McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Uncategorized topics .
| 51. |
Calculate the discharge over rectangular Weir of 3 metres length under the head of 400mm.Use Francis formula. |
| A. | 1.268 m3/s |
| B. | 1.396 m3/s |
| C. | 1.475 m3/s |
| D. | 1.528 m3/s |
| Answer» B. 1.396 m3/s | |
| Explanation: francis formula for discharge q = 1.84 lh3/2. | |
| 52. |
converts mechanical energy into hydraulic energy. |
| A. | dynamo |
| B. | pump |
| C. | turbine |
| D. | generator |
| Answer» B. pump | |
| Explanation: a pump is a mechanical device which converts the mechanical energy into hydraulic energy. the hydraulic energy is in the form of pressure energy. the pumps are generally used for lifting liquid from a lower level to a higher level. | |
| 53. |
Torsional sectional modulus is also known as |
| A. | polar modulus |
| B. | sectional modulus |
| C. | torsion modulus |
| D. | torsional rigidity |
| Answer» A. polar modulus | |
| Explanation: the ratio of polar moment of inertia to radius of section is called polar modulus or torsional section modulus. its units are mm3 or m3 (in si). | |
| 54. |
is a measure of the strength of shaft in rotation. |
| A. | torsional modulus |
| B. | sectional modulus |
| C. | polar modulus |
| D. | torsional rigidity |
| Answer» C. polar modulus | |
| Explanation: the polar modulus is a measure of the strength of shaft in rotation. as the value of polar modulus increases torsional strength increases. | |
| 55. |
What are the units of torsional rigidity? |
| A. | nmm2 |
| B. | n/mm |
| C. | n-mm |
| D. | n |
| Answer» A. nmm2 | |
| Explanation: the product of modulus of rigidity (c) and polar moment of inertia (j) is called torsional rigidity. torsional rigidity is a torque that produces a twist of one radian in a shaft of unit length. | |
| 56. |
The angle of twist can be written as |
| A. | tl/j |
| B. | cj/tl |
| C. | tl/cj |
| D. | t/j |
| Answer» C. tl/cj | |
| Explanation: the angle of twist = tl/cj | |
| 57. |
The power transmitted by shaft SI system is given by |
| A. | 2πnt/60 |
| B. | 3πnt/60 |
| C. | 2πnt/45 |
| D. | nt/60 w |
| Answer» A. 2πnt/60 | |
| Explanation: in si system, power (p) is measured in watts (w) ; p = 2πnt/60 where t = average torque in n.m | |
| 58. |
Area of catchment is measured in |
| A. | mm3 |
| B. | km2 |
| C. | km |
| D. | mm |
| Answer» B. km2 | |
| Explanation: catchment area can be defined as the area which contributes the surplus water present over it to the stream or river. it is an area which is responsible for maintaining flow in natural water bodies. it is expressed in square kilometres. | |
| 59. |
catchment area is a sum of free catchment area and intercepted catchment area. |
| A. | total |
| B. | additional |
| C. | combined |
| D. | overall |
| Answer» C. combined | |
| Explanation: combined catchment area is defined as the total catchment area which contributes the water in to stream or a tank. combined catchment area = free catchment area + intercepted catchment area. | |
| 60. |
has steep slopes and gives more run off. |
| A. | intercepted catchment area |
| B. | good catchment area |
| C. | combined catchment area |
| D. | average catchment area |
| Answer» B. good catchment area | |
| Explanation: good catchment area consists of hills or rocky lands with steep slopes and little vegetation. it gives more run off. | |
| 61. |
Trend of rainfall can be studied from |
| A. | rainfall graphs |
| B. | rainfall records |
| C. | rainfall curves |
| D. | rainfall cumulatives |
| Answer» B. rainfall records | |
| Explanation: rainfall records are useful for calculating run off over a basin. by using rainfall records estimate of design parameters of irrigation structures can be made. the maximum flow due to any storm can be calculated and predicted. | |
| 62. |
Runoff coefficient is denoted by |
| A. | p |
| B. | n |
| C. | k |
| D. | h |
| Answer» C. k | |
| Explanation: the runoff coefficient can be defined as the ratio of runoff to rainfall. | |
| 63. |
is a graph showing variations of discharge with time. |
| A. | rising limb graph |
| B. | crest graph |
| C. | hydraulic graph |
| D. | gauge graph |
| Answer» C. hydraulic graph | |
| Explanation: hydrograph is a graph showing variations of discharge with time at a particular point of the stream. the hydrograph shows the time distribution of total run off at a point of measurement. | |
| 64. |
Calculate the torque which a shaft of 300 mm diameter can safely transmit, if the shear stress is 48 N / mm2. |
| A. | 356 knm |
| B. | 254 knm |
| C. | 332 knm |
| D. | 564 knm |
| Answer» B. 254 knm | |
| Explanation: given, the diameter of shaft d | |
