# Grand Test - 1 - PDF Flipbook

Grand Test - 1

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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
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.
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

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
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

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

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
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.
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

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

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

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

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.

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

Solution:

Using integration by parts

∫ = − ∫ 1 . = −

11

13. l i →m0 1 1− − 5 = ____
10 1− −

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] = _____

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
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
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

18. A bar, L metre long and having its area of cross-section A, is

stored in the bar is given by

a) 2

b) 2
2

c)
2

d)

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

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

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.

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 _______

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
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
a) Increase with time
b) Decrease with time
c) Remain unchanged
d) Fluctuate

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

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
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
a) 3.75 cm
b) 0.375 cm
c) 37.5 cm
d) 0.0375 cm
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

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%

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) −

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 _______

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

Solution:

Tresaguet construction was started in 1764 AD in France. Telford

construction was started in 1803 AD in London (England).

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

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
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

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

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

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
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)

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%

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 ______

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 _____

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 ______

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

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____

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

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
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

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

− 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

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 ______

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 _______

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

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 _______

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 _______

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
______

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

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_____

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

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

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 ________

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

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

Congress specifications should be

a) 4.5%

b) 4.75%

c) 5.0%

d) 5.25%

Solution:

Given R = 50 m

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

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

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

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_____

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