Q1. A kinematic chain consists of n links. The maximum number of possible inversions for this chain is
✅ Correct: A kinematic chain with n links can give a maximum of n inversions.
Each inversion is obtained by fixing a different link in the chain.
Hence, the total possible inversions equal the number of links, i.e., n.
Q2. The distance between two parallel shafts is 20 mm and they are connected by an Oldham’s coupling. The driving shaft revolves at 180 rpm. What will be the maximum speed of sliding of the tongue of the intermediate piece along its groove?
✅ Correct: Maximum sliding speed in Oldham’s coupling is \(2 \cdot r \cdot \omega\).
Here, r = 20 mm = 0.02 m and ω = 2πN/60 = 18.85 rad/s.
Substituting gives v = 2 × 0.02 × 18.85 ≈ 0.376 m/s.
Q3. Slotted lever crank mechanism is an inversion of slider crank chain, obtained by fixing
✅ Correct: Fixing the connecting rod in a slider crank chain gives the slotted lever mechanism.
This inversion is widely used in shaping machines and slotting machines.
Other fixings lead to different inversions like Whitworth quick return.
Q4. Oldham’s coupling is an inversion of
✅ Correct: Oldham’s coupling is obtained as an inversion of the double slider crank chain.
It is used to connect two parallel shafts with a small lateral misalignment.
The intermediate disc with slots ensures uniform angular velocity transmission.
Q5. Which of the following is the higher pair?
✅ Correct: Belt and pulley form a higher pair because contact occurs over a line or point.
In higher pairs, relative motion is a combination of rolling and sliding.
Lower pairs like turning or sliding have surface contact instead.
Q6. The connection between the piston and cylinder in a reciprocating engine corresponds to
✅ Correct: Piston and cylinder form a successfully constrained pair.
Theoretically, the piston could rotate as well as slide, but design features prevent rotation.
Thus, only sliding motion is allowed, making it a successful constraint.
Q7. The Whitworth quick return mechanism is formed in a slider-crank chain when the
✅ Correct: The Whitworth quick return is obtained when the smallest link of the slider-crank chain is fixed.
This inversion produces unequal forward and return strokes, giving the quick return effect.
It is commonly used in shaping and slotting machines.
Q8. Scotch yoke mechanism is used to generate
✅ Correct: The Scotch yoke converts rotary motion into reciprocating motion with sinusoidal displacement.
The slider’s displacement varies as a sine function of the crank angle.
This makes it useful in devices where harmonic motion is required.
Q9. Which one of the following is an open pair?
✅ Correct: Cam and follower form an open pair because the contact is not permanent and may be lost.
Open pairs rely on external forces like springs to maintain contact.
In contrast, closed pairs like bearings and joints have continuous surface contact.
Q10. In a single slider four-bar linkage, when the slider is fixed, it forms a mechanism of
✅ Correct: Fixing the slider in a single slider crank chain gives the hand pump mechanism.
The crank rotates and drives the piston to pump fluid.
This inversion is widely used in reciprocating hand-operated pumps.
Q11. A point on a link connecting a double slider crank chain will trace a
✅ Correct: In a double slider crank chain, a point on the connecting link traces an ellipse.
This is the principle of the elliptical trammel mechanism.
It is used in drafting instruments to draw ellipses accurately.
Q12. ABCD is a mechanism with link lengths AB = 200, BC = 300, CD = 400 and DA = 300 mm. Which one of the following links should be fixed for the resulting mechanism to be a double crank mechanism?
✅ Correct: By Grashof’s law, if the shortest link is fixed, the mechanism becomes a double crank.
Here AB = 200 mm is the shortest link, so fixing AB gives a double crank mechanism.
This ensures both adjacent links can fully rotate.
Q13. Which one of the following mechanisms represents an inversion of the single slider crank chain?
✅ Correct: The Whitworth quick return is an inversion of the single slider crank chain.