| 65. |
What is the bending moment at end supports of a simply supported beam? |
| A. | maximum |
| B. | minimum |
| C. | zero |
| D. | uniform |
| Answer» C. zero | |
| Explanation: at the end supports, the moment (couple) developed is zero, because there is no distance to take the perpendicular acting load. as the distance is zero, the moment is obviously zero. | |
| 66. |
What is the maximum shear force, when a cantilever beam is loaded with udl throughout? |
| A. | w×l |
| B. | w |
| C. | w/l |
| D. | w+l |
| Answer» A. w×l | |
| Explanation: in cantilever beams, the maximum shear force occurs at the fixed end. in the free end, there is zero shear force. as we need to convert the udl in to load, we multiply the length of the cantilever beam with udl acting upon. for maximum shear force to obtain we ought to multiply load and distance and it surely occurs at the fixed end (w×l). | |
| 67. |
What is the maximum bending moment for simply supported beam carrying a point load “W” kN at its centre? |
| A. | w knm |
| B. | w/m knm |
| C. | w×l knm |
| D. | w×l/4 knm |
| Answer» D. w×l/4 knm | |
| Explanation: we know that in simply supported beams the maximum bm occurs at the central span. | |
| 68. |
How do point loads and udl be represented in SFD? |
| A. | simple lines and curved lines |
| B. | curved lines and inclined lines |
| C. | simple lines and inclined lines |
| D. | cant represent any more |
| Answer» C. simple lines and inclined lines | |
| Explanation: according to bis, the standard symbols used for sketching sfd are | |
| 69. |
curve is formed due to bending of over hanging beams. |
| A. | elastic |
| B. | plastic |
| C. | flexural |
| D. | axial |
| Answer» A. elastic | |
| Explanation: the line to which the longitudinal axis of a beam bends or deflects or deviates under given load is known as elastic curve on deflection curve. elastic curve can also be known as elastic line or elastic axis. | |
| 70. |
The relation between slope and maximum bending moment is |
| A. | directly proportion |
| B. | inversely proportion |
| C. | relative proportion |
| D. | mutual incidence |
| Answer» B. inversely proportion | |
| Explanation: the relationship between slope and maximum bending moment is inversely proportional because, for example in simply supported beams slope is maximum at supports and zero at midspan of a | |
| 71. |
What is the SF at support B? |
| A. | 5 kn |
| B. | 3 kn |
| C. | 2 kn |
| D. | 0 kn |
| Answer» D. 0 kn | |
| Explanation: total load = 2×2 = 4kn shear force at a = 4 kn ( same between a and c ) | |
| 72. |
Where do the maximum BM occurs for the below diagram. |
| A. | -54 knm |
| B. | -92 knm |
| C. | -105 knm |
| D. | – 65 knm |
| Answer» C. -105 knm | |
| Explanation: moment at b = 0 moment at c = – (10 × 3) × (3/2) | |
| 73. |
The ratio of maximum deflection of a beam to its is called stiffness of the beam. |
| A. | load |
| B. | slope |
| C. | span |
| D. | reaction at the support |
| Answer» C. span | |
| Explanation: the stiffness of a beam is a measure of it’s resistance against deflection. the ratio of the maximum deflection of a beam to its span can be termed as stiffness of the beam. | |
| 74. |
Stiffness of the beam is inversely proportional to the of the beam. |
| A. | slope |
| B. | support reaction |
| C. | deflection |
| D. | load |
| Answer» C. deflection | |
| Explanation: stiffness of a beam is inversely proportional to the deflection. smaller the deflection in a beam due to given external load, greater is its stiffness. | |
| 75. |
The maximum should not exceed the permissible limit to the span of the beam. |
| A. | slope |
| B. | deflection |
| C. | load |
| D. | l bending moment |
| Answer» B. deflection | |
| Explanation: the maximum deflection of a loaded beam should not exceed the permissible limit in relation to the span of a beam. while designing the beam the designer | |
| 76. |
In cantilever beam the deflection occurs at |
| A. | free end |
| B. | point of loading |
| C. | through out |
| D. | fixed end |
| Answer» A. free end | |
| Explanation: deflection can be defined as | |
| 77. |
The maximum deflection in cantilever beam of span “l”m and loading at free end is “W” kN. |
| A. | wl3/2ei |
| B. | wl3/3ei |
| C. | wl3/4ei |
| D. | wl2/2ei |
| Answer» B. wl3/3ei | |
| Explanation: maximum deflection occurs at free end distance between centre of gravity of bending moment diagram and free end is x = | |
| 78. |
fluids are practical fluids |
| A. | ideal |
| B. | real |
| C. | vortex |
| D. | newtonian |
| Answer» B. real | |
