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Answer Key (Test Code : 308)
1. (d) 2. (d) 3. (b) 4. (b) 5. (d) 6. (a)
7. (b) 8. (d) 9. (b) 10. (c) 11. (a) 12. (b)
13. (d) 14. (c) 15. (c) 16. (c) 17. (a) 18. (a)
19. (b) 20. (d) 21. (a) 22. (c) 23. (d) 24. (c)
25. (b) 26. (b) 27. (a) 28. (c) 29. (c) 30. (a)
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1. (d)
Explanation:
The reaction provided from the water will be
udl.
Therefore the situation can be analysed as:
Let the reaction has w udl intensity, then
W + W = wL
w =
Therefore, Bending Moment at center
= W ×
– w ×
×
=
-
×
=
-
= 0.
2. (d)
Explanation:
Strain Energy stored per unit volume, U =
As σ =
U =
= U P
2
U = kP2 (k = Constant)
Therefore, U1 = kp12 and U2 = kP2
2
Therefore, U = k(P1 + P2)2
U = kP12 + kP2
2 + 2kP1P2
U = U1 + U2 + 2kP1P2
U – 2kP1P2 = U1 + U2
⇒ U > U1 + U2.
3. (b)
Explanation: Major Principal Stress,
=
√( )
Hence, σx = 8 MPa, σy = 6MPa and
τxy =4 MPa
=
√( )
= 7 √
= 7 √ = 11.123 MPa.
4. (b)
Explanation:
Bending stresses are given by
=
σ =
× y
Now, σ at N.A = 0 and σ will be maximum on
the extreme fibre of shaded region.
Therefore, σmax =
× ymax
=
= 6 N/mm
2
Therefore, σavg =
=
= 3 N/mm
2
Force = σavg × Area = 3×40×80 = 9.6 kN.
5. (d)
Explanation:
The individual free body diagrams are:
l/4 l/2 l/4
W W
0.8 m
50 kN 50 kN
1.2 m
35 kN 35 kN
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Then the total elongation = Δ1 + Δ2 + Δ3
=
+
+
=
[
] 106 mm
=
= 1.27 mm.
6. (a)
Explanation:
Material A is more brittle than material B, as
the strain or deformation observed in material
B is more for same stress. But the ultimate
strength of A is more than that of B, as it can
bear more stress.
7. (b)
Explanation:
Cut the frame in such a way that each part of
the frame looks like a stable cantilever frame.
DS = 3 × (Number of cuts required for making
stable cantilever) – (Number of reactions
required for making stable cantilever).
Number of cuts required = 4
So, DS = 3 × 4 – 0 = 12.
8. (d)
Explanation:
The ILD of force in member U1 L1 is, when
load is exactly at L1. U1 L1 will be having zero
force because all the force will be taken by the
support.
If the applied unit load is moving from L2 to
L6. We can consider the equilibrium of the left
part.
So, the ILD of force in U1 L1 is same as that of
the vertical reactions V1
So, the full ILD of the member U1 L1.
So, ordinate at L2 = 0.8.
1.0 m
45 kN 45 kN
1
2
3
4
20 m 5 m
V2 V1
FV = 0
1
V1
FU L
FU L = v1
=
= 0.8
FU L
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9. (b)
Explanation:
Number of cuts required = 4
Number of reactions required = 3
Number of additional equations because of
hinges = 2 (at A) + 3 (at B) + 1 (at C)
= 6
Since two members joined at a point by an
internal hinge produce one additional
equation, similarly three members joined at a
point by internal hinge produce two additional
equations.
Also four members at a joint by an internal
hinge produce three additional equations.
Thus,
DS = 3 × 4 – 3 – 6 = 3.
10. (c)
Explanation:
ILD for B.M. at a distance of 4 m from left
support is:
Now, maximum B.M. will be, when load is
distributed in such a way on both side of the
cross section that W
= W
(l1 – x)
⇒
(4 – x) =
(16 – 5 + x)
⇒
=
⇒ 12 – 3x = 2 + 2x
⇒ 10 = 5x
⇒ x = 2 m.
So, max value of B.M. will be when:
Total area = Area of ABCD + Area of CDEF
= ½ [2.4 + 1.2] × 2 + ½ [2.4 + 1.2] × 3
= 3.6 + 5.4
= 9.0 m2.
Thus, B.M. = 10 kN/m × 9 m2 = 90 kNm.
11. (a)
Explanation:
At joint B, two non – collinear member meet.
So, force in each member should be zero.
So, FAB = 0. Also, FCB = 0.
12. (b)
Explanation:
At joint D, member BD and CD is collinear.
So, force in member AD should be zero.
A
B
C
b a
= 2.4
l1
=
x 6 m
(5-x)
4 m
F
A
B
C
E
1.2
2.4
6 m 4 m
2 m 3 m
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13. (d)
Explanation:
For calculating depth of neutral axis xu, we will
equate:
Total compressive force = Total Tensile Force.
Area of stress diagram × width = Tensile Force.
*
+ × 0.67 fck × 300 = 0.87 fy Ast
*
+ × 0.67 fck × 300 = 0.87 × fy × 4 ×
× 20
2
xu × 0.67 × 20 × 300 = 0.87 × 415 × 4 ×
× 20
2
xu = 158 mm.
14. (c)
Explanation:
Calculation of xulim:
xulim = 0.48 × 450 = 216 mm
Calculation of xu:
0.36 × fck b xu = 0.87 fy Ast
0.36 × 25 × 230 × xu = 0.87 × 415 × 5 ×
× 16
2
Xu = 175.346 mm
Since xu < xulim, the beam is under reinforced.
