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Civil Engineering | Page - 6 | Multiple Choice Questions (MCQ)

Q 07bb7d2122-sec2 -
A tensile test is conducted on a mild steel bar. The following data was obtained from the test: diameter of the steel bar= 3cm, Gauge Length of the bar = 20cm, Load at Elastic limit= 250kN, Extension at a load of 150kN= 0.21 mm, Maximum Load = 380kN, Total Extension= 60mm, Diameter of the rod at failure=2.25cm, determine the percentage decrease in Area?
A. 23.25%
B. 32.75%
C. 41.75%
D. 43.75%
Answer - D
Detail - Percentage decrease in Area = ((Original Area - Area at failure)/(original Area)) x 100 = ((0.785 x 30 x 30 - 0.785 x 22.5 x 22.5)/(0.785 x 30 x 30)) x 100= 43.75%

Q f346541fde-sec2 -
A tensile test is conducted on a mild steel bar. The following data was obtained from the test: diameter of the steel bar= 3cm, Gauge Length of the bar = 20cm, Load at Elastic limit= 250kN, Extension at a load of 150kN= 0.21 mm, Maximum Load = 380kN, Total Extension= 60mm, Diameter of the rod at failure=2.25cm, determine percentage elongation?
A. 10%
B. 20%
C. 30%
D. 40%
Answer - C
Detail - percentage elongation = (total increase in length/ original length) x 100 % = (60/200) x 100 = 30%

Q 6be0910b3c-sec2 -
A tensile test is conducted on a mild steel bar. The following data was obtained from the test: diameter of the steel bar= 3cm, Gauge Length of the bar = 20cm, Load at Elastic limit= 250kN, Extension at a load of 150kN= 0.21 mm, Maximum Load = 380kN, Total Extension= 60mm, Diameter of the rod at failure=2.25cm, determine the stress at elastic limit?
A. 3.54 x 10^5 N/sq(mm)
B. 353.68N/Sq(m)
C. 353.68GN/Sq(m)
D. 0.353 N/Sq(mm)
Answer - C
Detail - stress at elastic limit = load at elastic limit/area = (250x 1000)/(0.785 x 30 x 30)= 353.68N/sq(mm) = 353.68 x 10^6 N/Sq(m) =353.68 GN/Sq(m)

Q 3109cc6af4-sec2 -
A tensile test is conducted on a mild steel bar. The following data was obtained from the test: diameter of the steel bar= 3cm, Gauge Length of the bar = 20cm, Load at Elastic limit= 250kN, Extension at a load of 150kN= 0.21 mm, Maximum Load = 380kN, Total Extension= 60mm, Diameter of the rod at failure=2.25cm, determine young's modulus?
A. 202 GN/sqm
B. 2.02 x 10^6 N/sq(mm)
C. 0.202 x 10^4N/sq(mm)
D. 202MN/sqm
Answer - A
Detail - from the data given Area of bar = 0.785 x 30 x 30 =706.5 sq(mm),now,, within elastic limit load is 150kN, stress = 150000/706.5= 212.314N/sq(mm)strain = 0.21/200 = 0.00105 young modulus = stress/strain = 212.314/0.00105 = 2.02 x 10^5 N/sq(mm) =2.02x 10^11 N/sq(m) = 202 x 10^9 N/sq(m) = 202GN/Sq(m)

Q fa69f5d7e1-sec2 -
consider three dimensional system subjected to stress in x, y, z directions, then total strain in x direction is?
A. =longitudinal strain in x direction - lateral strain in y direction - lateral strain in z direction
B. =longitudinal strain in x direction + lateral strain in y direction + lateral strain in z direction
C. =longitudinal strain in x direction - longitudinal strain in y direction -longitudinal strain in z direction
D. =longitudinal strain in x direction + longitudinal strain in y direction + longitudinal strain in z direction
Answer - B
Detail - lateral strain = - poisson ratio x longitudinal strain

Q 1c558f8d52-sec2 -
Find the young Modulus of a brass rod of diameter 25 mm and of length 250mm which is subjected to a tensile load of 50kN when the extension of the rod is equal to 0.3 mm.?
A. 80526
B. 86134
C. 84925
D. 82520
Answer - C
Detail - stress = force/area = 50000N/0.785 x 25 x 25 = 101.91 N/sq(mm)strain = extension of rod/original length of rod = 0.3/250 = 0.0012now,modulus of elasticity = stress/strain = 101.91/0.0012 = 84925 N/sq(mm)

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)


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