct
answer using the codes given below the lists:
List-I
A. Leaping weir
B. Gutter inlet
C. Inverted syphon
D. Catch basin
List-II
1. To prevent grit, sand, debris, etc. from entering the storm
sewer.
2. To carry the sewer below a stream or railway line
3. To drain rain water from roads to the storm sewer
13
4. To separate storm water and the sanitary sewage
Codes:
ABCD
a) 4 3 1 2
b) 4 3 2 1
c) 3 4 2 1
d) 3 4 1 2
Answer: b
18. A bar, L metre long and having its area of cross-section A, is
subjected to gradually applied tensile load W. The strain energy
stored in the bar is given by
a) 2
b) 2
2
c)
2
d)
Answer: b
Solution:
Strain energy stored = 12P ×
= 1 × W × WL
2 AE
= W2L
2AE
14
19. If the depth is 8.64 cm on a field over a base period of 10 days,
then the duty is
a) 10 hectares per cum/s
b) 100 hectares per cum/s
c) 864 hectares per cum/s
d) 1000 hectares per cum/s
Answer: d
Solution:
D × ∆ = 864 × B
∴ D = 864 ×10 = 1000 ha/m3/s
8.64
20. Consider the following statements:
Irrigation water has to be supplied to the crops when the moisture
level falls
1. below field capacity
2. to wilting point
3. below wilting point
Which of these statements is/are correct?
a) 1
b) 2 only
c) 3 only
d) 2 and 3
Answer: a
Solution:
15
The soil moisture is not allowed to be depleted upto the wilting
point, as it would result inconsiderable fall in crop yield. The
optimum level up to which the soil moisture may be allowed to
be depleted in the root zone without fall in crop yield has to be
worked out for every crop and soil by experimentation. The
irrigation water should be supplied as soon as the moisture falls
up to this optimum level and its quantity should be just sufficient
to bring the moisture content up to its field capacity.
21. The discharge capacity required at the outlet to irrigate 3000 ha
of sugarcane having a kor depth of 173 mm and a kor period of
30 days is _______ m3/s.
Answer: 2.0 m3/s
Solution:
Required discharge capacity
= = 3000 × 104 ×173 × 10−3 = 2 m3/s
30 ×24 ×60 ×60
22. The total number of independent equations that form the Lacey's
regime theory is _______
Answer: 3
Solution:
A rigid bed canal has one degree of freedom i.e. for a given
channel a change in discharge would cause a change in the depth
only.
A man-made alluvial channel has three degrees of freedom i e.
for a given channel, change in discharge can cause changes in
width, depth and bed slope Thus three equations independent of
16
each other are needed, as given by Lacey, to represent three
degrees of freedom viz. depth, width and gradient. A natural
alluvial river has four degrees of freedom as its planiform can also
alter.
23. The stress level, below which a material has a high probability
not failing under reversal of stress, is known as
a) Elastic limit
b) Endurance limit
c) Proportional limit
d) Tolerance limit
Answer: b
Solution:
The stress which can be withstood for some specified number of
cycles is the fatigue strength of material. The stress level which
can be withstood for an infinite number of cycles, without failure
is called endurance limit.
24. The stresses in concrete in a reinforced concrete element under
sustained load due to creep
a) Increase with time
b) Decrease with time
c) Remain unchanged
d) Fluctuate
Answer: b
17
25. Limit state of serviceability for deflection including the effects
due to creep, shrinkage and temperature occurring after erection
of partitions and application of finishes as applicable to floors and
roofs is restricted to
a)
150
b)
200
c)
250
d)
350
Answer: c
Solution:
The deflection shall generally be limited to the following:
i) The final deflection due to all loads including the effects of
temperature, creep and shrinkage and measured from the
as-cast level of the support of floors, roofs and all other
horizontal members shall not normally exceed span/250.
ii) The deflection including the effects of temperature, creep
and shrinkage occurring after the erection of partitions and
the application of finishes should not normally exceed
span/350 or 20 mm whichever is less.
