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Disclaimer :
The questions have been shared by students only. Since the questions are
purely memory based, , does not take any responsibility
regarding the correctness of questions.
In case there is any kind of discrepancy in questions please write to us at
or call us at :
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(Morning Session)
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1. Ranjan was not happy that Sajan decided to do the project on his own on observing his unhappiness,
sajan explained to Rajan that he preferred to work independently
Which one of the statement below is logically valid and can be inferred from the above
sentences ?
1 mark
(a) Rajan has decided to work only is group.
(b) Rajan and Sajan were formed into a group against their wishes.
(c) Sajan decided to give into Rajans request to work with him.
(d) Rajan had believed that Sajan and he would be working together.
Sol. (d)
2. I f y = 5x2+3 than the tangent at x = 0, y = 3
1 mark
(a) passes through x = 0, y = 0
(b) has a slope +1
(c) is paralled to x-axis
(d) has a slope of 1
Sol. : (c)
y =5x2+ 3
dy
dx=10x
tangent at x =0, y = 3 is equal to zero
tangent is parallel to x-axis.
3. A student is required to demonstrate a high level of comprehension the subject, espesially in the
social sciences.
The word closest in meaning to comprehension is
1 mark
(a) understanding (b)Meaning
(c) Concentration (d) Stability
Sol. (a)
4. Choose the most appropirate word from the option given below to complete the following sentences,
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one of his biggest ................... was his ability to forgive
1 mark
(a) vice (b) virtues
(c) choices (d) strength
Sol. (b)
5. A boundary has a fixed daily cost of Rs 50,000 whenever it operates and variable cost of Rs. 8000
Q, where Q is the daily production in tonnes. What is the cost of production in Rs. per tonne for
a daily production of 100 tones.
1 mark
Sol. :
Total cost for 100 tonn
=8000 100 + 50000
=8,50,000
Cost per tonn =850000
Rs.8500/tonn.=100
6. Find the odd one in the following group ALRVX, EPVZB, ITZDF, OYEIX
(a) ALRVX (b) EPVZB
(c) ITZDF (d) OYEIX
2 marks
Sol. (d)
7. One percent of the people of country X are taller than 6ft, 2 percent of the people of country Y
are taller thatn 6ft. There are thrice as many people in country X as in country Y. Taking both
countries together. What % of people taller than 6ft.
(a) 3 (b)2.5
(c) 1.5 (d)1.25
2 marks
Sol. (d)
Let the population of x = Nxand population of y = Ny
Nx= 3Ny
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ESE - 2014
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particular topic appears invarious tests is based onits importance in exam.(See schedule of topicsbelow)
2. The test has beendesigned in such a waythat it covers all conceptsinvolved in any chapter insome test or the other.
3. Immediately after each teststudents will be asked tofill up self evaluationperforma so that they donot missout on thedifficulties they faced inparticular tests and whatthey should have done toovercome it.
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Total no. of people taller than 6ft =yx
2NN 1
100 100
% of people taken than 6ft =
y y
y y
3N 2N100
1003N N
=1.25
8. The sum of eigen value matrix (M) is
When [ M] =
215 650 795
655 150 835
485 355 550
2 marks
(a) 915 (b) 1355
(c) 1640 (d) 2180
Sol. (a)
9. The smallest angle of a triangle is equal to two third of the smallest angle of a quadrilateral. The
ratio between the angle of the quadritateral is 3 : 4 : 5 : 6. The largest angle of the triangle is
twice its smallest angle what is the sum, in degrees of the second largest angle of the triangle
and the largest angle of the quadritateral.
2 marks
Sol. :
3
2
1
3
5
4
6 18 360=
20=
largest angle of quadrilateral =120
smallest angle of quadrilateral = 60
Smallest angle of triangle = 2
40=3 203
Largest angle of triangle =2 40 = 80
Third angle of triangle =60
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Sum of largest angle of quadrilateral and second largest angle of triangle.
=120 + 60 = 180
10. The monthly rainfall chart based on 50 years of rainfall in Agra in shown in the following figure
which of the following are true ? (K percentile is the value such that K % of data fall below that
value)
800
700
600
500
400
300
200
100
Feb Mar Apr May une uly Aug Sep Oct Nov Dec
650
600
400
200
100
50
Rainfall (mm)
650
(i) On average it rains more in J uly than in Dec.
