Q1. In a radial cam mechanism, the follower primarily moves in which direction relative to the cam center?

A. Along a line passing through the cam center B. Along a line tangent to the cam C. Only in a circular path D. Perpendicular to the follower axis

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: A radial (disc) cam normally drives a follower whose line of motion passes through, or is offset from, the cam center. The cam profile converts rotary motion into reciprocating or oscillating follower motion. The follower itself does not move tangentially around the cam.

Q2. Which follower is most suitable when very high-speed operation requires low sliding friction at the cam-follower contact?

A. Knife-edge follower B. Roller follower C. Flat-faced follower D. Spherical follower

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: A roller follower replaces sliding contact with rolling contact, greatly reducing friction and wear. Knife-edge followers have high sliding and contact stresses, while flat-faced followers also experience substantial sliding.

Q3. For a simple harmonic motion follower rise, the follower velocity is zero at which positions?

A. Only at the middle of the rise B. Only at the beginning C. At the beginning and end of the rise D. It is never zero

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: In simple harmonic motion, displacement follows a cosine-type law and velocity is proportional to sine of the cam angle. Hence velocity is zero at the two ends of the stroke and maximum near the middle.

Q4. The main purpose of a dwell period in a cam profile is to:

A. Increase follower velocity continuously B. Reverse cam rotation C. Eliminate the base circle D. Keep the follower stationary while the cam continues to rotate

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: During dwell, the follower displacement remains constant even though the cam rotates. Dwell is therefore used whenever a mechanism requires the output to remain at a fixed position for a specified cam angle.

Q5. Which cam follower has a line contact ideally at the point of contact and therefore can sustain comparatively high loads?

A. Roller follower B. Knife-edge follower C. Point-contact spherical follower D. Flexible follower

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: A roller follower has a finite contact width and develops line contact with the cam surface under idealized conditions. This permits lower contact stress than an ideal knife-edge point contact and is suitable for higher loads.

Q6. Pressure angle in a cam mechanism is the angle between the follower direction of motion and the:

A. Tangent to the cam profile B. Normal to the cam profile at the contact point C. Cam shaft axis D. Base circle radius only

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: The pressure angle is measured between the follower line of motion and the common normal at the cam-follower contact. A large pressure angle increases the side thrust on the follower and is generally undesirable.

Q7. A translating roller follower is offset from the cam center. The offset mainly affects the:

A. Camshaft rotational speed only B. Follower material density C. Pressure-angle characteristics and required cam profile D. Base-circle diameter only

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Follower offset changes the geometry between the follower line of motion and the cam center. Consequently it alters pressure angle, contact geometry, and the exact profile required to produce the desired motion.

Q8. Which follower motion law is generally preferred for moderate-speed operation when acceleration discontinuities at the ends are to be minimized compared with uniform acceleration?

A. Uniform velocity motion B. Instantaneous step motion C. Constant displacement motion D. Cycloidal motion

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: Cycloidal motion gives zero velocity and zero acceleration at the beginning and end of a stroke. This avoids the infinite acceleration associated with idealized instantaneous changes and provides good dynamic characteristics.

Q9. For a cam with a circular-arc follower motion, the follower displacement is controlled primarily by the:

A. Geometry of the cam profile and follower linkage B. Mass of the camshaft alone C. Lubricant viscosity alone D. Bearing diameter alone

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: Cam mechanisms are kinematic devices: the follower displacement is determined by the cam profile and the follower geometry. Mass and lubrication affect dynamics and losses but do not define the prescribed geometric displacement law.

Q10. Which statement about a knife-edge follower is correct?

A. It always has rolling contact B. It has very small contact area and therefore high contact stress C. It eliminates wear completely D. It cannot transmit any force

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: A knife-edge follower theoretically contacts the cam at a point. The small contact area causes high contact stress and wear, so knife-edge followers are generally limited to low-speed or lightly loaded applications.

Q11. A cam is rotating clockwise and its follower is required to rise while maintaining minimum pressure angle. A common design approach is to:

A. Increase the follower mass indefinitely B. Use zero clearance in every bearing C. Select an appropriate base circle and follower offset/profile geometry D. Reduce the camshaft speed to zero

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Pressure angle depends strongly on cam geometry, follower line of motion, and rise specification. Base-circle size and offset can be selected to obtain acceptable pressure angles while satisfying the required displacement law.