26. It is required to produce a small-scale map of an area in
magnetic zone by directly plotting and checking the work in the
field itself. Which one of the following surveys will be most
appropriate for purpose?
a) Chain
18
b) Theodolite
c) Plane Table
d) Compass
Answer: c
Solution:
Chain survey is used for moderately small area sand measurement
of ill-defined details e.g. edge of a marsh or for filling in detail
between already established points. Chain is used to measure the
length of the line and tape is employed for measurement of
offsets.
Compass survey is conducted mainly for angular measurements.
It is done for large areas with rough ground having many details.
However, magnetic zone affects the accuracy of compass by
introducing local attraction errors. The details can be plotted in
the office only.
Theodolite can be used for horizontal and vertical angle
measurement.
Plane table survey is most suitable for small and medium scale
mapping. The observations and plotting of details can be done
simultaneously. The errors and mistakes in plotting can be
checked by drawing check lines.
19
27. What is the slope correction for a length of 300 m along a
gradient of 1 in 20?
a) 3.75 cm
b) 0.375 cm
c) 37.5 cm
d) 0.0375 cm
Answer: a
Solution:
Slope correction = ℎ2
2
Difference in elevation between points,
h = 1 × 30
�(20)2+1
≃ 1.50 m
∴ Slope correction = (1.50)2
2 ×30
= 0.0375 m = 3.75 cm
28. If L is the length of the chain, W is the weight of the chain and
T is the tension, the sag correction for the chain line is
a) 2 2
24 3
b) 2 2
24 2
c) 2 2
24 2
20
d) 2 3
24 3
Answer: b
29. A clay sample has a void ratio of 0.50 in dry state and specific
gravity of solids = 2.70. Its shrinkage limit will be
a) 12%
b) 13.5%
c) 18.5%
d) 22%
Answer: c
Solution:
At shrinkage limit, soil is fully saturated.
WS = × 100 = 0.5 × 100
2.7
= 18.5%
30. A clayey soil has liquid limit = WL Plastic limit = Wp and natural
moisture content = W. The consistency index of the soil is given
by
a) −
−
b) −
−
c) −
−
d) −
−
Answer: a
21
31. The plasticity index and the percentage of grain size finer than
2 microns of a clay sample are 25 and 15, respectively. Its activity
ratio is _______
Answer: 1.67
Solution:
Activity
= Per cent Plasticity Index than 2μm = 25 = 1.67
of particles finer 15
As activity is more than 1.25 so it is active soil.
32. Which one of the following is the chronological sequence in
regard to road construction/design/development?
a) Telford, Tresaguet, CBR, Macadam
b) Tresaguet Telford, Macadam, CBR
c) Macadam CBR Tresaguet, Telford
d) Tresaguet, Macadam, Telford, CBR
Answer: b
Solution:
Tresaguet construction was started in 1764 AD in France. Telford
construction was started in 1803 AD in London (England).
Macadam construction was started in .1815 AD in England. CBR
construction was started in 1928 AD in USA.
33. A 3% downgrade curve is followed by a 1% upgrade curve and
rate of change of grade adopted is 0.1% per 20 m length. The
length of the respective vertical curve is ______ m
Answer: 800 m
22
34. Which of the following is the correct sequence to analyze a
project for implementation?
a)Time-cost study, Network, WBS, Scheduling with resource
allocation
b)Network, Time-cost study, Scheduling with resource
allocation, WBS
c)WBS, Network, Scheduling with resource allocation, Time-
cost study
d)WBS, Time-cost study, Network, Scheduling with resource
allocation
Answer: c
Solution:
In analyzing a project, the first step is always work Breakdown
Structure (WBS). From WBS, network can be prepared.
Thereafter scheduling with resource allocation follows. Finally,
time – cost study is carried out.
35. A point of C on a straight link OA moves with a velocity u from
O to A. OA itself turns with a velocity ω rad./s with plane Oxy.