(ii) Every year, the amount of rainfall in August is more than that in J anuary.
(iii) J uly rainfall can be estimated with better confidence than Feb. rainfall.
(iv) In Aug, there is at least 500 mm of rainfall.
(a) (i) and (ii) (b) (i) and (iii)
(c) (ii) and (iii) (d) (iii) and (iv)
Sol. (b)
11. Some of the non-toxic metal normally found in natural water are :
(a) Arsenic, Lead, Mercury (b)Calcium, Sodium, Silver
(c) Cadmium, curomium, copper (d) Iron, Mangnese, Magnesium
Sol. (d)
12. An incompressible homogeneous fluid flowing steadily in a variable dia pipe having the large and
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14. The minimum value of 15 minutes peak hour factor on a section of a road is
(a) 0.1 (b) 0.2
(c) 0.25 (d) 0.33
1 mark
Sol. : (c)
15 min peak hr factor is used for traffic intersection design
PHF =
15
V/4V
V =Peak hourly volumeveh.
inhr
V15 =Max 15 Min Vol. within the peak hr (veh.)
Max value is 1.0 & min value is 0.25
Normal range is 0.7 0.98
Ans is 0.25
15. the ultimate collapse load (Pu ) is terms of plastic moment Mpby kinematic approach for a propped
cantilever of length L with P acting at its mid span as shown in fig would be
L/2 L/2
(a) P =p2M
L(b)P =
p4M
L
(c)p6M
L(d)
p8M
L
1 mark
Sol. : (c)
2
L/2 L/2
MP
Pu
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L
2 =
External work done =load defleciton
= uP
= uL
P2
Internal work done = p pM M 2
= p3M
By principle of virtual work,
u
LP
2
= p3M
uP =p
6M
L
16. for a saturated cohesive soil, a triaxial test yeilds the angle of interval friction ( ) as zero. The
conducted test is
(a) Consolidated Drained (CD) test
(b) Consolidated undrain (CU) test
(c) Unconfined compresion (UC) test
(d) Uncosolidated undrain (UU) test
1 mark
Sol. : (d)
17. The degree of static indeterminacy of a rigid jointed frame PQR supported as shown is
1 mark
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y
R
EI
Q
90
S
Px
EI
145
Sol. : (d) (Unstable)
18. The action of negative friction on the pile is to
(a) Increase the ultimate load on the pile
(b) Reduce the allowable load on the pile
(c) Maintain the working load on the pile
(d) Reduce the settlement
1 mark
Sol.(b)
19. In a beam of length L four influence line diagram for shear at a section located at a distance
of L /4 from the left support marked (P, Q, R, S) are shown below the correct, I LD is .....
(a)
0.75
L/4 3 L/4
P0.25
(b)
0.6
L/4
0.6
3 L/4
Q
(c)
0.5
L/4 3L/4
0.5R
(d) L/4 3 L/4S
A (P), B (Q) C (R) D (S)
Sol. : (a)
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y
x
L/4
3L/4
xL
4
=y
3L
4
x =y
3
also x+y =1
y
y3
=1
4y
3=1
y =3
4
x = 14
Ans is (a)
20. A conventional flow duration curve is a plot between 1-marks
(a) Flow and % time flow is exceeded
(b) Duration of flooding and ground level elevation
(c) Duration of water supply in a city and proportion of area recurring supply exceeding this
duration.
(d) Flow rate and duration of time taken to empty of a reservoir at that flow rate.
1-marks
Sol. (a)
21. The following statements are related to temperature stress developed in concrete pavement slab
with for edge (without any restrain)
P : The temperature stress will be zero during both day and night time if the pavement slab
is considered weight less.
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Q. : The temp stress will be compressive at the bottom of the slab during night time if the self
weight of the pavement slab is considered.
R : The temp stress will be compressive at the bottom of the slab during day time if the self
weight of the pavement slab is considered.
1 mark
The Ture statements is/are :
(a) P only (b) Q only
(c) P and Q only (d) P and R only.
