CE GATE 2015 (Test 3)



Tag: ce gate 2015
Q.1
The combined correction due to curvature and refraction (in m) for distance of 1 km on the surface of Earth is
A. 0.0673
B. 0.673
C. 7.63
D. 0.763
Answer : Option A
Explaination / Solution:

C = 0.0673d2 = 0.0673×1

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Q.2
A 6 m high retaining wall having a smooth vertical back face retains a layered horizontal backfill. Top 3 m thick layer of the backfill is sand having an angle of internal friction, φ = 30° while the bottom layer is 3 m thick clay with cohesion, c = 20 kPa. Assume unit weight for both sand and clay as 18 kN/m3. The totalactive earth pressure per unit length of the wall (in kN/m) is: 
A. 150
B. 216
C. 156
D. 196
Answer : Option A
Explaination / Solution:
No Explaination.


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Q.3
A column of size 450 mm × 600 mm has unsupported length of 3.0 m and is braced against side sway in both directions. According to IS 456:2000, the minimum eccentricities (in mm) with respect to major and minor principle axes are
A. 20.0 and 20.0
B. 26.0 and 21.0
C. 26.0 and 20.0
D. 21.0 and 15.0
Answer : Option B
Explaination / Solution:



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Q.4
A guided support as shown in the figure below is represented by three springs (horizontal, vertical and rotational) with stiffness kx, ky and kθ respectively. The limiting values of kx, ky and kθ are

A. ∞, 0, ∞
B. ∞, ∞, ∞
C. 0, ∞, ∞
D. ∞, ∞, 0
Answer : Option A
Explaination / Solution:

As rotation and horizontal deflection in zero as per given figure. Therefore its stiffness is '∞' as deflection = 0. stiffness= Force/deflection and stiffness is zero in y direction

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Q.5
A steel member ‘M’ has reversal of stress due to live loads, whereas another member ‘N’ has reversal of stress due to wind load. As per IS 800:2007, the maximum slenderness ratio permitted is
A. less for member ‘M’ than that of member ‘N’
B. more for member ‘M’ than for member ‘N’
C. same for both the members
D. not specified in the Code
Answer : Option A
Explaination / Solution:

M – due to live load 
N – due to wind load 
As per IS800. 
M - λ - 180
N - λ - 350     M<N

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Q.6
Net ultimate bearing capacity of a footing embedded in a clay stratum
A. increases with depth of footing only
B. increases with size of footing only
C. increases with depth and size of footing
D. is independent of depth and size of footing
Answer : Option D
Explaination / Solution:



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Q.7
Prying forces are
A. shearing forces on the bolts because of the joints
B. tensile forces due to the flexibility of connected parts
C. bending forces on the bolts because of the joints
D. forces due the friction between connected parts
Answer : Option B
Explaination / Solution:
No Explaination.


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Q.8
 For the plane stress situation shown in the figure, the maximum shear stress and the plane on which it acts are 

A. –50 MPa, on a plane 45° clockwise w.r.t. x-axis
B. –50 MPa, on a plane 45° anti-clockwise w.r.t. x-axis
C. 50 MPa, at all orientations
D. Zero, at all orientations
Answer : Option D
Explaination / Solution:



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Q.9
Surcharge loading required to placed on the horizontal backfill of a smooth retaining vertical wall so as to completely eliminate tensile crack is:
A. 2c
B. 2cka
C. 2cka
D. 2c/ka
Answer : Option D
Explaination / Solution:
No Explaination.


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Q.10
Match the information related to test on aggregates given in Group-I with that in Group-II. 
Group-I                                                         Group-II 
P. Resistance to impact                                1. Hardness 
Q. Resistance to wear                                  2. Strength 
R. Resistance to weathering action              3. Toughness 
S. Resistance to crushing                            4. Soundness
A. P-1, Q-3, R-4, S-2
B. P-3, Q-1, R-4, S-2
C. P-4, Q-1, R-3, S-2
D. P-3, Q-4, R-2, S-1
Answer : Option B
Explaination / Solution:

Resistance to impact →Toughness Resistance to wear → Hardness Resistance to weathering → Soundness Resistance to crushing → Strength

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