Q12. The base circle of a radial cam is the:

A. Largest circle that passes through every profile point B. Pitch circle of the gear C. Circle through the follower center D. Smallest circle centered at the cam axis and tangent to the cam profile reference

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: The base circle is the smallest circle centered on the cam rotational axis and tangent to the cam profile. It is a fundamental reference used for defining lift and profile geometry.

Q13. A dwell in a cam displacement diagram is represented by a segment with:

A. Zero slope B. Maximum slope C. Negative infinite slope D. Constant positive slope

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: Follower velocity is proportional to the slope of the displacement-versus-cam-angle curve. During dwell, displacement is constant, so the slope and follower velocity are both zero.

Q14. If the cam speed is doubled while the cam angle for a prescribed motion remains unchanged, the time available for the motion becomes:

A. Double B. Half C. Four times D. Unchanged

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: Time for a given cam angle is inversely proportional to angular speed. Doubling the cam speed therefore halves the time available for rise, return, or dwell over the same angular interval.

Q15. In a cam mechanism, the pitch curve of a roller follower is the locus of the:

A. Cam center B. Point on the base circle C. Roller center as the cam rotates D. Follower pivot only

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: The pitch curve is obtained by tracing the center of the roller relative to the cam as the cam rotates. The actual cam profile is offset from this pitch curve by the roller radius along the normal direction.

Q16. For a helical compression spring, the spring index is defined as:

A. Wire diameter divided by free length B. Free length divided by mean diameter C. Pitch divided by wire length D. Mean coil diameter divided by wire diameter

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: Spring index C = D/d, where D is mean coil diameter and d is wire diameter. It is an important geometric parameter affecting manufacturability, stress concentration, and spring performance.

Q17. A compression spring primarily stores mechanical energy in the form of:

A. Elastic strain energy B. Kinetic energy only C. Thermal energy only D. Chemical energy

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: When a spring is compressed, elastic deformation stores strain energy. For a linear spring, the stored energy is U = 1/2 kx², where k is stiffness and x is deflection.

Q18. Which spring is commonly used when a force is required to act along the spring axis and the spring is subjected to compression?

A. Leaf spring only B. Helical compression spring C. Torsion bar only D. Spiral clock spring only

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: A helical compression spring is specifically designed to resist axial compressive loads and return to its original length after unloading, within its elastic limit.

Q19. The stiffness of a close-coiled helical compression spring is increased by:

A. Increasing the number of active coils B. Reducing wire diameter C. Increasing wire diameter while keeping other dimensions appropriate D. Increasing mean coil diameter

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: For a helical compression spring, k is proportional to d⁴/(D³n). Thus increasing wire diameter has a strong positive effect on stiffness, whereas increasing active coils or coil diameter reduces stiffness.

Q20. A Belleville spring is best described as a:

A. Helical tension spring B. Leaf spring with constant radius C. Rubber torsion spring D. Conical disc spring

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: A Belleville spring is a conical disc spring capable of producing relatively high force over a small axial deflection. Stacking arrangements can tailor its load-deflection behavior.

Q21. A spring used to return a valve or lever to its initial position after actuation is functioning mainly as a:

A. Restoring actuator B. Speed governor C. Flywheel D. Rigid link

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: The spring stores energy during displacement and releases it to return the mechanism toward its original position. Hence it acts as a restoring element or actuator.

Q22. The Wahl correction factor is associated with the design of:

A. Flat belts B. Helical springs C. Gear teeth only D. Flywheels

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: The Wahl factor accounts for the combined effects of direct shear and curvature-induced stress concentration in helical spring wire. It is widely used in spring stress calculations.

Q23. Which spring type is most suitable for storing energy while allowing large angular rotation?

A. Compression spring only B. Disc spring only C. Spiral spring D. Leaf spring with no curvature

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Spiral springs are wound strips designed to store energy through bending and are useful when a relatively large angular displacement is required, such as in clocks and recoil mechanisms.

Q24. A tension spring generally has hooks or loops at its ends because it:

A. Must always carry compressive load B. Must prevent any deformation C. Works only in torsion D. Must be connected to carry an axial tensile load

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: Extension or tension springs are stretched by axial tensile loading. End hooks or loops provide convenient attachment points so the spring can transmit the tensile force to the connected members.