Sol. : (c)
22.x
x sinxlim
x
is equal to
1 marks
(a) (b) 0
(c) 1 (d)
Sol. : (a)
x
x sinxlim
x
=x
sinxlim 1
x
=1
23. A steel section is subjected to a combination of shear and bending action. The applied shear force
is V and shear capacity of the section is Vs. For such a section, high shear force (as per IS 800
- 2007) is defined as
1 mark
(a) V > 0.6Vs (b) V > 0.7Vs
(c) V > 0.8 Vs (d) V > 0
Sol. : (a)
Clause 9.2.1 IS 800:2007
24. The dimension for kinematic viscosity is :
(a)L
MT(b) 2
L
T
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(c)
2L
T (d)ML
T
Sol. : (c)
SI unit is m2/s
25. In reservoir with an uncontrolled spillway the peak of the plotted outflow hydrogroph
1 mark
(a) Lies outside the plotted inflow hydrograph.
(b) Lies on the recession limb of the plotted inflow hydrograph.(c) Lies on the peak of the inflow hydrograph.
(d) is higher than peak of the plotted inflow hydrograph.
Sol. : (b)
inflow hydrograph
outflow hydrograph
Recession limb
26. For a sample of water with the ionic composition shown below, the Carbonate and Non-carbonate
hardness concentration (in mg/l as CaCO3) respectively are.
meq/l
meq/l
0 4 5 7
Ca2+ Mg
2+Na
+
HCO3 SO42
3.5 7
2 marks
(a) 200 and 500 (b) 175 and 75
(c) 75 and 175 (d) 50 and 200
Sol. :
Carbonate hardness =3.5 103g eq [if NCH is present sodium alkalimitry will be
absent i.e. NaHCO3absent]
=3.5 10350g
las CaCO3
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=175 mg/l as CaCO3
Non carbonate hardness = total hardness - carbonate hardness
Total hardness = 5 50 mg/l as CaCO3[total hardness is due to Ca2+and Mg2+]
= 250 mg/l as CaCO3
NCH = 250 175 = 75 mg/l as CaCO3
27. The perception - reaction time for a vechile travelling at 90 km/h, given the co-efficienc of
longitudinal friction of 0.35 and the stopping sight distance of 170 m(assume g = 9.81 m/s2) is
...... seconds.
2 marks
Sol. :
SSD =2V
Vt2g
170 =
25
905 18
90 t18 2 9.81 0.35
t =3.1594 sec.
28. A traffice surveying conducted on a road yield an average daily traffic count of 5000 vehicle. The
axle load distribution on the same road is given in the following table.
Axle Load(tones)
Frequency ofTraffic (f)
18141086
1020351520
the design period of the road is 15 years. The yearly Traffice growth rate is 7.5% the load safety
factor (LSF) is 1.3. I f the vehicle damage factor (VDF) is calculated from he above data, the design
traffic (In million standard axle load MSA) is
2 marks
Sol. :
Calculation of vehicle damage factor :
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VDF =
4 y 4 4 4
3 51 2 41 2 3 4 5
s 3 s s s
1 2 3 4 5
W WW W WV V V V VW W W W W
V V V V V
where Ws= standard axle load = 80 kN = 8.2 tonn
VDF =
4 4 4 4 418 14 10 8 6
10 20 35 15 208.2 8.2 8.2 8.2 8.2
10 20 35 15 20
=4.988.
N =15365 5000 (1.075) 1
4.9880.075
=237.758 MSA
Note : that LSF is used to calculate streses not for MSA calculation.
29. Mathematical idealization of a crane has three bar with their vertices arranged as shown with
load of 80 kN hanging vertically. The co-ordinate of the vertices are given in parenthesis. The
forces in members QR is ..................