| Explanation: these fluids possess properties such as viscosity, surface tension. they are compressible in nature. the certain amount of resistance is always offered by the fluids, they also possess shear stress. they are also known as practical fluids. | |
| 79. |
The inverse of specific weight of a fluid is |
| A. | specific gravity |
| B. | specific volume |
| C. | compressibility |
| D. | viscosity |
| Answer» B. specific volume | |
| Explanation: specific volume is the volume of the fluid by unit weight it is the reciprocal of specific weight is denoted by “v”. si units are m3/n. | |
| 80. |
Specific gravity of water is |
| A. | 0.8 |
| B. | 1 |
| C. | 1.2 |
| D. | 1.5 |
| Answer» B. 1 | |
| Explanation: the specific gravity is also called as relative density. it is dimensionless quantity and it has no units. the specific gravity of water is the ratio of specific weight of fluid to specific weight of water, as both | |
| 81. |
Compute the maximum deflection at free end of a cantilever beam subjected to udl for entire span of l metres. |
| A. | wl4/8ei |
| B. | wl4/4ei |
| C. | wl3/8ei |
| D. | wl2/6ei |
| Answer» A. wl4/8ei | |
| Explanation: the slope at free end = a/ei = wl3/6ei | |
| 82. |
Which of the following is not an example of Malleability? |
| A. | wrought iron |
| B. | ornamental silver |
| C. | torsteel |
| D. | ornamental gold |
| Answer» C. torsteel | |
| Explanation: torsteel is an example of mechanical property ductility. the ductility is a property of a material by which material can be fractured into thin wires after undergoing a considerable deformation without any rupture. | |
| 83. |
Which of the following is not correct? |
| A. | angles and t section are strong in bending |
| B. | channels can be used only for light loads |
| C. | i sections are most efficient and economical shapes |
| D. | i section with cover plates are provided when large section modulus is required |
| Answer» A. angles and t section are strong in bending | |
| Explanation: angles and t section are weak in bending. channels can be used only for light loads. i sections (rolled and built-up) are most efficient and economical shapes. i section with cover plates are provided when large section modulus is required. generally, islb or ismb are provided in such cases. | |
| 84. |
Local buckling can be prevented by |
| A. | limiting width-thickness ratio |
| B. | increasing width-thickness ratio |
| C. | changing material |
| D. | changing load on member |
| Answer» C. changing material | |
| Explanation: local buckling of compression members of beam causes loss of integrity of | |
| 85. |
Which of the following is true? |
| A. | in case of rolled section, less thickness of plate is adopted to prevent local buckling |
| B. | for built-up section and cold formed section, longitudinal stiffeners are not provided to reduce width to smaller sizes |
| C. | local buckling cannot be prevented by limiting width-thickness ratio |
| D. | in case of rolled section, high thickness of plate is adopted to prevent local buckling |
| Answer» D. in case of rolled section, high thickness of plate is adopted to prevent local buckling | |
| Explanation: in case of rolled section, higher thickness of plate is adopted to prevent local buckling. local buckling cannot be prevented by limiting width-thickness ratio. for built-up section and cold formed section, longitudinal stiffeners are provided to reduce width to smaller sizes. | |
| 86. |
Which of the following is not true? |
| A. | only plastic section can be used in intermediate frames |
| B. | slender sections are preferred in hot rolled structural steelwork |
| C. | compact sections can be used in simply supported beams |
| D. | semi-compact sections can be used for elastic designs |
| Answer» B. slender sections are preferred in hot rolled structural steelwork | |
| Explanation: only plastic section can be used in intermediate frames which form collapse mechanism. compact sections can be used in simply supported beams which fail after reaching mp at one section. semi- compact sections can be used for elastic designs where section fails after reaching my at extreme fibres. slender sections are not preferred in hot rolled structural steelwork, but they are extensively used in cold formed members. | |
| 87. |
As per IS specification, the beam sections should be |
| A. | not symmetrical about any principal axes |
| B. | at least symmetrical about one of the principal axes |
| C. | symmetrical about all principal axes |
| D. | unsymmetrical about all principal axes |
| Answer» B. at least symmetrical about one of the principal axes | |