Q25. Spring surge is undesirable because it can cause:

A. Dynamic oscillations and excessive stresses B. Permanent elimination of vibration C. Zero deflection under load D. Infinite stiffness

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: Spring surge is a longitudinal resonance phenomenon in which different parts of a spring oscillate dynamically. It can cause noise, loss of force control, and high cyclic stresses.

Q26. A spring actuator is called nonlinear when its force-deflection relation is:

A. Exactly proportional at every point B. Not proportional to deflection over the operating range C. Always zero D. Independent of displacement

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: For a linear spring, F = kx and stiffness is constant. A nonlinear spring has a variable slope dF/dx, which may be achieved through geometry, material behavior, or progressive coil contact.

Q27. In a torsion spring, the principal deformation of the wire is:

A. Pure axial compression B. Pure axial extension C. Bending due to applied moment D. Rolling contact

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Although called a torsion spring, the wire in a close-coiled torsion spring is predominantly subjected to bending as the spring arms rotate. The applied torque produces a bending moment in the wire.

Q28. Which factor generally increases the fatigue life of a spring?

A. Increasing stress concentration B. Operating continuously beyond yield C. Introducing sharp surface defects D. Reducing operating stress amplitude

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: Fatigue damage is strongly related to cyclic stress amplitude and surface condition. Lowering alternating stress and improving surface quality generally increases the number of cycles to failure.

Q29. The energy stored in a linear spring is proportional to the:

A. Square of deflection B. First power of deflection only C. Cube of mass D. Inverse of stiffness only

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: For a linear spring, U = 1/2 kx². Therefore, for a fixed stiffness, doubling deflection increases stored energy by a factor of four.

Q30. A leaf spring in an automobile suspension primarily acts as a:

A. Rigid gear pair B. Flexible load-carrying and energy-storing member C. Cam follower only D. Fluid actuator

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: A leaf spring supports load, deflects elastically, stores strain energy, and can also locate the axle in some suspension designs. It therefore combines spring and structural functions.

Q31. In a simple gear pair, the velocity ratio is determined primarily by the:

A. Module alone B. Face width alone C. Numbers of teeth on the gears D. Gear material only

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: For two external gears, speed ratio magnitude is inversely related to tooth counts: N1/N2 = T2/T1. Module and material affect size, strength, and durability but not the ideal speed ratio.

Q32. A simple gear train is used when:

A. Several gears must be rigidly compounded on one shaft B. All gears must be planetary C. The input shaft must not rotate D. Each shaft carries at most one gear in the train

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: In a simple gear train, each shaft carries only one gear. Idler gears may be inserted to alter shaft spacing or direction without changing the overall magnitude of the speed ratio.

Q33. The main function of an idler gear is to:

A. Change direction or spacing without changing the overall speed ratio magnitude B. Always increase speed C. Always decrease speed D. Increase torque without changing speed

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: An idler gear introduces another gear mesh. For an ideal simple train, its tooth count cancels from the overall ratio, so it mainly changes rotation direction and permits greater center distance.

Q34. A compound gear train is characterized by:

A. Only one gear on each shaft B. Two or more gears rigidly mounted on the same shaft C. No fixed shafts D. Only internal gears

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: In a compound train, at least one shaft carries two or more gears rigidly together. This allows larger speed ratios to be obtained in a compact arrangement.

Q35. A reverted gear train has the input and output shafts:

A. Always perpendicular B. Always parallel but widely separated C. Coaxial D. At random positions

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: A reverted gear train is a compound gear train in which the first and last gears are coaxial. It is useful where a large speed ratio is needed while keeping input and output shafts on the same axis.

Q36. In an epicyclic gear train, at least one gear:

A. Must remain stationary permanently B. Has no teeth C. Must be a bevel gear D. Has an axis that moves relative to the frame

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: In an epicyclic or planetary train, one or more gear axes move relative to the fixed frame, carried by an arm or carrier. This produces compact arrangements with multiple possible ratios.

Q37. The sun gear in a typical planetary gear train is:

A. The central gear around which planet gears revolve B. The outer stationary ring in every design C. A belt pulley D. The carrier itself

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: The sun gear is the central gear. Planet gears mesh with the sun and usually also with a ring gear, while the carrier supports the planet gear axes.