The velocity of the point in the Oxy when it is at a distance r from
O is
a) +ω
b) √ 2 + ω 2 2
23
c) √ 2 − ω 2 2
d) √ 2 + ω 2
Answer: b
24
Two Marks Questions
36. The following systems of equations
1 + 2 + 3 = 3 has
1 − 3 = 0
1 − 2 + 3 = 3
a) Unique solution
b) No solution
c) Infinite number of solutions
d) Only one solution
Answer: b
Solution:
Given AX = B
1 1 1 1 3
�1 0 1� � 2� = �0�
1 −1 1 3 1
Consider the augmented matrix [ | ]
1 1 13
[ | ] = �1 0 1 �0�
1 −1 1 1
2 ⟶ 2 − 1; 3 ⟶ 3 − 1
1 1 13
∽ �0 −1 0 �−3�
0 −2 0 −2
3 ⟶ 3 − 2 2
1 1 13
∽ �0 −1 0 �−3�
0 0 04
( ) = 2, ( | ) = 3
25
, ( ) ≠ 2, ( | )
∴Solutions does not exist
37. The derivative of f(x, y) at point (1, 2) in the direction of vector
̅ + ̅ is 2√2 and in the direction of the vector – 2 ̅ is -3. Then the
derivative of f(x, y) in direction −��� − 2 ̅ is
a) 2√2 + 3
2
b) − 7
√5
c) −2√2 − 3
2
d) 1
√5
Answer: b
Solution:
Directional derivative of f in the direction of i + j = 2√2
⇒ ∇ . � = 2√2 ⇒ � ̅ + ̅ + � �
| � |
( +̅ ̅) = 2√2 ⇒ + = 4 ……. (1)
√2
Similarly,
∇ . � = – 3 ⇒ � ̅ + ̅ + � �
| � |
(−2 ̅) = – 3 ⇒ − 2 = –6 ⇒ = 4 ……. (2)
2
Sub. (2) in (1), we obtain, = 1
∴ ∇ . ̅ = � ̅ + ̅ + � � ∙ − −̅ 2 ̅ = −7
| |̅ √5 √5
26
38. f(x, y) = (x2 + xy) � + (y2+xy) � . Its line integral over the
straight line from (x, y) = (0, 2) to (x, y) = (2, 0) evaluates to
a) – 8
b) 4
c) 8
d) 0
Answer: d
Solution:
f(x, y) = (x2 + xy) � + (y2 + xy) �
∫ F�. d ̅ = ∫ ( 2 + ) + ( 2 + )
= x+ y = 2 ⟹ dx = – dy
∫02[ 2 + (2 − )] + [(2 − )2 + (2 − )](− )
=0
39. l i →m0 ( −1)+2( −1) = _____
( −1)
Answer: 1
Solution:
Using L’ Hospitals Rule
= l i →m0 ( −1)+ . −2 �00�
(1− )+ ( )
Again, using L’ Hospitals Rule
27
= l i →m0 + + . −2 �00�
+ +
Again, using L’ Hospitals Rule
= l i →m0 + + + . +2 �00�
+ + −
= 1+1+1+0+0 = 1
1+1+1−0
40. A town has an existing horizontal flow sedimentation tank with
an overflow rate of 17 m3/day/m2, and it is desirable to remove
particles that having settling velocity of 0.1 mm/sec. Assuming
the tank is an ideal sedimentation tank, the percentage of particles
removal is approximately equal to
a) 30%
b) 50%
c) 70%
d) 9%
Answer: b
Solution:
η = × 100
= 0.1 × 100 = 50.8%
�2147 ××13060000�
41. A straight 100 m long raw water gravity main is to carry water
from an intake structure to the jack well of a water treatment
plant. The required flow through this water main is 0.21 m3/s.
Allowable velocity through the main is 0.75 m/s. Assume f =
0.01, g = 9.81 m/s2. The minimum gradient (in cm/ 100m length)
28
to be given to this gravity main so that the required amount of
water flows without any difficulty is ______
Answer: 4.7 - 4.9
Solution:
Q = A.V
0.21 = d2 × 0.75
4
d = 0.597 m, L = 100 m
ℎ = 2 = 0.01 ×100 × (0.75)2
2 2 ×9.81 ×0.597
= 0.048 m
= 4.8 cm
42. Effluent from an industry 'A' has a pH of 4.2. The effluent from
another industry 'B' has double the hydroxyl (OH–) ion
concentration than the effluent from industry 'A'. pH of effluent
from the industry 'B' will be _____
Answer: 4.5
Solution:
(PH)A = 4.2 ⟹ (H+)A = 10 – 4.2
(OH –)A = 10 – 9.8 mol/lit
(OH –)A = 2 (OH –)A
= 2 × 10 – 9.8 mol/lit
= 3.169 × 10 – 10 mol/lit
(H+)B = 10−14 = 3.154 × 10−5 mol/lit
3.169 × 10−10
(PH)B = log10 1
(H+)B
29
= log10 3.154 1 10−5 = 4.5.
