MCQOPTIONS
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MCQOPTIONS
This section includes 439 Mcqs, each offering curated multiple-choice questions to sharpen your SRMJEEE knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
A cantilever is a beam whose |
| A. | Both ends are supported either on rollers or hinges |
| B. | One end is fixed and other end is free |
| C. | Both ends are fixed |
| D. | Whose both or one of the end has overhang |
| Answer» C. Both ends are fixed | |
| 102. |
Shear force of following diagram |
| A. | Rectangle |
| B. | Square |
| C. | Circle |
| D. | Trapezoidal |
| Answer» B. Square | |
| 103. |
SI units of Bending moment is |
| A. | kN |
| B. | kN2 |
| C. | kNm |
| D. | km |
| Answer» D. km | |
| 104. |
Shear stress causes . |
| A. | Deformation |
| B. | Elongation |
| C. | contraction |
| D. | None of above |
| Answer» E. | |
| 105. |
Net force acting on the cross section of beam in bending is |
| A. | Tensile |
| B. | Compressive |
| C. | Shear |
| D. | None |
| Answer» E. | |
| 106. |
Bending equation is applicable to a beam of |
| A. | Heterogeneous material |
| B. | Homogeneous material |
| C. | Alloy |
| D. | None |
| Answer» C. Alloy | |
| 107. |
Moment of inertia of a circular section of 2 cm diameter, about an axis through its centre of gravity, is . |
| A. | π/64 |
| B. | π/4 |
| C. | π/16 |
| D. | π/2 |
| Answer» C. π/16 | |
| 108. |
The stress induced in a body due to suddenly applied load compared to when it is applied gradually is |
| A. | same |
| B. | half |
| C. | two times |
| D. | four times |
| Answer» D. four times | |
| 109. |
For any part of the beam, between two concentrated load Shear force diagram is a |
| A. | Horizontal straight line |
| B. | Vertical straight line |
| C. | Line inclined to x-axis |
| D. | Parabola |
| Answer» B. Vertical straight line | |
| 110. |
What is the unit of section modulus? |
| A. | mm |
| B. | mm2 |
| C. | mm3 |
| D. | mm4 |
| Answer» D. mm4 | |
| 111. |
The modulus of rigidity and poisson‟s ratio of a material are 80 GPa and 0.3 respectively. Its young‟s modulus will be |
| A. | 160 GPa |
| B. | 208 GPa |
| C. | 120 GPa |
| D. | 104 GPa |
| Answer» E. | |
| 112. |
When the bending moment is parabolic curve between two points, it indicates that there is |
| A. | No loading between the two points |
| B. | Point loads between the two points |
| C. | U.D.L. between the two points |
| D. | Uniformly varying load between the two points |
| Answer» D. Uniformly varying load between the two points | |
| 113. |
The length of a column, having a uniform circular cross-section of 7.5 cm diameter and whose ends are hinged, is 5 m. If the value of E for the material is 2100 kN/cm2, the permissible maximum crippling load will be |
| A. | 1.288 kN |
| B. | 12.88 kN |
| C. | 128.8 kN |
| D. | 288.0 kN |
| Answer» E. | |
| 114. |
In bending, neutral axis always is |
| A. | Perpendicular to the centroidal axis |
| B. | Coincides with the centroidal axis |
| C. | Parallel to the centroidal axis |
| D. | None |
| Answer» C. Parallel to the centroidal axis | |
| 115. |
Moment of Inertia is the integration of the square of the distance of the centroidand the del area along the whole area of the structure. |
| A. | True |
| B. | False |
| C. | none |
| D. | all |
| Answer» B. False | |
| 116. |
A sudden increase or decrease in shear force diagram between any twopoints indicates that there is |
| A. | No loading between the two points |
| B. | Point loads between the two points |
| C. | U.D.L. between the two points |
| D. | None of these |
| Answer» C. U.D.L. between the two points | |
| 117. |
In a cantilever carrying a uniformly varying load starting from zero at the free end, the Bending moment diagram is |
| A. | A horizontal line parallel to x-axis |
| B. | A line inclined to x-axis |
| C. | Follows a parabolic law |
| D. | Follows a cubic law |
| Answer» E. | |
| 118. |
Shear stress in the beam acting on the cross section is |
