UPSSSC-JE’21 CE: Daily Practice Quiz. 01-November-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. For a shaft subjected to combined bending moment M and twisting moment T, the ratio of the maximum shear stress to the maximum bending stress is
(a) M/T
(b) 2M/T
(c) T/M
(d) T/2M

Q2. The Poisson’s ratio for concrete is approximately
(a) 0.15
(b) 0.3
(c) 0.05
(d) 0.5

Q3.The maximum deflection of a cantilever due to pure, moment M at its free end is
(a) ML2/2EI
(b) ML2/3EI
(c) ML2/4EI
(d) ML2/6EI

Q4.In case of biaxial stress, the maximum value of shear stress is given by
(a) Difference of the normal stresses
(b) Half the difference of the normal stresses
(c) Sum of the normal stresses
(d) Half the sum of the normal stresses

Q5. If dry density, water density and specific gravity of solids of a given soil sample are 1.60 g/cc, 1.00 g/cc and 2.56 respectively, the porosity of the soil sample, is
(a) 0.375
(b) 0.370
(c) 0.380
(d) 0.390

Q6. For a soil deposit having porosity (n) = 33% and specific gravity (G) = 2.60, the critical gradient is:
(a) 1.0
(b) 1.05
(c) 1.07
(d) 1.10

SOLUTION

S1. Ans.(d)
Sol. Maximum shear stress in solid circular shaft (τ_max )=16T/(πD^3 )
Maximum bending stress in solid circular shaft (σ_max ) = 32M/(πD^3 )
τ_max/σ_max =(16T/(πD^3 ))/(32M/(πD^3 ))
▭(τ_max/σ_max =T/2M) or ▭(σ_max/τ_max =2M/T)
Where,
T = Applied Torque
M = Bending Moment

S2. Ans.(a)
Sol.

S3. Ans.(a)
Sol. The maximum deflection in a cantilever beam due to pure bending M at free end→

▭(Slope (θ)=ML/EI)
▭(Deflection (δ)=(ML^2)/2EI)

S4. Ans.(b)

Max. value of shear stress = Radius of Mohr circle
▭(τ_max=(σ_1-σ_2)/2)
Hence, half the difference of the normal stress.

S5. Ans.(a)
Sol. Given, dry density (γ_d ) = 1.6 g/cc
Water density (γ_ω )=1.00 g\/cc
Specific gravity (G) = 2.56
Porosity (n) = ?
Now,
γ_d=(Gγ_ω )/(1+e)
1.6=(2.56×1.00)/(1+e)
▭(e=0.6 )
n=e/(1+e)
0.6/(1+0.6)
▭(n=0.375 )

S6. Ans.(c)
Sol. Given, Porosity (n) = 33%
Specific gravity (G) = 2.60
Critical hydraulic gradient (I_C ) = ?
▭(I_C=(G-1)/(1+e))
e=n/(1-n)=0.33/(1-0.33)=0.49
I_C=(2.6-1)/(1+0.49)
▭(I_C=1.073 )

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