×
(PH) of effluent from industry B (PH)B = 4.5
43. A suspension of sand like particles in water with particles of
diameter 0.10 mm and below is flowing into a settling tank at 0.10
m3/s. Assuming g = 9.81 m/s2, specific gravity of particles = 2.65,
and kinematic viscosity of water = 1.0105 × 10-2 cm2/s. The
minimum surface area (in m2) required for this settling tank to
remove particles of size 0.06 mm and above with 100% efficiency
is ______
Answer: 31.214
Solution:
Vs = ( −1) 2 = 9.81[2.65−1]�0.06 × 10−3�
18 1
18 ×1.0105 × 10−2 × (100)2
= 3.20 × 10−3 m/sec
For 100% removal VS = Vo
Surface area = = 3.2 0.1 = 31.214 m2
× 10−3
44. The static Indeterminacy of the structure shown below is
a) Unstable
b) Stable, determinate
c) Stable, 5th degree indeterminate
d) Stable, 3rd degree indeterminate
Answer: d
30
Solution:
Dse = 6 – 3 = 3
Dsi = 3C = 3 × 1 = 3
Force releases at C = 3 – 1 = 2
Force releases at D = 2 – 1 = 1
Ds = Dse + Dsi – number of releases
= 3 + 3 – (2 + 1) = 3
45. Member 'AB' of the truss shown below has a lack of fit of mm.
B if E = 2 x 105 MPa, area of c/s = 20 mm2. The force in 'AB'
is____
Answer: 0
Solution:
The truss ‘ABC’ shown is statically determinate. determinate
structures are not subjected to stresses due to lack of fit,
temperature changes etc.
Hence force in AB = zero.
46. A uniform beam (EI = constant) PQ in the form of a quarter-
circle of radius R is fixed at end P and free at the end Q, where a
load W is applied as shown. The vertical downward
displacement, , at the loaded point Q is given by: =
� 3�. Find the value of (correct to 4-decimal places).
______
31
Answer: 0.7854
Solution:
Consider a Radial Vector which makes an angle ′ ′ with vertical
Moment at 'X ', Mx = W R sin
=
= 1 ∫0 �2 . .
= 1 ∫0 �2( )
= 3 ∫0 �2 2 = . 3
4
∴ =
4
∴ = 0.7854
32
47. In a cultivated area, the soil has porosity of 45% and field
capacity of 38%. For a particular crop, the root zone depth is 1.0
m, the permanent wilting point is 10% and the consumptive use
is 15 mm/d. If the irrigation efficiency is 60%, what should be the
frequency of irrigation such that the moisture content does not fall
below 50% of the maximum available moisture?
a) 5d
b) 6d
c) 9d
d) 15d
Answer: b
Solution:
To estimate the equivalent depth of water, the dry mass specific
gravity � = � is required. Since it is not given, assume mass
specific gravity of soil S = 1.3
Max. available depth of water, y
y = S.d[F.C – PWP]
= 1.3 × 1 [0.38 – 0.10]
y = 0.364 m
As the water content should not fall below 50% the depth of water
required to be supplied during irrigation,
= 50% of y = 0.182 m
Since irrigation efficiency is given it is accounted as follows.
Irrigation efficiency,
33
η =
0.60 = ℎ
0.182
∵ depth of water stored = 0.109 m or 109 mm
Frequency of irrigation,
f = ℎ
∵ Frequency of irrigation = 109 = 7.28 days
15
Nearest answer is 6 days
Another Method:
Since the required mass specific gravity � = � is not given, it
is assumed that the soil is fully saturated at its field capacity.