Q38. A differential gear mechanism is used primarily to:

A. Force both wheels to rotate at exactly equal speed B. Permit two output shafts to rotate at different speeds while transmitting torque C. Eliminate all torque D. Convert rotation to pure translation

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: An automotive differential allows the left and right driven wheels to rotate at different speeds during cornering while transmitting drive torque to both. The average output speed is related to the carrier speed.

Q39. For involute gears, the pressure angle is measured between the common normal at contact and the:

A. Gear shaft axis B. Tooth root only C. Common tangent to the pitch circles D. Addendum circle radius

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: The pressure angle is the angle between the line of action, which is normal to the involute tooth surfaces, and the common tangent to the pitch circles. It affects radial force and tooth geometry.

Q40. The module of a metric gear is defined as:

A. Number of teeth divided by pitch diameter squared B. Outside diameter divided by face width C. Circular pitch divided by tooth number D. Pitch circle diameter divided by number of teeth

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: Metric module m = d/T, where d is pitch circle diameter and T is number of teeth. A larger module means larger teeth for a given tooth count.

Q41. Which gear pair is normally used to transmit motion between intersecting shafts?

A. Bevel gears B. Spur gears only C. Rack and pinion D. Worm and wheel only

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: Bevel gears have conical pitch surfaces and are designed primarily for intersecting shafts. Spur gears are for parallel shafts, while worm gears generally connect non-parallel, non-intersecting shafts.

Q42. A worm gear drive is particularly useful when:

A. Only a 1:1 ratio is required B. A large speed reduction is required in a compact arrangement C. Shafts must intersect at 90° exactly in all cases D. No sliding is permitted

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: Worm drives can provide large speed reductions in one stage and can be compact. Their high sliding action, however, requires appropriate lubrication and thermal consideration.

Q43. Gear backlash is the:

A. Distance between two shafts B. Tooth height above the pitch circle C. Clearance between mating tooth flanks in the circumferential direction D. Thickness of the gear rim

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Backlash is the intentional clearance between mating tooth surfaces measured along the pitch circle. Some backlash is required to accommodate manufacturing errors, thermal expansion, and lubrication.

Q44. A rack and pinion mechanism converts:

A. Only rotary motion into rotary motion B. Only linear motion into angular oscillation C. Heat into mechanical energy D. Rotary motion into linear motion or vice versa

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: A pinion is a gear that meshes with a straight rack. Rotation of the pinion produces linear rack motion, while linear rack motion can drive the pinion in the reverse direction.

Q45. Gear efficiency is reduced mainly by:

A. Friction, sliding, and other power losses B. Increasing tooth accuracy C. Using suitable lubrication D. Reducing unnecessary sliding

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: Real gears lose power through tooth friction, bearing losses, windage, and other effects. Proper lubrication and accurate manufacture reduce these losses, but they cannot make a real transmission perfectly lossless.

Q46. A four-bar linkage consists of:

A. Three links and one gear B. Four rigid links connected by four lower pairs C. Four gears only D. Two links and two springs

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: A four-bar mechanism has four rigid links connected by four turning pairs. One link is fixed, and the remaining links form the input, coupler, and output members.

Q47. In a four-bar mechanism, the link fixed to the frame is called the:

A. Coupler link B. Follower pin C. Ground link D. Crank pin

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: The ground link is the fixed reference member. The other three links move relative to it and are commonly designated as crank, rocker, and coupler depending on the configuration.

Q48. A four-bar linkage has at least one link capable of complete rotation when the:

A. All links have equal lengths only B. All joints are welded C. Coupler length is zero D. Grashof condition is satisfied and the appropriate link is fixed

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: For a four-bar mechanism, Grashof's criterion s + l ≤ p + q indicates that at least one link can rotate completely. Which link becomes the crank depends on which member is fixed.

Q49. A crank-rocker mechanism has:

A. One link rotating continuously and another oscillating B. Two continuously rotating cranks only C. Only translational motion D. No fixed link

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: In a crank-rocker mechanism, the shortest link can rotate fully when an appropriate link is fixed, while the opposite rocker oscillates through a limited angle. The coupler transfers motion between them.

Q50. A double-crank or drag-link mechanism occurs when:

A. The longest link is always fixed B. The shortest link is fixed and Grashof's condition permits rotation C. No link can rotate D. All four links are equal and welded

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: If the shortest link is fixed in a Grashof four-bar, both links connected to the fixed link can rotate completely. This is known as a double-crank or drag-link mechanism.

