UPSSSC-JE'21 CE: Daily Practice Quiz. 12-October-2021 |_00.1
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UPSSSC-JE’21 CE: Daily Practice Quiz. 12-October-2021

Know your strengths and practice your concepts with this quiz on UPSSSC-JE Recruitment 2021. This quiz for UPSSSC-JE Recruitment 2021 is designed specially according to UPSSSC-JE Syllabus 2021.

Quiz: Civil Engineering
Exam: UPSSSC-JE
Topic: MISCELLANEOUS

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes

Q1. A flat road has a curve segment with a radius of 150 m. While negotiating this curve, vehicle slipped on its tyres as well as tried to roll over at a particular speed. This speed assuming a friction coefficient of 0.4 is :
(a) 18 m/s
(b) 25 m/s
(c) 32 m/s
(d) 20 m/s

Q2. For linear elastic systems, the type of displacement function for the strain energy is
(a) linear
(b) quadratic
(c) cubic
(d) quartic

Q3. Muller Breslau principle in structural analysis is used for
(a) drawing ILD for any force function
(b) writing virtual work equation
(c) superposition of load effects
(d) none of these

Q4. Calculate the curvature correction (in m.) if distance between the instrument and staff is 1000 m.
(a) –0.0785
(b) 0.0785
(c) –0.0673
(d) 0.0673

Q5. Consider the following equipments
1. Odometer
2. Tacheometer
3. Passometer
4. Perambulator
Which of the above equipments can be employed for measurement of horizontal distance?
(a) 1 and 3 only
(b) 1 and 2 only
(c) 1, 2, 3 and 4
(d) 2 and 3 only

Q6. what is the percentage of the fine aggregate of fineness modulus 2.6 to be combined with coarse aggregate of fineness modulus 6.8 for obtaining combine aggregate of fineness modulus 5.4?
(a) 60%
(b) 40%
(c) 50%
(d) 33%

SOLUTION

S1. Ans.(b)
Sol. In case of road turn the centrifugal force should be equal to friction force

Fr = mv²/r
μ RN = mv²/r
0.4 × mg = mv²/150

V² = 150 × 0.4 g
V² = 588
V = 24.24 m/sec
≈25 m/sec

S2. Ans.(b)
Sol. Strain energy (U) =1/2×stress×strain
=1/2×σ×∈
→ for linear elastic system.
U=1/2×E×∈×∈
=1/2 E ∈^2
▭(U α ∈^2 )→Quadratic equation.

S3. Ans.(a)
Sol. Muller Breslau principle in structure analysis is used for drawing influence line diagram for any force function such as shear force, bending moment or any reactive force and moment.

S4. Ans.(a)
Sol. Given,
Distance between the instrument and staff (d)= 1000 m. = 1 km
Correction due to curvature is given by
C_c = –0.0785 d²
C_c = –0.0785 (1)²
▭(C_c=-0.0785 m.)

S5. Ans.(c)
Sol. Passometer, Perambulator, Pedometer, Measuring wheel, speedometer, Tacheometer, Odometer etc. are used in reconnaissance surveying for measuring horizontal distance.
S6. Ans.(d)
Sol. Given,
Fineness modulus of fine aggregate (FM)_fine=2.6
Fineness modulus of coarse aggregate (FM)_coarse=6.8
Fineness modulus of combine aggregate (FM)_combine=5.4
Percentage of fine aggregate (R) = ?

The percentage of fine aggregate will be given by –
R=((FM)_coarse-(FM)_combine)/((FM)_coarse-(FM)_fine )×100
=(6.8-5.4)/(6.8-2.6)×100
▭(R=33%)

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