2 marks
P (0,4)
80kN22.84
104.03
(3,0)
R
55.13(1, 0)
Q
x
y
Sol. :
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P (0,4)
80 kN22.84
104.03
(3,0)R
55.13
(1, 0)
Q
R
RR
3m
4m
1m
Q RR R 80
80 3 = RQ 2
RQ= 120 kN
RR= 40 kN
tan =1 1 4
sin , cos4 17 17
tan =3 4
Sin , cos4 5 5
120
A
A40
P
80
120
FPR
FQR
taking moment of all forces about P.
pM 0
120 1, + FQR 4 =0
FQR = 30 kN
30. The reinforced concrete section, the stress at extreme fibre in compression is 5.8 MPa. The depth
of Neutral Axis in the section is 58 mm and grade of concrete is M25. Assuming Linear elastic
behavior of the concrete, the effective curvature of the section (in per mm) is
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(a) 2 106 (b) 3 106
(c) 4 106 (d)5 1062 mark
Sol. :
5.8 MPa
58 mm
M25 conrature
M f E
I y R , E = 5000 25 25000 N/mm2
26
2
1 f 5.8N / mm4 10 per mm
R yE 58mm 25000N /mm
curvature = 4 106per mm
31. An isolated three-phase traffic signal is designed by websters method. The critical flow ratio for
three phase are 0.2, 0.3 and 0.25 respectively and lost time per phase is 4 second. The optimum
cycle length (in sec.) is ........
2 mark
Sol. :
Total lost time in a cycle L = 4 3 = 12 sec.
C =1.5L 5
1 y
:
1.5 12 592sec1 (0.2 0.3 0.25)
32. Group - I Group - II
P : Curve J (i) No apporent heaving of soil around the footing.
Q Curve K (ii) Rankine passive zone develops imperfectly
R : Curve L (ii i) Well defined slip surface extends to ground surface.
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Load
K L
Settlement
(a) P -Q, Q - 3, R - 2
(b) P - 3, Q - 2, R - 1
(c) P - 1, Q-2, R - 3
Sol. : (d)
L General sheer failre
K local shear failre
J Punching sheer failre
33. 16 MLD of water is flowing through a 2.5 km long pipe of diameter 45 cm. The chlorine at the
rate of 32 kg/d is applied at the entry of this pipe so that disinfected water is obtained at the exit.
These is a proposal to increase the flow through the pipe to 22 ML D from 16 mLD. Assume the
dilution co-efficient n = 1. The minimum amount of chlorine (in kg per day) to be applied to
achieve the same degree of disinfection for the enhanced flow.
(a) 60.5 (b) 4.4
(c) 38 (d) 23.27
2 mark
Sol. : In the disinfection process we have the relationship,
tCn =K
where t = time required to kill all organism.
c = cocentration of disinfactant
n = dilution coefficient
k = constant
n1 1t C =n
2 2t C
in our case n = 1
1 1 2 2t C t C
t1 =1
L
v , L = length of pipe ; V1= velocity of flow
t1 =1
L
Q / A
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t1
=1
LA
Q
C1 =1
1
W
Q , where W1= weight of disinfactant per day ; Q1= discharge per day
1
1 1
WLA
Q Q =
2
2 2
WLA
Q Q
2W =22
121
QW
Q =
222
32kg/day16
= 60.5 kg/day
Ans. (a)
34. A levelling is carried out to established the reduced level (RL) of point R with respect to the bench
mark (BM) at P. The staff reading taken are given below
Staff station BS IS FS RL
P 1.655
Q 0.95 1.5
R 0.75
I f RL of P is + 100 m then RL (m) of R is
(a) 103.355 (b) 103.155
(c) 101.455 (d) 100.355 2 mark
Sol. : (c)
station BS FS
P 1.655
Q 0.95 1.5
R 0.75
RL of P =100
RL of Q =100+1.655+1.5
=103.155
RL of R =103.155 (0.95 + 0.75)
=103.155 1.70
RL of R =101.455
35. The flow net constructed for a dam is shown in the figure below. Taking coefficient of permeability
as 3.8 106 m/s, the quantity of flow (in cm3/sec) under the dam per m is
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50 m
6.3 m
1.6 m
17.2 m
9.4 m
2 mark
Sol. :
Q =
f
d
NKH
N
Nd =10
Nf =3
H =6.3 m
K =3.810-6m/s
Q =6 m 3
3.8 10 6.3ms 10
=3
6 m /s7.182 10
m
=3
6 6 cm /s7.182 10 10
m
Q =7.1823
cm /s
m
36. The tension and shear force (both in kN) in each both of the joint as shown below respectively
are
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4
53
P = 250 kNu
(a) 30.3 and 20 (b)30.33 and 25
(c) 33.33 and 20 (d)33.33 and 25
2 mark
Sol. : tan =3
4
cos =4
5
u4Pu
P cos5
Pu
u
3PuP sin
5
sin =3
5
Tension in each bolt =u4P
5 6
=4 250
6 5
= 33.33 kN
Shear in each bolt =u3P
5 6= 25 kN
Ans. (d)
37. An incompressible fluid is flown at steady rate in horizontal pipe. From a section the pipe divides
into two horizontal parallel pipes of diameter (d1and d2) that run, for a distance of L each and
then again join back to a pipe of the original size. For both the pipes, assume the head loss due
to friction only and the Darcy-weilbach friction factor to be same. The velocity ratio betweenbigger and smaller branched pipe is.