| A. | Normal to the cross section |
| B. | Tangential to the cross section |
| C. | Neither normal nor tangential |
| D. | None |
| Answer» C. Neither normal nor tangential | |
| 119. |
Shear force diagram is representation of shear force plotted as ordinate. |
| A. | Scalar |
| B. | Aerial |
| C. | Graphical |
| D. | Statically |
| Answer» D. Statically | |
| 120. |
stress at which extension of material takes place more quickly as comparedto increase in load is called |
| A. | elastic point of the material |
| B. | plastic point of the material |
| C. | breaking point of the material |
| D. | yielding point of the material |
| Answer» E. | |
| 121. |
A uniformly distributed load of 20 kN/m acts on a simply supported beam of rectangular cross section of width 20 mm and depth 60 mm. What is the maximum bending stress acting on the beam of 5m? |
| A. | 5030 Mpa |
| B. | 5208 Mpa |
| C. | 6600 Mpa |
| D. | Insufficient data |
| Answer» C. 6600 Mpa | |
| 122. |
Deformation per unit length in the direction of force is known as |
| A. | Strain |
| B. | lateral strain |
| C. | linear strain |
| D. | linear stress |
| Answer» E. | |
| 123. |
The moment of inertia of a triangular section of base „b‟ and height „h‟ about an axis passing through its base is ……. times the moment of inertia about an axis passing through its C.G. and parallel to the base |
| A. | 9 |
| B. | 4 |
| C. | 2 |
| D. | 3 |
| Answer» E. | |
| 124. |
Point of contra-flexure is a |
| A. | Point where Shear force is maximum |
| B. | Point where Bending moment is maximum |
| C. | Point where Bending moment is zero |
| D. | Point where Bending moment=0 but also changes sign from positive to negative |
| Answer» E. | |
| 125. |
A simply supported beam 6 m long and of effective depth 50 cm, carries a uniformly distributed load 2400 kg/m including its self weight. If the lever arm factor is 0.85 and permissible tensile stress of steel is 1400 kg/cm2, the area of steel required, is |
| A. | 14 cm2 |
| B. | 15 cm2 |
| C. | 16 cm2 |
| D. | 17 cm2 |
| Answer» D. 17 cm2 | |
| 126. |
A composite section of R.C.C. column 300mm×300mm in section having 20mm diameter 4 bars, one at each corner. Strength of concrete is 5 N/mm2 and modular ratio Es/Ec=9. Calculate load taken by column. |
| A. | 150 KN |
| B. | 200 KN |
| C. | 400 KN |
| D. | 500 KN |
| Answer» E. | |
| 127. |
What is modular ratio? |
| A. | ratio of deflection in each material |
| B. | ratio of modulus of elasticity of bot h material |
| C. | ratio of load acting in each section |
| D. | all of above |
| Answer» D. all of above | |
| 128. |
For any part of the beam, between two concentrated load Shear force diagramis a |
| A. | Horizontal straight line |
| B. | Vertical straight line |
| C. | Line inclined to x-axis |
| D. | Parabola |
| Answer» B. Vertical straight line | |
| 129. |
What is the formula of radius of gyration? |
| A. | k2 = I/A |
| B. | k2 = I2/A |
| C. | k2 = I2/A2 |
| D. | k2 = (I/A)1/2 |
| Answer» B. k2 = I2/A | |
| 130. |
Tensile strength of a material is obtained by dividing the maximum loadduring the test by the |
| A. | area at the time of fracture |
| B. | original cross-sectional area |
| C. | average of (a) and (b) |
| D. | minimum area after fracture |
| Answer» C. average of (a) and (b) | |
| 131. |
Young's modulus is defined as the ratio of |
| A. | volumetric stress and volumetric strain |
| B. | lateral stress and lateral strain |
| C. | longitudinal stress and longitudinal strain |
| D. | shear stress to shear strain |
| Answer» B. lateral stress and lateral strain | |
| 132. |
What is the maximum shear force, when a cantilever beam is loaded withudl throughout? |
| A. | w×l |
| B. | w |
| C. | w/l |
| D. | w+l |
| Answer» B. w | |
| 133. |
Which of the following are statically determinate beams? |
| A. | Only simply supported beams |
| B. | Cantilever, overhanging and simply supported |
| C. | Fixed beams |
| D. | Continuous beams |
| Answer» C. Fixed beams | |
| 134. |