| Explanation: the beam sections should be at least symmetrical about one of the principal axes as per is specification. angle and t- sections are inherently weak in bending while channels can only be used for light loads. | |
| 88. |
A beam is said to be of uniform strength, if |
| A. | b.m. is same throughout the beam |
| B. | shear stress is the same through the beam |
| C. | deflection is the same throughout the beam |
| D. | bending stress is the same at every section along its longitudinal axis |
| Answer» D. bending stress is the same at every section along its longitudinal axis | |
| Explanation: beam is said to be uniform strength if at every section along its longitudinal axis, the bending stress is same. | |
| 89. |
Which stress comes when there is an eccentric load applied? |
| A. | shear stress |
| B. | bending stress |
| C. | tensile stress |
| D. | thermal stress |
| Answer» B. bending stress | |
| Explanation: when there is an eccentric load it means that the load is at some distance from the axis. this causes compression in one side and tension on the other. this causes bending stress. | |
| 90. |
What is the expression of the bending equation? |
| A. | m/i = σ/y = e/r |
| B. | m/r = σ/y = e/i |
| C. | m/y = σ/r = e/i |
| D. | m/i = σ/r = e/y |
| Answer» A. m/i = σ/y = e/r | |
| Explanation: the bending equation is given by m/i = σ/y = e/r | |
| 91. |
On bending of a beam, which is the layer which is neither elongated nor shortened? |
| A. | axis of load |
| B. | neutral axis |
| C. | center of gravity |
| D. | none of the mentioned |
| Answer» B. neutral axis | |
| Explanation: when a beam is in bending the layer in the direction of bending will be in compression and the other will be in tension. one side of the neutral axis will be shortened and the other will be elongated. | |
| 92. |
Consider a 250mmx15mmx10mm steel bar which is free to expand is heated from 15C to 40C. what will be developed? |
| A. | compressive stress |
| B. | tensile stress |
| C. | shear stress |
| D. | no stress |
| Answer» D. no stress | |
| Explanation: if we resist to expand then only stress will develop. here the bar is free to expand so there will be no stress. | |
| 93. |
The motion of the earth about its axis is periodic and simple harmonic. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: the earth takes 24 hours to complete its rotation about its axis, but the concept of to and fro motion is absent, and hence the rotation of the earth is periodic and not simple harmonic. | |
| 94. |
An object of mass 0.2kg executes simple harmonic motion along the x-axis with a frequency of (25/π)Hz. At the position x = 0.04, the object has kinetic energy of 0.5J and potential energy 0.4J. The amplitude of oscillation is? |
| A. | 6cm |
| B. | 4cm |
| C. | 8cm |
| D. | 2cm |
| Answer» A. 6cm | |
| Explanation: total energy, e=2π2 mv2 a2 0.5+0.4=2π2×0.2×(25/π)2 a2 a2=0.9/(0.4×252) a=3/(2×25)=3/50 m=6cm. | |
| 95. |
A spring of force constant 800N/m has an extension of 5cm. The work done in extending it from 5cm to 15cm is? |
| A. | 8j |
| B. | 16j |
| C. | 24j |
| D. | 32j |
| Answer» A. 8j | |
| Explanation: at x1 = 5 cm, u1=1/2×k(x1)2=1/2×800×0.052=1j | |
| 96. |
A simple pendulum is attached to the roof of a lift. If the time period of oscillation, when the lift is stationary is T, then the frequency of oscillation when the lift falls freely, will be |
| A. | zero |
| B. | t |
| C. | 1/t |
| D. | ∞ |
| Answer» A. zero | |
| Explanation: in a freely falling lift, g=0 | |
| 97. |
A lightly damped oscillator with a frequency v is set in motion by a harmonic driving force of frequency v’. When v’ is lesser than v, then the response of the oscillator is controlled by |
| A. | spring constant |
| B. | inertia of the mass |
| C. | oscillator frequency |
| D. | damping coefficient |
| Answer» A. spring constant | |
| Explanation: frequency of driving force is lesser than frequency v of a damped oscillator. the vibrations are nearly in phase | |
| 98. |
What is the time period of a pendulum hanged in a satellite? (T is the time period on earth) |
| A. | zero |
| B. | t |
| C. | infinite |
| D. | t/√6 |
| Answer» C. infinite | |
| Explanation: in a satellite, g= 0 t=2π√(l/g)=2π√(l/0)=∞. | |
| 99. |
Which of the following functions represents a simple harmonic oscillation? |
| A. | sinωt-cosωt |
| B. | sinωt+sin2ωt |
| C. | sinωt-sin2ωt |
| D. | sin2 ωt |
| Answer» A. sinωt-cosωt | |
| Explanation: y=sinωt-cosωt dy/dt=ωcosωt+ωsinωt | |
| 100. |
The ratio of velocity of sound in hydrogen and oxygen at STP is |
| A. | 16:1 |
| B. | 8:1 |
| C. | 4:1 |
| D. | 2:1 |
| Answer» C. 4:1 | |
| Explanation: vh/vo = √(mo/mh) | |
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