Mu = 0.36 fck b Xu (d – 0.42 Xu) × 10-6
Mu = 136.60 kNm.
15. (c)
Explanation:
Water content (w) =
w =
=
For soil A,
w =
= 0.8
= 0.8 ---------- (I)
+ = 1 ---------- (II)
From equation (I) and (II)
= 0.56 kg, = 0.44 kg
For soil B,
w =
= 0.6
= 0.6 ---------- (III)
+ = 1 ---------- (IV)
From equation (I) and (II)
= 0.625kg, = 0.375kg
For mixed soil, Mw = 0.44 + 0.375 = 0.815 kg
Md = 0.56 + 0.625 = 1.185 kg
Therefore, w =
= 0.6877 ≈ 68.8%
16. (c)
Explanation:
Void ratio, e =
For soil X,
0.4 =
= 0.4
+ = 1.4 (Given)
Therefore, = 1m3, = 0.4m
3
For soil Y,
w = 0.8 =
Stress
Variation
Strain
Variation
300 mm
Xu
Xu
0.67 fck
0.0020
0.0035
500 mm
4 bars of diameter 20 mm
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= 0.8
+ = 1.8 (Given)
Therefore, = 1 m3, = 0.8 m
3
For soil Z,
e =
=
= 0.6
n =
= 0.375
17. (a)
Explanation:
From figure, Total stress at x-x
= (6 × 9.81) + (6 × 22)
= 190.86 kN/m2
Pore water pressure at x-x
= 9.81 × 12 = 117.72 kN/m2
Effective stress at x-x
= 190.86 – 117.72 = 73.14 kN/m2.
18. (a)
Explanation:
Present Population = 25000
Design Population = 45000
Increase in Population = 45000-25000 = 20000
Increase per year =
= 1000
Now, increase in discharge required to reach
design discharge = 5800 – 4200 = 1600 m3/d.
At present per head water consumption is =
= 0.168 m
3/d
Increase in water consumption = 0.168 × 1000
= 168 m3/d
Therefore, number of years required to reach
design discharge =
= 9.5 years.
19. (b)
Explanation:
Treatment of water volume = 4.2 million litres
= 4.2×106×10
-3 m
3
= 4.2×103 m
3
Treatment discharge =
m
3/s
= 0.04861 m3/s
Settling velocity, Vs = 0.27 mm/s
= 0.27×10-3
m3/s
Surface area of tank required =
m2
= 180 m2.
20. (d)
Explanation:
Given speed V = 60 kmph =
m/s
Coefficient of friction, f = 0.35
Braking distance =
= (
)
= 40.45 m.
21. (a)
Explanation:
For A, RF =
1 cm on map = 5000 cm on ground.
For D, RF =
Decreasing order of scale,
=
>
>
>
.
X X
6m
6m Sand Bed (γsat = 22 kN/m
3)
River (γw = 9.81 kN/m3)
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22. (c)
Explanation Shrunk Scale (SS) = Shrinkage Factor (SF) ×
Original Scale (OS).
Now, SF =
=
L
SS = 9.6 =
L × 10
(On map) OL =
= 10.41 cm
OL on ground = 10.41 × 150 = 1561.5 cm
(Since 1 cm = 150 cm scale is given)
OL (ground) = 15.615 m.
23. (d)
Explanation:
SS = SF × OS =
×
(Since SF =
L)
=
SS =
.
24. (c)
Explanation: PB – 0.7 × 0.88 × 9.81 + 0.2 × 0.88 × 9.81 + 0.8 ×
13.6 × 9.81 – 0.4 × 13.6 × 9.81 – 0.3 × 9.81 = PA
PA – PB = 46.107 kPa ≈ 46.11 kPa.
25. (a)
Explanation:
Equation of streamline is given by:
=
=
ln x = - ln y + ln c, where c is constant.
xy = c
y =
At x = 3, y = 1
1 =
c = 3
Therefore, xy = c
xy = 3.
26. (b)
Explanation
ax =
+
+
= u + v + w
u = 2y2, v = 3x
2, w = 0
ax = 2y2 × 0 + 3x
2 × 4y
ax = 12x2y
Acceleration at (x, y) (1, 1)
ax = 12 (1) (1) = 12 m/s2.
27. (a)
Explanation CD = CV × CC
CC =
= Area of Vena – Contracta.
= Area of Orifice.
CC = (
)
CD = (
)
× 0.97 = 0.35.
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© Copyright: CIZ Campus, Dainik Bhaskar Building, Sector 25 D, Chandigarh. (Corp. Off.: Lucknow) www.competeindiazone.com [email protected] | 98143-76777, 98144-76777, 0172-4346477
T H I R D Y E A R ( C E ) | S C H O L A R S H I P T E S T R E S U L T
Result of CIZ Scholarship Test (3rd Year CCET CE) held on 11.09.2018
MM: 90
S.No. Name Marks Rank Scholarship (%)
1 Manish Kumar 42 1
50%
2 Saurabh Jha 22 2
40%
3 Rohit Bamola 18 3
40%
4 Jaskarn Singh 14 4
25%
5 Somnath Mishra 11 5
25%
6 Pulkit malik
10 6
25%
7 Abhishek Dhiman 9 7
10%
8 Akash Bailwal 8 8
10%
9 Rishi Kumar 6 9
10%
10 Shubham Singh Thapa 5 10
10%
11 Vivek Neupane 4 11
10%
12 Tashi Negi 1 12
10%
13 Arshit Goel -4 13
10%
14 Anshul Puri -4 13
10%
15 Suryakant -10 15
10%
16 Sanjay -10 15
10%
17 Amanpreet Singh -10 15
10%
18 Prince Sood -13 18
10%