From this assumption, the mass specific gravity is estimated as
fallows.
FC = ℎ
ℎ
− Vv.γw
V.γd
= = �Since porosity, n = VVv�
0.38 = 0.45
S = 1.1
Max. available depth of water,
y = S.d[FC – PWP]
= 1.18 × 1[0.38 – 0.10]
= 0.33 m
Depth of water to be supplied during irrigation = 50% of y = 0.165 m
34
η =
0.6 =
0.165
Water stored = 0.09 m = 99 mm
Frequency of irrigation, f
= 99 = 6.6 days say 6 days.
15
[Note: Actually, the soil is partially saturated at its field capacity.
However, in the above case it is assumed to be saturated]
48. For the retaining wall shown in the given figure if the stress at
the heel is zero, then the maximum storage 'H' will be
Specific gravity of the material of the wall = 2.25
a) 7.5 m
b) 5 m
c) 4 m
d) 3 m
Answer: a
Solution:
The resultant of the hydrostatic pressure and weight of the dam
should pass through the outer third point.
B = H
√S−C
Considering zero uplift
35
Hmax = B√S = 5 × √2.25 = 7.5 m
49. A canal was designed to supply the irrigation needs of 1200
hectares of land growing rice of 140 days base period having a
Delta of 134 cms. If this canal water is used to irrigate wheat of
base period 120 days having a Delta of 52 cm, the area (in
Hectares) that can be irrigated is ______
Answer: 2650
Solution:
Duty, D = 8.64B∆
Discharge, Q = A = A.∆
D 8.64B
Discharge for Rice, QR = AR.∆R
8.64 BR
Discharge for wheat, Qw = Aw.∆w
8.64 Bw
Given condition is QR = Qw
∴ AR.∆R = Aw.∆w
BR Bw
1200 ×1.34 = Aw ×0.52
140 120
Area of wheat, Aw = 2650 hect.
50. Irrigation water is to be provided to a crop in a field to bring the
moisture content of the soil from the existing l8% to the field
capacity of the soil at 28%. The effective root zone of the crop is
70 cm. If the densities of the soil and water are 1.3 g/cm3 and 1.0
g/cm3 respectively, the depth of irrigation water (in mm) required
for irrigating the crop is _______
Answer: 91
36
Solution:
Present water content of soil, w = 18%
Field capacity, FC = 28%
Depth of root zone, d = 70 cm
Dry mass specific gravity of soil,
S = = 1.3 = 1.3
1
Depth of water required, y = S.d[FC – w]
= 1.3 × 70[0.28 – 0.18] = 9.1 cm
= 90 mm
51. Oil in a hydraulic cylinder is compressed from an initial, volume
2m3 to 1.96 m3. If the pressure of oil in the cylinder changes from
40 MPa to 80 MPa during compression, the bulk modulus of
elasticity of oil is ______ MPa
Answer: 2000 MPa
Solution:
K = (P2− P1) = P2− P1 = P2− P1
�−vd1v� −�v2v−1v1�
v1 − v2
v1
K = 80 − 40 = 2000 Mpa
�2−12.96�
52. For the state of plane stress (in MPa) shown in the figure
below, the maximum shear stress (in MPa) is _______
Answer: 5
37
Solution:
Maximum shear stress,
= �� −2 �2 + 2
= ��(−22)−4�2 + 42 = 5 MPa
53. A uniform beam weighing 1800N is supported at E and F by
cable ABCD. Determine the tension (in N) in segment AB of this
cable (correct to I-decimal place). Assume the cables ABCD, BE
and CF to be weightless _______
Answer: 1310 to 1313N
Solution:
∑ = 0
VA (4) = 1800 (2.5)
VA = 1125 N
∑ = 0
VA + VD = 1800N
VD = 675N
∑ = 0
38
(resultant reaction at D (FCD) should be along cable line CD).