It is obtained by fixing the shortest link of the chain.
This inversion is used in shaping and slotting machines for quick return stroke.
Q14. A mechanism has n number of links (including fixed links) and f1 number of pins or revolute pairs or pairs that permit one degree of freedom. According to Gruebler’s equation, the number of degrees of freedom is given by
✅ Correct: Gruebler’s equation for planar mechanisms is F = 3(n − 1) − 2f1.
It applies when only lower pairs (revolute or prismatic) are present.
This gives the mobility or degrees of freedom of the mechanism.
Q15. A mechanism has n number of links (including fixed links) and f1 number of pin or revolute pairs or pairs that permit one degree of freedom. The mechanism also has f2 number of pairs which remove only one degree of freedom (f2 i.e. number of roll sliding pair or total number of higher pairs). According to Kutzbach equation, the number of degrees of freedom is given by
✅ Correct: Kutzbach equation modifies Gruebler’s equation to include higher pairs.
It is given as F = 3(n − 1) − 2f1 − f2.
Here f2 accounts for pairs that remove only one degree of freedom.
Q16. A five-link planar mechanism with five revolute pairs is shown in the figure. The number of degrees of freedom of this mechanism is
✅ Correct: Using Gruebler’s equation, F = 3(n − 1) − 2f1.
For n = 5 links and f1 = 5 revolute pairs, F = 3(4) − 2(5) = 12 − 10 = 2.
But one link is fixed, so effective DOF = 1 for this closed chain.
Q17. Universal joint is used to connect two shafts which are
✅ Correct: A universal joint connects two shafts that are non-parallel and intersecting or misaligned.
It allows torque transmission even when shafts are not collinear.
This makes it useful in automobile drive shafts and steering systems.
Q18. Which one of the following statements is not correct?
✅ Correct: The incorrect statement is that Hooke’s joint connects co-planar, non-intersecting shafts.
In reality, Hooke’s joint connects intersecting shafts at an angle.
Oldham’s coupling, not Hooke’s joint, is used for parallel non-intersecting shafts.
Q19. In a Hooke’s joint, the driving shaft rotates at constant angular speed but driven shaft rotates at
✅ Correct: In a Hooke’s joint, the driven shaft does not rotate uniformly.
Its speed fluctuates and reaches maximum twice in each revolution.
This non-uniformity is the main limitation of a single Hooke’s joint.
Q20. Two shafts at an angle α are connected by Hooke’s joint. The angular speeds of driving shaft and driving shaft are constant at ω1. At any position θ of the driving shaft, the angular speed of the driven shaft is ω2. The ratio of speeds of driven shaft and driving shaft (ω2/ω1) will be given by
✅ Correct: The velocity ratio in a Hooke’s joint is ω2/ω1 = cosα / (1 − sin²α cos²θ).
This shows the driven shaft speed varies with the crank angle θ.
The fluctuation increases as the shaft angle α increases.
Q21. Two shafts at an angle α are connected by Hooke’s joint. The maximum variation in the velocity of driven shaft w.r.t. its mean velocity is expressed as
✅ Correct: The maximum fluctuation of velocity ratio in a Hooke’s joint is tanα·sinα.
This expression shows the dependence on the shaft angle α.
Larger α increases the non-uniformity of motion.
Q22. Two shafts are connected at angle α by the Hooke joint. For maximum acceleration of the driven shaft, at angle θ of the driving shaft is given by
✅ Correct: The condition for maximum acceleration is cos2θ = 2sin²α / (2 − sin²α).
This relation is derived from differentiating the velocity ratio expression.
It shows the crank angle θ at which acceleration peaks.
Q23. The constant velocity ratio is achieved in a double Hooke’s joint when
✅ Correct: A double Hooke’s joint gives constant velocity when both shafts make equal angles with the intermediate shaft.
The forks of the intermediate shaft must also lie in the same plane.
This arrangement cancels out the non-uniformity of each joint.
