Civil Engineering | Page - 2 | Multiple Choice Questions (MCQ)

Q 7313c7c859-sec2 -
Find the Minimum diameter of a steel wire, which is used to raise a load of 4000 N if the stress is not exceed 95MN/sq(m)?
A. 0.732mm
B. 0.732cm
C. 0.732 m
D. 7.32 mm
Answer - B
Detail - Stress = 95 x 10^6 N/sq(m) = 95 N/sq(mm) = force/Area alsoArea = Force/Stress = 4000/95 =42.105 sq(mm)alsoArea = 0.785xdiameterxdiameterhencediameter = (42.105/0.7850^0.5 =(53.64)^0.5 =7.32 mm = 0.732 cm

Q 8d79ed5b58-sec2 -
A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20kN. If the modulus of elasticity of the material of the rod is 2 x 10^5 N/Sq(mm) ; determine : the elongation in rod?
A. 0.477cm
B. 0.00477cm
C. 0.0477cm
D. 0.000477cm
Answer - C
Detail - elongation of rod = original length x strain, we know that original length is equal to 150cmstrain = stress/modulus of elasticity = force/ Area x modulus of elasticity = 20000/(0.785 x 20 x 20 x 200000) = 0.000318 by putting value of strain in elongation formula we getelongation of rod = strain x original length = 0.000318 x 150 = 0.0477 cm

Q 58bb4c50be-sec2 -
A rod of 150cm long and of diameter 2.0 cm is subjected to an axial pullof 20kN. If the modulus of elasticity of the material of the rod is 2 x 10^5 N/sq(mm), determine strain?
A. 0.000318
B. 0.00318
C. 0.0000318
D. 0.0318
Answer - A
Detail - we know that, Modulus of elasticity is stress/strain,stress is equal to force/area, hence by putting value of Area = 0.785 x diameter x diameter, and force is equal to 20000 N, we get stress equal to 63.662 N/sq(mm),now, from the equation of hookes law,strain = stress/modulus of elasticity = 63.662/200000 = 0.000318

Q 0caf6cbdda-sec2 -
A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20kN. If the modulus of elasticity of the material of the rod is 2 x 10^5 N/Sq(mm) ; determine : the stress?
A. 63.600 N/sq(mm)
B. 63.500 N/sq(mm)
C. 63.662 N/sq(mm)
D. 63.562 N/sq(mm)
Answer - C
Detail - Area of Rod:- (π/4)(radius^2) :- 100π sq(mm),now,Stress= Force/Area = 20000/100π = 63.662 N/sq(mm)

Q 0bd476f6b4-sec2 -
what is lateral strain?
A. change in length at right angle to the direction of applied axial load to the respective length of body
B. change in the length in the direction of applied load to the respective length of body
C. change in the volume to the original volume of body
D. change in the shape to the original shape of body
Answer - A
Detail - No Details

Q eb5c401150-sec2 -
what is longitudinal strain?
A. change in length at right angle to the direction of applied axial load to the respective length of body
B. change in the length in the direction of applied load to the respective lenght of body
C. change in the volume to the original volume of body
D. change in the shape to the original shape of body
Answer - B
Detail - No Details

Q c5fd530e49-sec2 -
what is poisson ratio?
A. Ratio of the tensile stress to the tensile strain
B. Ratio of the shear stress to the shear strain
C. Ratio of the Lateral Strain to the Longitudinal strain
D. Ratio of the ultimate stress to the Permissible stress
Answer - C
Detail - No Details

Q 4928f893a7-sec2 -
what is Factor of safety?
A. Ratio of the tensile stress to the tensile strain
B. Ratio of the shear stress to the shear strain
C. Ratio of the Lateral Strain to the Longitudinal strain
D. Ratio of the ultimate stress to the Permissible stress
Answer - D
Detail - No Details

Q 997c8830f1-sec2 -
what is modulus of rigidity or shear modulus?
A. Ratio of the tensile stress to the tensile strain
B. Ratio of the shear stress to the shear strain
C. Ratio of the Lateral Strain to the Longitudinal strain
D. Ratio of the ultimate stress to the Permissible stress
Answer - B
Detail - No Details

Q d4f70e0d8d-sec2 -
what is modulus of elasticity or young modulus?
A. ratio of tensile stress to tensile strain
B. ratio of shear stress to shear strain
C. ratio of ultimate stress to permissible stress
D. ratio of lateral strain to longitudinal strain
Answer - A
Detail - No Details


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