Q51. A double-rocker mechanism is obtained when:

A. The shortest link is always fixed B. There is no fixed link C. The link opposite the shortest link is fixed in a Grashof chain D. The coupler is removed

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: In a Grashof four-bar, fixing the link opposite the shortest link results in a double-rocker arrangement: the two links connected to the fixed frame oscillate rather than making complete rotations.

Q52. The coupler link of a four-bar mechanism is the link that:

A. Is always fixed B. Always rotates at constant speed C. Contains the fixed frame D. Connects the input and output moving links

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: The coupler joins the crank and rocker and transmits motion between them. Its motion is generally complex, combining translation and rotation rather than being a simple pure rotation.

Q53. Toggle positions in a four-bar linkage occur when:

A. The relevant links become collinear, producing a limiting configuration B. All links are perpendicular C. The frame is removed D. The coupler is weightless

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: At a toggle or dead-center configuration, two relevant links become collinear. Mechanical advantage can become very large theoretically, but the mechanism may have poor force-transmission characteristics and may lock or reverse depending on the application.

Q54. The transmission angle in a four-bar mechanism is commonly used to assess:

A. Gear tooth hardness B. How effectively force is transmitted through the linkage C. Spring fatigue only D. Bearing temperature only

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: Transmission angle is a useful kinematic indicator of force-transmission quality. Values near 90° are generally favorable for many crank-rocker applications, whereas very small or very large deviations can increase joint forces.

Q55. A Watt's straight-line mechanism is an example of a:

A. Pure gear train B. Spring actuator C. Approximate straight-line linkage D. Cam with roller follower

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Watt's linkage uses a coupler point that follows an approximately straight path over a limited range. It is a classic example of using link geometry to approximate straight-line motion.

Q56. A four-bar mechanism used to generate a prescribed output oscillation is classified as a:

A. Force sensor B. Flywheel C. Rigid shaft coupling D. Function-generating mechanism

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: Function generation means relating input motion to a desired output motion. Four-bar linkages can be synthesized so that the output rocker angle approximates a specified function of crank angle.

Q57. The instantaneous center method is useful in four-bar analysis because it can determine:

A. Instantaneous velocity relationships B. Only static stress C. Material hardness D. Gear tooth temperature

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: Instantaneous centers provide a graphical method for determining velocity relationships between links. For planar mechanisms, relative velocities can be related through the instantaneous centers of the links.

Q58. If all four links of a four-bar mechanism are equal, the mechanism is commonly called a:

A. Worm drive B. Parallelogram linkage C. Slider-crank D. Geneva mechanism

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: With all four links equal, opposite links remain parallel during motion, forming a parallelogram linkage. This arrangement can be used to preserve orientation of a coupler-related member.

Q59. The angular velocity of the coupler in a general four-bar mechanism is:

A. Always equal to input speed B. Always zero C. Generally variable during a cycle D. Always constant and independent of geometry

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Because the four-bar has changing geometry, the coupler's instantaneous center and velocity change continuously. Therefore its angular velocity generally varies even if the input crank rotates uniformly.

Q60. A mechanism in which the input and output shafts rotate continuously in the same general direction can be obtained using a suitable:

A. Single fixed link only B. Knife-edge cam C. Compression spring D. Four-bar drag-link arrangement

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: A drag-link or double-crank four-bar can transmit continuous rotation between the input and output links when its geometry and fixed-link selection satisfy the required Grashof condition.

Q61. A single slider-crank mechanism converts:

A. Rotary motion into reciprocating motion or vice versa B. Only translation into translation C. Only rotary motion into rotary motion D. Heat into rotation

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: The crank rotates while the connecting rod transfers motion to the slider, producing reciprocating linear motion. The process is reversible, as in reciprocating engines and compressors.

Q62. The piston mechanism of a conventional reciprocating engine is a:

A. Double-crank four-bar only B. Single slider-crank mechanism C. Geneva mechanism D. Worm mechanism

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: The crank, connecting rod, and piston form the moving members of a slider-crank chain, with the cylinder/frame providing the fixed guide. It converts between crank rotation and piston reciprocation.

Q63. The connecting rod of a slider-crank mechanism is a:

A. Ternary link with three joints B. Fixed link C. Binary link D. Gear pair

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: A conventional connecting rod has two pin joints: one at the crank and one at the slider. A link with two joints is called a binary link.