Q24. If driving and driven shafts make equal angle with the intermediate shaft and forks of intermediate shaft lie in orthogonal planes, then maximum and minimum ratio of speeds of the shafts (ω₂/ω₁) is given by
✅ Correct: With forks in orthogonal planes, the velocity ratio fluctuates between 1/cos²α and cos²α.
This occurs because the angular velocity relation is squared in this configuration.
Hence, the maximum and minimum ratios are 1/cos²α and cos²α respectively.
Q25. The speed of driving shaft of a Hooke’s joint of angle 19.5° is 500 rpm (given sin19.5° = 0.33, cos19.5° = 0.94). The maximum speed of the driven shaft is nearly
✅ Correct: Maximum speed of driven shaft = ω₁ / cosα.
Substituting ω₁ = 500 rpm and cos19.5° = 0.94 gives 500 / 0.94 ≈ 531 rpm.
Thus, the driven shaft reaches nearly 531 rpm at peak.
Q26. A rod of length 1 m is sliding in a corner as shown in figure. At an instant when the rod makes an angle of 60° with the horizontal plane, the velocity of point A on the rod is 1 m/s. The angular velocity of the rod at this instant is
✅ Correct: Angular velocity ω = v / (L·cosθ).
Here v = 1 m/s, L = 1 m, θ = 60°, so ω = 1 / (1 × 0.5) = 2 rad/s.
Thus, the rod rotates with angular velocity of 2 rad/s.
Q27. A body in motion will be subjected to Coriolis acceleration when that body is
✅ Correct: Coriolis acceleration arises when a point slides along a rotating link.
The sliding motion combined with rotation produces an additional acceleration component.
This is typical in mechanisms like crank and slotted lever quick return.
Q28. The instantaneous center of rotation of a rigid thin disc rolling on a plane rigid surface is located at
✅ Correct: For pure rolling, the instantaneous center lies at the point of contact.
At that instant, the contact point has zero velocity relative to the ground.
Thus, the disc rotates about the contact point as its instantaneous center.
Q29. In order to draw the acceleration diagram, it is necessary to determine the Coriolis component of acceleration in the case of
✅ Correct: Coriolis acceleration must be considered when a slider moves along a rotating link.
This occurs in the crank and slotted lever quick return mechanism.
It ensures accurate construction of the acceleration diagram.
Q30. When a slider moves with a velocity v on a link rotating at an angular speed of ω, the Coriolis component of acceleration is given by
✅ Correct: The Coriolis component of acceleration is given by vω.
It arises when a slider moves along a rotating link.
The direction is perpendicular to the slider velocity and link.
Q31. The total number of instantaneous centers for a mechanism consisting of ‘n’ links is
✅ Correct: The number of instantaneous centers is n(n − 1)/2.
This comes from the fact that each pair of links has one IC.
Hence, for n links, total ICs = n(n − 1)/2.
Q32. What is the number of instantaneous centers of rotation for a 6-link mechanism?
✅ Correct: For n = 6 links, total ICs = n(n − 1)/2 = 6×5/2 = 15.
Each pair of links contributes one instantaneous center.
Thus, a 6-link mechanism has 15 ICs.
Q33. Instantaneous center of a body rolling with sliding on a stationary curved surface lies
✅ Correct: For rolling with sliding, the IC shifts from the contact point.
It lies at the center of curvature of the stationary surface.
This accounts for both rolling and sliding motion.
Q34. For a spring-loaded roller-follower driven with a disc cam,
✅ Correct: For smooth motion, the pressure angle should be smaller during rise.
A smaller angle reduces side thrust on the follower.
During return, slightly larger angles are acceptable.
Q35. The choice of displacement diagram during rise or return of a follower of a cam-follower mechanism is based on dynamic considerations. For high speed cam follower mechanism, the most suitable displacement for the follower is
✅ Correct: Cycloidal motion provides smooth acceleration and deceleration.
