Quiz: Civil Engineering
Exam: OPSC
Topic: MISCELLANEOUS
Each question carries 1 mark
Negative marking: 1/3 mark
Time: 10 Minutes
Q1. If the sides of a slab simply supported on edges and spanning in two directions are equal, the maximum bending moment is multiplied by…………
(a) 0.2
(b) 0.3
(c) 0.4
(d) 0.5
Q2. If the maximum bending moment of a simply supported slab is M kg – cm, the effective depth of the slab is
(a) M/100Q
(b) M/(10√Q)
(c) √(M/Q)
(d) √(M/(100 Q))
Q3. From a flow net which of the following information can be obtained?
1. Rate of flow
2. Pore water pressure
3. Exit gradient
4. Permeability
Select the correct answer using the codes given below:
(a) 1, 2, 3 and 4
(b) 1, 2 and 3 only
(c) 2, 3 and 4 only
(d) 1 only
Q4. Consider the following statements.
1. Seepage force is applied by flowing water to the soil skeleton through frictional drag.
2. The magnitude of seepage force per unit volume of soil at any point is equal to γw li, where γw is the unit weight of water and I is he hydraulic gradient at that point.
3. In a soil mass subjected to upward flow of water, quick and condition develops when pore pressure is equal to the total stress
Which of these statements is/are correct?
(a) 1 and 2
(b) 1 and 3
(c) 1 only
(d) 2 and 3
Q5. In case of principal axis of a section
(a) sum of moment of inertia is zero
(b) difference of moment of inertia is zero
(c) product of moment of inertia is zero
(d) None of these
Q6. The shear stress at a point in a shaft subjected to a torque is
(a) Directly proportional to the polar moment of inertia and to the distance of the point form the axis
(b) Directly proportional to the applied torque and inversely proportional to the polar moment of inertia
(c) Directly proportional to the applied torque and the polar moment of inertia
(d) Inversely proportional to the applied torque and the polar moment of inertia
Q7. The allowable stress in a long column can be increased by increasing the
(a) radius of gyration
(b) eccentricity
(c) slenderness ration
(d) length of the column
Q8. Which of the following is the correct sequence of plasticity of minerals in soil in an increasing order?
(a) Kaolinite, silica, montmorillonite, illite
(b) Kaolinite, silica, illite, montmorillonite
(c) Silica, kaolinite, illite, montmorillonite
(d) Silica, kaolinite, montmorillonite, illite
Q9. In a plate load test, minimum size of plate to be used is
(a) 250 mm × 250 mm
(b) 300 mm × 300 mm
(c) 400 mm × 400 mm
(d) 450 mm × 450 mm
Q10. A flownet of a coffer dam foundation has 6 flow channels and 18 equipotential drops. The head of water lost during seepage is 6 m. if the coefficient of permeability of foundation is 4 × 10–5 m/mm, then the seepage loss per metre length of dam will be.
(a) 2.16 × 10–2 m³/day
(b) 6.48 × 10–2 m³/day
(c) 11.52 × 10–2 m³/day
(d) 34.56 × 10–2 m³/day
SOLUTION
S1. Ans.(d)
Sol.
▭(ly=lx)
M_x=(r^4/(1+r^4 )) (WL_x^2)/8
▭(r=ly/lx=1 )
Multiplied factor = 1^4/(1+1^4 )=0.5
S2. Ans.(d)
Sol. moment of resistance (M) = Qd² (N-mm)
M/(10×10) (kg-cm)=Qd^2
▭(d=√(M/100Q))
S3. Ans.(b)
Sol. Seepage pressures, uplift pressures, exit gradient and pore-water pressure can be obtained from a flownet.
S4. Ans.(b)
Sol. seepage force is due to frictional drag and its magnitude is given by iγ_w per unit volume of soil.
For quick condition during upward flow.
Effective stress = total stress – pore pressure
So statement 1 and 3 is correct
S5. Ans.(c)
Sol. An object rotating about an axis, the resistance of a body to accelerate is called inertia of mass. In case of principal axis of a section the product of moment of Inertia is zero.
S6. Ans.(b)
Sol. (by Torsion equation)/(T/J=τ/R=Gθ/L)
Torque (τ) x Applied torsion (T)
τ α 1/J
J= Polar moment of Inertia
S7. Ans.(a)
Sol. The allowable stress in a long column can be increased by increasing it’s radius of gyration and radius of gyration depends on moment of inertia.
R=√(I/A)
R- radius of gyration
I- Moment of Inertia
S8. Ans.(c)
Sol. increasing order of plasticity in clay mineral –
Silica → kaolinite → illite → montmorillonite
S9. Ans.(b)
Sol. in plate load test-
→ minimum size of plate = 30 cm.
→ maximum size of plate = 90 cm.
S10. Ans.(c)
Sol. Seepage loss per metre length
q=k×N_f/N_d ×H
Given,
k = 4 × 10-5 m/min
= 4 × 10-5 × 24 × 60 m/day
= 0.0576 m/day
∴ q=0.0576×6/18×6
= 0.1152 m³/day
= 11.52 × 10-2 m³/day
Nf is number of flow channels = 6.0
Nd is number of equipotential drops = 18