Q64. In an offset slider-crank mechanism, the line of slider motion is:

A. Always coincident with the crank center B. Perpendicular to the crank plane only C. Circular D. Offset from the crank centerline

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: In an offset slider-crank, the slider guide does not pass through the crank center. This offset changes the kinematics and causes unequal forward and return strokes for a given crank rotation arrangement.

Q65. The slider-crank mechanism is a special case of a:

A. Four-bar mechanism with one sliding pair B. Gear train C. Cam mechanism D. Spring mechanism

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: A slider-crank can be viewed as a four-link mechanism in which one of the turning pairs is replaced by a sliding pair. It therefore belongs to the class of four-bar-derived mechanisms.

Q66. For a slider-crank with a very long connecting rod compared with crank radius, the piston motion approaches:

A. Uniform circular motion B. Simple harmonic motion C. Pure constant velocity motion D. Random motion

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: As connecting-rod length becomes very large compared with crank radius, the angularity of the rod becomes small. The slider displacement then approaches a sinusoidal or simple harmonic relation with crank angle.

Q67. In a slider-crank mechanism, dead centers occur when the crank and connecting rod are:

A. Perpendicular B. Parallel to the cylinder wall C. Collinear D. At 45° exactly

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: At inner and outer dead centers, the crank and connecting rod lie along the line of slider motion and are collinear. The slider is at an extreme position and instantaneous velocity is zero.

Q68. A Whitworth quick-return mechanism is used to:

A. Obtain equal stroke times only B. Eliminate reciprocation C. Convert translation into gear rotation D. Obtain a slower working stroke and faster return stroke

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: Quick-return mechanisms are designed so the cutting or working stroke takes more time than the return stroke. This improves productivity in machines such as shapers and slotters.

Q69. In a quick-return mechanism, the return stroke is called 'quick' because:

A. Its angular travel or time is smaller than that of the working stroke B. It has greater load C. It has zero displacement D. It always uses a spring

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: The defining feature is unequal time for the two strokes. The return stroke takes less time, so its average speed is higher than that of the working stroke when the stroke lengths are equal.

Q70. A crank and slotted lever quick-return mechanism is commonly used in a:

A. Centrifugal pump only B. Shaper machine C. Lathe chuck only D. Gear hobbing machine only

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: The crank and slotted lever mechanism is a standard quick-return mechanism for shaper machines. It provides a relatively slow cutting stroke and faster return stroke.

Q71. The Whitworth quick-return mechanism is derived from a:

A. Simple gear pair B. Four-bar parallelogram only C. Slider-crank inversion D. Cam and follower only

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: The Whitworth mechanism can be understood as an inversion of a slider-crank arrangement. Changing which link is fixed produces a mechanism with unequal timing of forward and return strokes.

Q72. In a crank-shaper mechanism, the ram usually performs:

A. Continuous rotary motion B. Pure oscillatory rotation C. Helical motion D. Reciprocating linear motion

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: The ram of a shaper moves back and forth along a straight line. The cutting action occurs during one stroke and the return stroke is made faster in quick-return designs.

Q73. The purpose of the quick-return ratio is to compare:

A. Time of cutting stroke with time of return stroke B. Gear tooth thickness with module C. Spring force with deflection only D. Bearing load with shaft speed

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: Quick-return ratio is commonly defined using the times or angular travel associated with the working and return strokes. It quantifies how much faster the return stroke is than the cutting stroke.

Q74. In a slider-crank, the slider velocity is maximum approximately when the crank is:

A. Exactly at both dead centers B. Near the mid-stroke region C. Always at the inner dead center D. Always at the outer dead center

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: At dead centers the slider velocity is zero. For a typical slider-crank, maximum speed occurs near the middle of the stroke, although the exact crank angle shifts with connecting-rod length.

Q75. Which mechanism is commonly used to convert rotary input into intermittent linear motion?

A. Simple rigid shaft B. Compression spring alone C. Geneva-type indexing combined with a suitable follower arrangement D. Flywheel alone

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Geneva mechanisms are primarily used for intermittent rotary motion, but indexed motion can be coupled to linear mechanisms when intermittent translation is required. The essential feature is controlled indexing rather than continuous motion.

Q76. A cam rotates at 300 rpm. How long does the cam take to rotate through 120°?