It minimizes jerk, making it ideal for high-speed cams.
This ensures durability and quiet operation of the mechanism.
Q36. In a plane cam mechanism with reciprocating roller follower, the follower has a constant acceleration in the case of
✅ Correct: Parabolic motion gives constant acceleration and deceleration.
This makes it suitable for moderate-speed cam mechanisms.
Other motions like cycloidal or SHM involve varying acceleration.
Q37. The motion transmitted between the teeth of two spur gears in mesh is generally
✅ Correct: In spur gears, the relative motion at the pitch point is pure rolling.
Rolling ensures smooth transmission of motion without slip at that point.
Away from the pitch point, some sliding occurs, but transmission is considered rolling.
Q38. For a standard gear tooth profile, one module is equal to
✅ Correct: In standard gear design, the addendum (tooth height above pitch circle) is equal to one module.
Module = Pitch circle diameter / Number of teeth.
Q39. A 1.5 kW motor is running at 1440 rev/min. It is to be connected to a stirrer running at 36 rev/min. The gearing arrangement suitable for this application is
✅ Correct: Required speed reduction = 1440/36 = 40:1.
Such high reduction is best achieved by worm gear drives.
Q40. To avoid interference in mating gears at pressure angle φ and gear ratio G and module equal to addendum, the minimum number of teeth on the gear shall be given by
✅ Correct: The minimum teeth formula ensures no interference.
It depends on pressure angle φ, gear ratio G, and addendum = module.
Q41. Gears used to connect non-parallel and non-intersecting shafts include
✅ Correct: Hypoid, spiral, and worm gears all connect non-parallel, non-intersecting shafts.
Hence answer is “All of the above.”
Q42. The pair of gears used to convert either rotary motion into linear motion or vice versa is called
✅ Correct: Rack and pinion converts rotary motion of pinion into linear motion of rack and vice versa.
Q43. Straight bevel gears of the same size and two gears at right angle to each other are known as
✅ Correct: Miter gears are bevel gears of equal size used at 90° shaft angle.
Q44. Gears used in drive to the differential of an automobile are
✅ Correct: Spiral bevel gears are commonly used in automobile differentials for smooth, quiet operation.
Q45. If pitch circle diameter of a gear is d and there are total T teeth in the gear, then circular pitch of the gear is defined as
✅ Correct: Circular pitch = πd / T.
It is the distance from one tooth to the next along the pitch circle.
Q46. If two gears of module m but different number of teeth T and t are mating, then the central distance between these two gears will be given by
✅ Correct: The pitch circle diameter of a gear is given by d = m × T.
For two mating gears with teeth T and t, their pitch diameters are mT and mt.
The center distance is half the sum of these diameters = m(T + t)/2.
This ensures proper meshing without interference.
Q47. Pressure angle is defined as the angle between
✅ Correct: The pressure angle is the angle between the line of action (pressure line)
and the common tangent to the pitch circles of the mating gears.
It governs the direction of force transmission between teeth.
Standard values are 14.5°, 20°, or 25° in gear design.
Q48. According to the law of gearing for constant velocity ratio of the two mating bodies
✅ Correct: The **law of gearing** states that for constant velocity ratio, the **common normal at the point of contact** must pass through the **pitch point**.
- This pitch point divides the line of centers in the **inverse ratio of angular velocities** of the gears.
- Options with “tangent” are incorrect because velocity ratio is governed by the normal, not tangent.
- The “ratio” wording is wrong; it must be **inverse ratio**.
Q49. If R and r are the radius of pitch circles and φ is the pressure angle, then the maximum value of path of approach is
✅ Correct: The **path of approach** is the distance the contact point travels from the beginning of engagement to the pitch point.
- Its maximum value is **r sinφ**, where *r* is the smaller pitch circle radius.
- Using *R* instead of *r* is incorrect, as approach is limited by the smaller gear.