Sol. :
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B
D
C
d1
A
Q1
Q2d2
L
L
1d = 24d
21
1
flv
2gd=
22
2
flv
2gd
21
1
v
d =
22
2
v
d
2122
v
v=
1
2
d
d = 4
1
2
v
v =2
Ans. 2
38. A box of weight 100 kN shown in the figure to be lifted without swinging. If all the forces are
coplanar, the magnitude and direction ( ) of force F w.r.t. x axis is
30
45
40 kNF
90 y
100 kN
X dir
(a) F = 56.389 kN and = 28.28
(b) F= 56.389 kN and = 28.28(c) F = 9.055, = 1.1414
(d) F = 9.055, = 1.1414
Sol. : For no swinging HorizontalF 0
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3045
40 kN F
x-axis
90kN
100 kN
90 cos 30 = 40 cos 45 + F cos
49.658 F cos
F cos from option (A) = 56.389 cos 28.28 = 49.658 kN
Ans. (a)
39. The speed-density (v k) relationship on a single lane road with unidirectional flow is
v = 70 0.7 K, where v is in km/hr and k is in veh/km. The capacity of the road (veh/hr) is
2 mark
Sol. :
u 70 0.7 K
km veh
hr km
100 =kJ
v =70f
u
k
Capacity =f jV K
4=
km Veh70 100hr km
4
= 1750 veh/hr
Ans. 1750 veh/hr
40. For a cantilever beam of span 3m as shown a concentrated load of 20 kN applied at the free end
causes a vertical displacement of 2 mm at a section located at a distance of 1m from the fixed
end. I f a concentrated vertically downward load of 10 kN is applied at the section located at a
distance of 1 m from the fixed end (with no other load on the beam) the maximum vertical
displacement in the same beam (in mm) is
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2mm
20 kN
1m2
2 mark
Sol. :
20kN
2mm
1m 2m
1m10kN
2m
20 =10 2 (From Betts Therorem
=1 mmAns = 1mm
41. For a beam cross section W = 230 mm, effective depth = 500 mm, the number of rebars of 12
mm diameter required to satisfy minimum tension reinforcement requirement specified by
IS -456:2000 (assume grade of steel as Fe500) is ..............
Sol. :
stminA
bd=
y
0.85
f
Ast =20.85 230 500mm
500
n =
st2
A
d
4
= 2
0.85 2302nos
(12)4
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42. A venturimeter having diameter of 7.5 cm at the throat and 15 cm at the enlarged end is installedin a horizontal pipline of 15 cm diameter. The pipe carries incompressible fluid at steady rate of
36 l/s. The difference of pressure head measured in terms of the moving fluid in between the
enlarged and the throat of the vent is observed to be 2.45 m. Taking the g = 9.8, the coefficient
of discharge of venturimeter (correct upto 2 decimal) is .......
2 mark
Sol. :
15 cm
7.5 cm
1 2
21
Q =30 l/s
1 21 2
P PZ Z
=2.45 m = h
g =9.81
Q =1 2
d2 21 2
a aC 2gh
a a
Cd =1 2
2 21 2
Q
a a2gh
a a
=
3 3
2 2
4 4
30 30 m / s
(0.15) (0.075)4 2 9.81 2.45
(0.15) (0.075)
Cd =0.95
43. A particles moves along a curve whose parametric equation are x = t3+ 2t, y = 3e2tand z
= 2 sin (5t), whre x, y and z show variation of the distance covered by the particles in (cm) with
time (t) (in second). The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is ......