VA (3) – HA (1) – 1800 (1.5) = 0
HA = 675N
= � 2 + 2
= √11252 + 6752 = 1311.96
(Range: 1310 to 1313N)
54. The beam of an overall depth 250 mm (shown below) is used in
a building subjected to two different thermal environments. The
temperatures at the top and bottoms surfaces of the beam are 36°C
and 72°C respectively. Considering coefficient of thermal
expansion ( )as l.50 × 10 –5 per °C, the vertical deflection of the
beam (in mm) at its mid-span due to temperature gradient is
______
Answer: 2.43
Solution:
From figure A1B1 = = 3 ( )
= � − �ℎ2�� = − 1 … … … (1)
2 2 = � + �ℎ2�� = + 2 … … . . (2)
(2) − (1)
ℎ( ) = ( 2 − 1)
39
1 1 = =
=
∴ ℎ � � = (∆ )
= ℎ = 250
(∆ ) (1.5×10−5)(72−36)
= 462.9
From geometry of circles
� 2 � . �2 � = (2 − ) (refer figure in question
number 2)
2 . − 2 = 2 ( 2)
4
= 2 = 32 = 2.43
8 8×462.9
Shortcut:
Deflection is due to differential temperature of bottom and top
(∆ =72o-36o=36o). Bottom temperature being more, the beam
deflects down.
As derived in the question 2
= � (∆8 ℎ ) 2� = 1.5×10−5×36×30002
8×250
= 2.43 ( )
55. Maximum strains in an extreme fiber in concrete and in the
tension reinforcement (Fe-415 grade and Es = 200 KN/mm2) in a
balanced section at limit state of flexure are respectively.
a) 0.0035 and 0.0038
b) 0.002 and 0.0018
40
c) 0.0035 and 0.0041
d) 0.002 and 0.0031
Answer: a
Solution:
= 0.0035
= + 0.002
1.15
= 415 + 0.002 = 0.0038
1.15×200×103
56. According to the concept of Limit State Design as per IS
456:2000, the probability of failure of a structure is_____
Answer: 0.0975
Solution:
The probability of failure of a structure as per IS:456 - 2000 is
0.0975
Probability of not failing under loads = 95% = 0.95
Probability of not failing under material strength criteria = 95%
= 0.95
Overall probability of not failing is = 0.95 × 0.95 = 0.9025
∴ Overall probability of failure = 1- 0.9025 = 0.0975
57. The local mean time at a place located in Longitude 90o 40' E
when the standard time is 6 hours and 30 minutes and the standard
meridian is 82o 30'E is?
a) 5 hrs, 2 min and 40sec
b) 5 hrs, 57 min and 20sec
c) 6 hrs, and 30 min
41
d) 7 hrs, 2 min and 40sec
Answer: d
Solution:
Local meridian = 90o 40ˈ E
Standard meridian = 82o 30ˈ E
Standard time = 6 to 30 min
Local meridian is a head of standard meridian by 8o 10ˈ E
For 360o → 24 hrs
For 8o 10ˈ → 8o10ˈ × 24 = 0 h 32 min 40 sec
360
Local mean time is ahead by 0 h 32 min 40 sec
∴ Local mean time
= 6 h 30 min + 0 h 32 min 40 sec
= 7 hrs 2 min 40 sec
58. The laboratory tests on a soil sample yields the following
results; natural moisture content = 18% liquid limit: 60%, plastic
limit: 25%, percentage of clay sized fraction 25%. The liquidity
index and activity (as per the expression proposed by Skempton)
of the soil, respectively, are
a) – 0.2 and 1.4
b) 0.2 and 1.4
c) – 1.2 and 0.714
d) 1.2 and 0.714
Answer: a
Solution:
Given
42
w = 18%, wL = 60%, wP = 60%, C = 25%
liquidity index,
IL = w− wP = 18 − 25 = b = – 0.2
WL− WP 60 − 25
IP = WL − WP = 60 – 25 = 35
Activity, A = IP = 35 = 1.4
C 25
59. For sand of uniform spherical particles, the void ratio in the
loosest state is ________
Answer: 0.91
Solution:
If each particle is assumed to be perfectly spherical and if they
occupy the loosest and densest states as shown in the above
figures, the void ratios will work out to be 0.91 and0.35
respectively. However, these values are not applicable to the
practical soils in which the particles are neither spherical nor they
occupy the states shown above.