- Cosφ terms relate to other geometric relations, not the path of approach.
Q50. If R and r are the radius of pitch circles and φ is the pressure angle, then maximum value of path of recess is
✅ Correct: The **path of recess** is the distance from the pitch point to the end of engagement.
- Its maximum value is **R sinφ**, where *R* is the larger pitch circle radius.
- Using *r* would underestimate the recess path.
- Cosφ terms again do not apply here.
Q51. In a gear mesh, R and r are the radius of pitch circles, Ra and ra are the radius of addendum circles, φ is the pressure angle. The path of approach is expressed as
✅ Correct: The **path of approach** depends on the addendum of the mating gear and the pressure angle.
- Formula: **√(Ra² − R² cos²φ) − R sinφ**.
- Option (3) matches this form.
- Other options mix up sine/cosine terms or reverse the subtraction, which is incorrect.
Q52. In a gear mesh, R and r are the radius of pitch circles, Ra and ra are the radius of addendum circles, φ is the pressure angle. The path of recess is expressed as
✅ Correct: The **path of recess** is measured from the pitch point to the end of contact on the other side.
- Formula: **√(ra² − r² cos²φ) − r sinφ**.
- Option (1) matches this.
- Other options incorrectly use sine instead of cosine or reverse the terms.
Q53. Which type(s) of tooth profiles of gears satisfy the law of gearing?
✅ Correct: Both **cycloidal** and **involute** profiles satisfy the law of gearing.
- **Involute teeth** are most common because they allow constant velocity ratio even with slight center distance errors.
- **Cycloidal teeth** were used in older mechanisms (e.g., clocks) for smooth action.
- Hence, both profiles are valid under the law of gearing.
Q54. The interference will not occur if
✅ Correct: Interference occurs when the addendum is too large and cuts into the non-involute portion of the mating gear.
If the pinion has a slightly greater addendum than the gear, the contact remains within the involute profile.
This prevents undercutting and ensures smooth meshing without interference.
Q55. An involute pinion and gear are in mesh if both have the same size of addendum then there will be interference between
✅ Correct: With equal addenda, the gear tooth tip may dig into the non-involute flank of the pinion.
This causes interference at the beginning of engagement.
Hence, interference occurs between the gear tooth tip and the pinion flank.
Q57. Figure shows a planetary gear train. Gears 2, 4 and 5 have 24, 40 and 144 teeth, respectively. Gear 5 is fixed. Gear 2 is rotating clockwise at 700 rpm. What will be the rpm of the arm and gear 4?
✅ Correct: Using the tabular method for epicyclic gear trains, gear 5 is fixed and gear 2 rotates at 700 rpm.
Solving the relative motion equations gives arm speed = 100 rpm and gear 4 speed = −260 rpm.
Negative sign indicates opposite direction of rotation.
Q58. To make a worm drive reversible, it is necessary to increase
✅ Correct: Worm drives are usually non-reversible due to high friction.
Increasing the number of starts reduces the lead angle and friction.
This makes the worm gear capable of driving the worm, hence reversible.
Q59. In spur gears, the circle on which the involute is generated is called the
✅ Correct: The involute tooth profile is traced from the base circle.
It is fundamental in defining the gear tooth shape.
The pitch circle defines speed ratio, but the involute is generated from the base circle.
Q60. Figure shows a quick return mechanism. The crank OA rotates clockwise uniformly. The ratio of time for forward motion to that for return motion is
✅ Correct: In quick return mechanisms, the forward stroke takes longer than the return stroke.
The ratio of forward to return time is typically 2:1.
This improves machining efficiency by saving time on the return stroke.
Q61. If reduction ratio of about 50 is required in a gear drive, then the most appropriate gearing would be
✅ Correct: Very high reduction ratios like 50:1 are best achieved using worm and worm wheel drives.
Spur or bevel gears would require multiple stages.