A. 0.0333 s B. 0.1000 s C. 0.2000 s D. 0.0667 s

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: At 300 rpm, one revolution takes 60/300 = 0.2 s. The angle 120° is one-third of a revolution, so time = 0.2/3 = 0.0667 s.

Q77. A follower rises 30 mm during 90° of cam rotation. If the cam rotates at 120 rpm, what is the average follower velocity during the rise?

A. 240 mm/s B. 120 mm/s C. 60 mm/s D. 480 mm/s

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: The cam rotates at 120 rpm = 2 revolutions/s. A 90° interval is one-quarter revolution, so rise time = 0.25/2 = 0.125 s. Average follower velocity = rise/time = 30/0.125 = 240 mm/s.

Q78. A linear spring has stiffness 500 N/m and is compressed by 40 mm. What force does it exert?

A. 12.5 N B. 20 N C. 50 N D. 200 N

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: Using Hooke's law, F = kx. Here k = 500 N/m and x = 0.04 m. Therefore F = 500 × 0.04 = 20 N.

Q79. A spring of stiffness 800 N/m is compressed by 50 mm. What elastic energy is stored?

A. 0.5 J B. 2.0 J C. 1.0 J D. 4.0 J

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: For a linear spring, U = 1/2 kx². With k = 800 N/m and x = 0.05 m, U = 0.5 × 800 × 0.0025 = 1.0 J.

Q80. A helical spring has mean coil diameter 40 mm and wire diameter 5 mm. What is its spring index?

A. 5 B. 10 C. 200 D. 8

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: Spring index C = D/d. Therefore C = 40/5 = 8. The spring index indicates the relative curvature of the wire and is important for stress concentration and manufacturability.

Q81. For a spring, stiffness is 2 kN/m. If the load is 100 N, the deflection is:

A. 50 mm B. 20 mm C. 100 mm D. 200 mm

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: The stiffness relation is F = kx. Thus x = F/k = 100/2000 = 0.05 m = 50 mm.

Q82. A 20-tooth gear drives a 60-tooth gear at 900 rpm. What is the speed of the 60-tooth gear?

A. 450 rpm B. 300 rpm C. 600 rpm D. 2700 rpm

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: For an ideal external gear pair, N1T1 = N2T2. Thus N2 = 900×20/60 = 300 rpm. The larger gear rotates more slowly and in the opposite direction.

Q83. A gear has 40 teeth and a module of 3 mm. What is its pitch circle diameter?

A. 13.33 mm B. 80 mm C. 120 mm D. 40 mm

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: For a metric gear, module m = d/T, so d = mT = 3×40 = 120 mm.

Q84. A 15-tooth pinion rotates at 1200 rpm and drives a 45-tooth gear. What is the gear speed?

A. 600 rpm B. 900 rpm C. 3600 rpm D. 400 rpm

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: N1T1 = N2T2. Therefore N2 = 1200×15/45 = 400 rpm. The speed reduction is 3:1.

Q85. Two external gears have 25 and 75 teeth. If the 25-tooth gear rotates at 1500 rpm, the 75-tooth gear rotates at:

A. 500 rpm B. 750 rpm C. 1000 rpm D. 4500 rpm

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: The speed ratio is inverse to tooth count. N2 = 1500×25/75 = 500 rpm. External gears rotate in opposite directions.

Q86. A compound gear train has a 20-tooth gear driving a 60-tooth gear on the second shaft; a 15-tooth gear on that shaft drives a 45-tooth output gear. The overall speed reduction is:

A. 3:1 B. 9:1 C. 6:1 D. 12:1

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: The first stage ratio is 60/20 = 3. The second stage ratio is 45/15 = 3. Multiplying the stages gives an overall reduction of 3×3 = 9:1.

Q87. A gear has 30 teeth and a pitch diameter of 90 mm. What is its module?

A. 0.333 mm B. 30 mm C. 3 mm D. 2700 mm

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Module m = pitch diameter/number of teeth = 90/30 = 3 mm.

Q88. A 24-tooth gear drives a 72-tooth gear. If the input torque is 20 N·m and losses are neglected, the output torque is:

A. 6.67 N·m B. 20 N·m C. 96 N·m D. 60 N·m

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: For an ideal gear pair, torque ratio equals the inverse speed ratio and equals tooth-count ratio for the driven to driver gear. Output torque = 20×72/24 = 60 N·m.