2 mark
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Sol. :
x =t3+2t
y =3 e2t
z =2 sin (5t)
dx
dt= 3t2+ 2
ax=2
2
d x
dt=6t
dydt
=3 e2t (2) = 6e2t
ay =2
2
d y
dt=12 e2t
dz
dt=5 2 cos (5t)
2
2
d z
dt=10 5 sin (5t) = 50 sin 5t
az =
2
2
d z
dt =50 Sin 5t
a
= x y z a i a i a k
a
at t =0 = 0i 12j 0k
a
= 12j
Meg. of acceleration at t = 0 = 12 cm/s2
Ans =12 cm/s2
44. The full data are given for laboratory sample 0 175kPa , 0 0 0e 1.1, 300kPa, e 0.9 .I f thickness of the clay specimen is 25 mm the value of coeffcient of volume compressibility is
...... 104m2/kN.
2 mark
Sol. :
mv =0 v
0
e
1 e a
1 e
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=(1.1 0.9)
125
2.1
=0.2
2.1 125
=7.619 104m2/kN
Ans is = 7.619
45. A given cohesionless soil has emax= 0.85, emin= 0.5. In the field, the soil is compasted to a mass
density of 1800 kg/m3at w/c of 8%. Take the mass density of water as 1000 kg/m3and Gs= 2.7.
The relative density (in %) of the soil is
(a) 56.43 (b) 60.25
(c) 62.87 (d)65.41
2 mark
Sol. : (d)
emax =0.85
emin =0.5
field =1800 kg/m3at w/c = 8%
w =1000 kg/m3
Gs =2.7
Relative density = Dr= ?
= wG 2.7 10001 w 1.08=
1 e 1 e
1 + e =2.7 1000 1.08
1800
e =0.62
Dr =Relative density = maxmax min
e e100e e
=0.85 0.62
1000.85 0.5
=65.714%
46. I f the following equaiton establishes equilibrium in slightly bent position,
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2
2d ydx
+ py 0EI
the mid-span deflection of a member shown in figure is
P
M
yEI
P
NL
y
x
[a is the amplitude contant for y]
(a) y =1 2
1 acosP L
x
(b) y =1 2
1 asinP L
x
(c) y = asinn x /L (d)y = acosn / Lx
2 mark
Sol. : (c)
47. Three rigid bucket are of idential height and base area. Further assume that each of these
buckets have negeligiuble mass and are full of water. The weight of water in these bucket are
denoted by W1, W2, W3respectively. Also let the force of water on the base of the bucket be F 1,
F2, F3respectively. Which of the following option are correct.
hhh
All these bucket have the same area
(a) W2= W1= W3and F2> F1> F2
(b) W2> W1> W3and F2> F1> F3
(c) W2= W1= W3and F1= F2= F3
(d) W2
> W1
> W3
and F2
= F1
= F3
2 mark
Sol. (d)
Bucket identical height
identical Base area
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h h h
AA
1 2 3
2 1 3W W W
Force on the base in each case will be equal to = w hA
Hence F1= F2= F3
48. A horizontal jet of water with its cross section area 0.0028m2hits a fixed vertical plate with a
velocity of 5 m/s. After impact the jet split symmetrically in a plane parallel to the plane of the
plate. The force of impact (in N) of the jet on the plate is
(a) 90, (b)80
(c) 70 (d)60
2 mark
Sol. : (c)
5m/s
area of jet =0.0028m2
Force on plate = waV V
= 2wav = 1000 0.0028 (5)
2
=70 N
49. A rectangular beam of 230 mm width and effective depth = 450 mm, is reinforced with 4 bars
of 12 mm diameter. The grads of cencrete is M 20, grade of steel is Fe 500. Given that for M20
grade of concrete, the ultimate shear strength uc = 0.36 N/mm2for steel percentage of = 0.25,
and 2uc 0.48N / m for steel percentage =0.5. For a factored shear force of 45 kN, the diameter
(mm) of Fe500 steel 2 legged stirrups to be used at spacing of 325 mm should be
(a) 8 (b)10
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(c) 12 (d)16
2 marks
Sol. (a)
230
450
4 12
uc =0.36 N/mm2 [For M 20] for
stA 100 0.25bd
uc =0.48 N/mm2 [For M 20] for
stA 100 0.5bd
Factored SF =45 kN = Vu
We have to calculate the dia of Fe500 2-L egged stirrup to be used at a spacing of 325 mm c/
c.
