Civil Engineering | Page - 1 | Multiple Choice Questions (MCQ)

Q 79a25ec25b-sec2 -
East Indian Railway Company was formed by .......................... for the construction of railway line?
A. Mr R.M. Stephenson
B. Mr G. Cluni
C. Mr T.R. Enderson
D. Mr Nicolas Cugnot
Answer - A
Detail - Mr RM Stephenson formed East Indian Railway Company in year 1844

Q b58c80550e-sec2 -
The first proposal for construction of railway line between which cities was conceived in year 1831-33, but it did not materilised ?
A. Bombay and Calcutta
B. Delhi and Calcutta
C. Madras and Bangalore
D. Madras and Bombay
Answer - C
Detail - No Details

Q d18cc1608c-sec2 -
The ultimate stress, for a hollow steel column which carries an axial load of 1.9 MN is 480N/Sq(mm). If the external diameter of the column is 200mm. determine the internal diameter. Take factor of safety as 4?
A. 120.12
B. 130.12
C. 140.12
D. 140.82
Answer - D
Detail - permissible stress = ultimate stess / factor of safety = 480/4 = 120N/Sq(mm),,also stress= force/area ,,, hence 120N/Sq(mm) = 1.9 x 1000000 N/ area of hollow steel column,, hence ,, Area of column= 1.9 x 1000000/120 = 15833.33 Sq(mm) ,, also Area = 0.785(external diameter^2 - internal diameter^2),,,hence by comparing area we get,,,15833.33 = 0.785 x (200 x 200 - internal diameter^2),, hence by solving we get,, 20169.85 =40000 - internal diameter^2,, also Internal diameter = (40000-20169.85)^0.5 =140.82 mm

Q b42faa6cda-sec2 -
The safe stress, for a Hollow steel column which carries an axial load of 2.1 x 10^3 kN is 125MN/sq(m). If the external diameter of the column is 30cm, determine the internal diameter?
A. 25.3cm
B. 26.2cm
C. 27.2cm
D. 28.1cm
Answer - B
Detail - given;- force = 2.1 x 10^3 kN = 2.1 x 10^6 N,,,,stress = 125 x 10^6 N/sq(m) =125 N/Sq(mm),,, now,, stress = force/Area,,, hence Area = Force/Stress = 2100000/125 =16800 Sq(mm),,,Also Area of Hollow cylinder = 0.785 x (external diameter^2 -internal diameter^2)= 0.785(30x 30 - Internal diameter^2),,,,hence by comparing Area = 16800= 0.785(300x300 - internal diameter^2),, hence,,, 90000- internal diameter^2 = 16800/0.785 =21401.274,,hence Internal diameter^2 = 90000-21401.274=68598.726 ,,hence internal diameter=(68598.726)^0.5 =261.91 mm = 26.2 cm

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)


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