Q89. A four-bar linkage has link lengths 20, 30, 40 and 50 mm. If 20 mm is the shortest and 50 mm is the longest, what is s+l?

A. 70 mm B. 60 mm C. 80 mm D. 90 mm

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: Grashof's criterion uses s+l compared with the other two links. Here s+l = 20+50 = 70 mm. The other two links total 30+40 = 70 mm, so the chain is on the Grashof boundary.

Q90. A crank rotates at 120 rpm. How many complete revolutions does it make in 15 seconds?

A. 15 B. 30 C. 60 D. 120

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: 120 rpm equals 2 revolutions per second. In 15 seconds, revolutions = 2×15 = 30.

Q91. A crank completes one revolution in 0.5 s. What is its rotational speed?

A. 60 rpm B. 30 rpm C. 120 rpm D. 240 rpm

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: One revolution takes 0.5 s, so revolutions per second = 1/0.5 = 2. Multiplying by 60 gives 120 rpm.

Q92. A rocker oscillates through 60° in 0.4 s during one half-cycle. What is its average angular speed in degrees per second?

A. 60°/s B. 100°/s C. 240°/s D. 150°/s

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: Average angular speed = angular displacement/time = 60/0.4 = 150°/s. This is an average value; instantaneous speed generally varies throughout a linkage cycle.

Q93. A slider-crank has a crank radius of 50 mm. What is the piston stroke for a centered mechanism?

A. 100 mm B. 50 mm C. 25 mm D. 200 mm

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: For a centered slider-crank, stroke = 2r. With r = 50 mm, stroke = 100 mm.

Q94. A crank rotates at 600 rpm. What is its angular speed in rad/s?

A. 31.42 rad/s B. 62.83 rad/s C. 10 rad/s D. 120 rad/s

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: Angular speed ω = 2πN/60. For N = 600 rpm, ω = 2π×600/60 = 20π ≈ 62.83 rad/s.

Q95. A shaper has a cutting stroke time of 8 s and return stroke time of 5 s. What is the quick-return ratio defined as cutting time/return time?

A. 0.625 B. 3.0 C. 1.6 D. 13

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Quick-return ratio = time of cutting stroke / time of return stroke = 8/5 = 1.6. A value greater than one indicates that the return stroke is completed faster.

Q96. A Scotch yoke has an eccentric radius of 30 mm. What is the total stroke of the reciprocating member?

A. 30 mm B. 15 mm C. 90 mm D. 60 mm

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: For a Scotch yoke, the displacement amplitude equals the eccentric radius. Therefore the total peak-to-peak stroke is 2r = 60 mm.

Q97. A slider moves through a 120 mm stroke in 0.6 s during a working stroke. What is its average speed?

A. 200 mm/s B. 72 mm/s C. 120 mm/s D. 300 mm/s

Check Answer

Correct Answer: A   |   Marks: 1

Explanation: Average speed = distance/time = 120/0.6 = 200 mm/s. This is average speed; the instantaneous slider speed is not constant in a crank mechanism.

Q98. A Geneva mechanism makes 6 indexing movements per complete revolution of its driving member. The output advances through how many degrees per index?

A. 30° B. 60° C. 90° D. 120°

Check Answer

Correct Answer: B   |   Marks: 1

Explanation: If the output makes 6 equal indexing positions per revolution, each index corresponds to 360/6 = 60° of output rotation.

Q99. A rack moves 120 mm when its pinion rotates 2 revolutions. What is the pinion pitch circumference?

A. 120 mm/rev B. 240 mm/rev C. 60 mm/rev D. 30 mm/rev

Check Answer

Correct Answer: C   |   Marks: 1

Explanation: Rack travel per pinion revolution equals the pitch circumference of the pinion. Thus 120 mm/2 = 60 mm per revolution.

Q100. An Oldham coupling transmits 5 kW at 600 rpm. Neglecting losses, what torque is transmitted approximately?

A. 15.9 N·m B. 47.7 N·m C. 159.2 N·m D. 79.6 N·m

Check Answer

Correct Answer: D   |   Marks: 1

Explanation: Power P = Tω. Thus T = P/ω, where P = 5000 W and ω = 2π×600/60 = 62.83 rad/s. T ≈ 5000/62.83 = 79.6 N·m.