v =uV
bd =
45 1000
230 450
= 0.4348 N/mm2
% tensile steel =
24 (12)4 100
230 450
=0.437 %
c =0.12
0.36 (0.437 0.25)0.25
=0.45 N/mm2
Since v c 0
Min shear reinforcement is required
Min sheer reinforcement is given by
sv
v
A
bS =0.4
0.87fy
Asv =v0.4 (S )(b)
0.87fy
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since we limit fyto 415 N/mm2hence,
Asv =22 ( )
4
=
20.4 325 230 82.814mm0.87 415
=7.26 mm
adopt =8 mm
50. A straight 100 m long raw water gravity main is to carry water from intake to the jackwell of
a water treatmented plant. . The required flow of w =0.25 m3/s. Allowable velocity through main
= 0.75 m/s. Assume f = 0.01, g = 9.81. The minimum gradient (in cm/100 length) required to
be given to this main so that water flow without any difficulty
Sol. :
Q =0.25 m3/s
Allowable velocity =0.75 m/s
f =0.01
g =9.81
2d
4
=
2Q 0.25 1mV 0.75 3
d =0.6515 m
2flv
2gd=hf=
20.01 100 (0.75)m
2 9.81 0.6515
= 0.044 m = 4.4 cm
Min gradent =fh 4.4cm
100m
l
hence ans is 4.4
51. Group I Group II
P Alidade
Q Arrow 1. Chain Survey
R Bubble tube 2. Levelling
S Stedia hair 3. Plant table surveying
4. Theodolite
(a) P 3 Q = 2, R1, S4.
(b) P2, Q4, R3, S-1
(c) P1, Q2, R4, S-3
(d) P3, Q1, R2, S-4
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Group I is 100/ ibs fruit colum group II is method of surv. Match.
Sol. (d)
P 3
Q 1
R 2
S 4
52. A traffic office impose on an average 5 number of penalties daily on traffic violators. Assume that
the number of penalties on different day is independent and follows a poisson distribution. The
probability that there will be less than 4 penalties in a day is .....
Sol. :
Mean = 5
P (x < 4) =p (x = 0) + p (x =1) + P (x = 2) + P (x = 3)
=5 0 5 1 5 2 5 3e s e 5 e 5 e 5
0! 1! 1! 3!
=5 25 125e 1 5
2 6
=5 118e
3
=0.265
53. The potable water is prepared from turbid surfaec water by adopting the foil treatment square.
(a) Turbid surface water Coagulation Flocculation Sedimentation Filteration disinfection storage and supply
(b) Turbid surface water disinfection Flocculation Sedimentation F ilteration Coagulation storage and supply
(c) Turbid surface water Filteration Sedimentation disinfection Flocculation Coagulation
(d) Turbid surface water Sedimentation Flocculation Coagulation disinfection Filteration.
Sol. (a)
54. A rectangular channel flow have bed slope of 0.0001 width = 3m coefficient n = 0.015, Q = 1 m3/
sec given that normal depth of flow ranges between 0.76 m and 0.8 m. The minimum width of
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throat (in m) that is possible at a given section while ensuring that the prevailing normal depth
does not exceed along the reach upstream of the contraction is approximately, equal to (assumenegligiable loss)
(a) 0.64 (b) 0.84
(c) 1.04 (d) 1.24
Sol. (b)
3mS = 0.00010
n = 0.015
Q = 1 m3/s
Normal depth of flow between 0.76 m to 0.8 m.
If prevailing normal depth of flow is not excreded, there must not be chocking of the section or
there must be just chocking.
Thus the width of the section should be such that for the prevailing sp. energy there should be
critical flow at the contracted section
i.e.