The Young‟s modulus of elasticity of a material is 2.5 times its modulus of rigidity. The Poisson‟s ratio for the material will be |
| A. | 0.25 |
| B. | 0.33 |
| C. | 0.50 |
| D. | 0.75 |
| Answer» B. 0.33 | |
| 135. |
Bending stresses in a beam are of |
| A. | Constant values |
| B. | Variable values |
| C. | Constant nature |
| D. | None |
| Answer» C. Constant nature | |
| 136. |
In a simple supported beam having length = l and subjected to a concentratedload (W) at mid-point. |
| A. | Maximum Bending moment = Wl/4 at the mid-point |
| B. | Maximum Bending moment = Wl/4 at the end |
| C. | Maximum Bending moment = Wl/8 at the mid-point |
| D. | Maximum Bending moment = Wl/8 at the end |
| Answer» B. Maximum Bending moment = Wl/4 at the end | |
| 137. |
In helicopters tail rotor is used to control |
| A. | Yaw |
| B. | Roll |
| C. | Pitch |
| D. | Lift |
| Answer» B. Roll | |
| 138. |
A hypersonic air breathing propulsion requires the following type of combustion chamber |
| A. | can type gas turbine combustor |
| B. | annular type gas turbine combustor |
| C. | ramjet combustor |
| D. | supersonic combustor |
| Answer» E. | |
| 139. |
μ is coefficient of friction. A wheeled vehicle travelling on a circular level track will slip and overturn simultaneously if the ratio of its wheel distance to the height of its centroid, is |
| A. | μ |
| B. | 2μ |
| C. | 3μ |
| D. | 1/2μ |
| Answer» C. 3μ | |
| 140. |
If a particle is projected inside a horizontal tunnel which is 554 cm high with a velocity of 60 m per sec, the angle of projection for maximum range, is |
| A. | 8° |
| B. | 9° |
| C. | 10° |
| D. | 11° |
| Answer» D. 11° | |
| 141. |
Work may be defined as |
| A. | force x distance |
| B. | force x velocity |
| C. | force x acceleration |
| D. | none of these. |
| Answer» B. force x velocity | |
| 142. |
The force which produces an acceleration of 1 m/sec² in a mass of one kg, is called |
| A. | dyne |
| B. | Netwon |
| C. | joule |
| D. | erg. |
| Answer» C. joule | |
| 143. |
The instantaneous centre of a member lies at the point of intersection of two lines drawn at the ends of the member such that the lines are inclined to the direction of motion of the ends at |
| A. | 30° |
| B. | 45° |
| C. | 60° |
| D. | 90° |
| Answer» E. | |
| 144. |
Power can be expressed as |
| A. | work/energy |
| B. | work/time |
| C. | work x time |
| D. | work/distance. |
| Answer» C. work x time | |
| 145. |
The maximum velocity of a body vibrating with a simple harmonic motion of amplitude 150 mm and frequency 2 vibrations/sec, is |
| A. | 188.5 m/sec |
| B. | 18.85 m/sec |
| C. | 1.885 m/sec |
| D. | 0.18845 m/sec. |
| Answer» D. 0.18845 m/sec. | |
| 146. |
The resultant of two forces acting at right angles is 5 kgf and if they act at an angle of 60°, it is √37 kgf. The magnitudes of the forces are : |
| A. | 2 kgf, 3 kgf |
| B. | 3 kgf, 4 kgf |
| C. | 4 kgf, 5 kgf |
| D. | 5 kgf, 3 kgf. |
| Answer» C. 4 kgf, 5 kgf | |
| 147. |
Angular acceleration of a particle may be expressed as |
| A. | radians/sec² |
| B. | degrees/sec² |
| C. | revolutions/sec |
| D. | all the above. |
| Answer» E. | |
| 148. |
A particle moving with a simple harmonic motion, attains its maximum velocity when it passes |
| A. | the extreme point of the oscillation |
| B. | through the mean position |
| C. | through a point at half amplitude |
| D. | none of these. |
| Answer» C. through a point at half amplitude | |
| 149. |
The resultant of two forces acting at right angles is √34 kg and acting at 60° is 70 kg. The forces are |
| A. | 1 kg and 4 kg |
| B. | 2 kg and 3 kg |
| C. | 3 kg and √5 kg |
| D. | 3 kg and 5 kg. |
| Answer» E. | |
| 150. |
The phenomenon of collision of two elastic bodies takes place because bodies |
| A. | immediately after collision come momentarily to rest |
| B. | tend to compress each other till they are compressed maximum possible |
| C. | attempt to regain its original shape due to their elasticities |
| D. | all the above. |
| Answer» E. | |