60. The difference between the free water levels in two wells 48 m
apart in an aquifer is 0.6 m. It took an interval of 8 hours between
detecting traces of tracer material at the two wells in succession.
The porosity of the aquifer is 25%, the coefficient of permeability
of the aquifer is ______ cm/sec
Answer: 3.33
43
Solution:
V = ki
= khL
= k�04.86�
Seepage velocity, Vs = V
n
n = 0.25
Vs = L
t
Vs = 48 × 102 cm/sec (given)
8 ×60 ×60
∴ 48 × 102 = 0.125.K�04.86�
8 ×60 ×60
∴ k = 3.33 cm/sec
61. While designing a hill road with a ruling gradient of 6%, if a
sharp horizontal curve of 50 m radius is encountered, the
compensated gradient at the curve as per the Indian Road
Congress specifications should be
a) 4.5%
b) 4.75%
c) 5.0%
d) 5.25%
Answer: a
Solution:
Given R = 50 m
Ruling gradient = 6%
Grade compensation = 30+R
R
44
= 30+50 = 1.6%
50
(or)
75 = 75 = 1.5%
R 50
Use minimum, i.e. = 1.5%
Hence compensated gradient = 6 – 1.5 = 4.5%
(as per IRC compensated gradient should not be less than 4%)
62. The radius of a horizontal circular curve on a highway is 120 m.
The design speed is 60 km/hour, and the design coefficient of
lateral friction between the tyre and the road surface is 0.15. The
estimated value of super elevation required (if full lateral friction
is assumed to develop), and the value of coefficient of friction
needed (if no super elevation is provided) will, respectively, be
a) 1 and 0.10
11.6
b) 1 and 0.37
10.5
c) 1 and 0.24
11.6
d) 1 and 0.24
12.9
Answer: c
Solution:
Design speed, R = 120 m
Designed speed, V = 60 km/hr (16.67 m/s)
Coefficient of lateral friction, f = 0.15
i) Superelevation for the development of full friction
e + f = v2 ⟶ e + 0.15 = 16.672
gR 9.87 ×120
45
e = 0.086 = 1 ≃ 1
11.63 11.6
ii) For no superelevation coefficient of friction required is
e + f = v2
gR
0 + f = 16.672
9.81 ×120
f = 0.236 = 0.24
63. A road is provided with a horizontal circular curve having
deflection angle to 55o and centre line radius of 250 m. A
transition curve is to be provided at each end of the circular curve
of such a length that the rate of gain of radial acceleration is 0.3
m/s3 at a speed of 50 kmph. Length of the transition curve
required at each of the ends
a) 2.57 m
b) 33.33 m
c) 35.73 m
d) 1666.67 m
Answer: c
Solution:
R = 250 m; C = 0.3 m/s3
v = 50 × 0.27 = 13.9 m/s.
length of transition curve based on comfort criteria.
LS = v3 = 13.93 = 35.73 m
CR 0.3 ×250
64. Equipment A has first cost of ₹ 1800 and annual operating cost
of ₹ 500. Equipment B has first cost of ₹ 1300 and annual
46
operating cost of ₹ X. The life of each equipment is 5 years.
Interest rate adopted is 8%, for which capital recovery factor for
5 years is 0.25. For indifference between the two pieces of
equipment, the value of X is
a)₹ 575
b)₹ 600
c)₹ 625
d)₹ 650
Answer: b
65. A particle of mass 2 kg is travelling at a velocity of 1.5m/s. A
force ( ) = 3 2( ) is applied to it in the direction of motion
for a duration of 2 seconds, where t denotes time in seconds. The
velocity (in m/s, up to one decimal place) of the particle
immediately after the removal of the force is_____
Answer: 5.5
Solution:
= 2kg; velocity = 1.5m/s = initial
f(t) = 3t2
f(t) = m dv ⇒ f(t)dt = mdV
dt
∫02 3t2 dt = m ∫1V.5 dV
�33t3 2 = m[V]1V.5 ⇒ 23 = 2 × [V − 1.5] ⟹ 8 = 2V − 3
�
0
V = 8+3 = 5.5m/s
2
47
Data Loading...