1/3
23 q2 g
=Ec= E initial
1/32
min
Q
B3
2 g
=E initial
Let is now calculate Einitial
Q =
2/3 1/2
0
1
AR Sn
1 =2/3
1/21 3y(3y) (0.0001)0.015 3 2y
y =0.78 m
E initial =y +2
2
q
2gy
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=
2
2
13
0.782 9.81 (0.78)
=0.7893 m
1/32
Q
Bmin3
2 g
=0.7893
2/3
2/31/3min
3 (1)
2g B=0.7893
Bmin =0.836 m
55. For the truss shown below, the member PQ is short by 3 mm. The magnitude of the vertical
displacement of joint R in mm is ............
QP
R
3m
4m 4m
Sol.
PQ is short by 3 mm
We have to find out vertical displacement of joint R in mm
R = u( )
Let in apply unit load at R as shown below
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ERQuPQP
R
1/21/2
uPR
R
uPR sin =1
2
uPQ+ uPR cos =0
uPQ = uPR cos
=1
.cos2sin
uPQ =1
cot2
=1 4/3
2
=2
3
R = PQ PQ2
u ( 3)3
= 2 mm upwards.
56. The probability density function of evaporation E on any day during a year in watershed is given
by
f(E) =
10 E 5 mm/day
50 otherwise
The probability that E lies in between 2 and 4 mm/day in a day in watershed is (in decimal)
Sol.
f(E) =
10 E 5 mm/day
50 otherwise
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P(2 < E < 4) =
4
2
f(E)dE = 4
42
2
1 1dE = E5 5
= 1
4 25
= 2/5
57. While designing for a steel column of Fe250 grade the base plate resting on a concrete pedestal
of M20 grade, the bearing strength of concrete (N/mm2) in LSM of design as per IS 456:2000 is
1 marks
Sol.
Permissible bearing stress
=0.45 fck
=0.45 20= 9 N/mm2
58. A long slope is formed in a soil with shear strength parameter C = 0, 34 .= F irm strate lies
below the slope and it is assumed that water table may occassionally rise to the surface, with
seepage taking place parallel to the slope. Use sat = 18 kN/m3 and w = 10 kN/m
3. The
maximum slope angle (in degree) to ensure the factor of safety 1.5. Assuming a potential failuresurface parallel to the slope would be
(a) 45.3 (b)44.7
(c) 12.3 (d)11.3
Sol.: (d)
3sat 18kN/m=
3w 10kN/m=
C 0=34=
F.O.S when water surface rises to surface is given by
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F.O.S = subsat
tantan
tan =sub
sat
tan
F.O.S
tan =sub
sat
tan
1.5
tan = 18 10 tan3418 1.5
=11.30
59. The amount of CO2generated in kg while completly oxidising one kg of CH4is .......
1 marks
Sol.( )
4 2 2 2O16g 44g
CH 2O CO 2H
16 g of CH4 when completely oxidised leads to 44 g of CO2
1 kg of CH4when completely oxidised leads to44
116
= 2.75 kg CO2
60. Given J =
13 2 1and K then=2 4 2 2
1 2 6 1
product KTJ K is
Sol.(a)
J =
3 2 12 4 21 2 6
, k =
121
KTJ K = 13 2 1
1 2 1 2 4 2 21 2 6 1
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= 1
6 8 1 21
=6 + 16 + 1
=23
61. With reference to the conventional cartesion (X, Y) the vertices of a triangle have x1, y1= 1, 0,
x2, y2= 2, 2, and x3, y3= 4, 3, the area of triangle is
(a)3
2(b)
3
4
(c)4
5(d)
5
2-1 marks
Sol. : (a)
A
B
a
c
C
(2, 2)
(4,3)
(1,0)
y
x
b
area of triangle is A = p(p a)(p p)(p c)
when P =a b c
2
a = 2 2(4 2) (2 1) 5
b = 2 2(4 1) (3) 3 2
c = 2 2(2 1) (2) 5
p =5 5 3 2 3
52 2
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A = 3 3 3 35 5 3 22 2 2 2
=
23 2 3 3
5 522 2 2
62. The degree of disturbances of a sample collected by sampler is expressed by a term called the area
ratio. I f outer diameter and inner dia of sample are D0and Direspectively, the area ratio is-
(a)
2 2
0 i2
i
D D
D
(b)
2 2
i 02
i
D D
D
(c)
2 2
0 i2
0
D D
D
(d)
2 2
i 02
0
D D
D
Sol. : (a)
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