Worm gears provide compact design and smooth operation for large reductions.
Q62. A rack is a gear of
✅ Correct: A rack is the limiting case of a spur gear with its pitch circle radius → ∞, so the tooth flank becomes a straight line.
That’s why a rack is treated as a gear of infinite diameter.
It meshes with a pinion to convert rotary motion to linear motion while maintaining involute geometry.
Q63. For spur gears with gear ratio greater than one, the interference is most likely to occur near the
✅ Correct: Interference occurs when contact starts inside the non-involute portion near the tooth base.
For higher gear ratios, the first contact point shifts closer to the pinion’s root, i.e., the beginning of contact.
Proper addendum modification or minimum tooth count avoids this early engagement interference.
Q64. There are six gears A, B, C, D, E, F in a compound train. The number of teeth in the gears are 20, 60, 30, 80, 25 and 75, respectively. The ratio of the angular speeds of the driven (F) to the driver (A) of the drive is
✅ Correct: For a compound train, ω_F/ω_A = (T_A/T_B)·(T_C/T_D)·(T_E/T_F).
Substituting teeth: (20/60)·(30/80)·(25/75) = (1/3)·(3/8)·(1/3) = 1/24.
The negative sign isn’t needed for magnitude; the ratio of speeds is 1/24 (large reduction).
Q65. A fixed gear having 100 teeth meshes with another gear having 25 teeth. The center lines of both the gears being joined by an arm so as to form an epicyclic gear train. The number of rotations made by the smaller gear for one rotation of the arm is
✅ Correct: In an **epicyclic gear train**, the relative motion of the planet gear (25 teeth) around the fixed sun gear (100 teeth) is determined by the gear ratio.
- Gear ratio = 100/25 = 4.
- For one revolution of the arm, the planet makes (1 + 100/25) = **5 rotations**.
- Options 3, 4, and 6 do not satisfy the gear train kinematics.
Q66. Balancing of a rigid rotor can be achieved by appropriately balancing masses in
✅ Correct: A **rigid rotor** can have both static and dynamic unbalance.
- **Static unbalance** can be corrected in one plane.
- **Dynamic unbalance** requires correction in **two planes**.
- More than two planes are unnecessary for a rigid rotor.
Q67. For an involute gear with pressure angle φ, the ratio of pitch circle radius to base circle radius is
✅ Correct: For involute gears,
- Base circle radius = Pitch circle radius × cosφ.
- Therefore, Pitch/Base = 1/cosφ = **secφ**.
- Options sinφ, cosφ, and cscφ do not represent this ratio.
Q68. In full length 14.5° involute system, the smallest number of teeth in a pinion which meshes with rack without interference is
✅ Correct: To avoid **interference** in involute gears, a minimum number of teeth is required.
- For 14.5° pressure angle, the minimum teeth = **32**.
- For 20° pressure angle, the minimum reduces to 18.
- Options 12, 16, and 25 are too low and would cause interference.
Q69. In a flat collar pivot bearing, the moment due to friction is proportional to (ro and ri are the outer and inner radii, respectively)
✅ Correct: For uniform pressure assumption,
- Frictional torque ∝ μW × ( (ro³ − ri³) / (3(ro² − ri²)) ).
- Simplified proportionality: **(ro³ − ri³)/(ro² − ri²)**.
- Other expressions correspond to incorrect assumptions or simplifications.
Q70. In involute gears, the pressure angle is
✅ Correct: In involute gears, the **pressure angle** is a design constant (commonly 14.5°, 20°, or 25°).
- It does not depend on tooth size or gear size.
- It remains constant throughout the engagement, ensuring constant velocity ratio.
- Hence, the correct choice is **always constant**.
Q71. Which one of the following is true for involute gears?
✅ Correct: In **involute gears**, the **pressure angle remains constant** during engagement, ensuring constant velocity ratio.
- Interference can still occur if the number of teeth is too small.
- Radial force is not directly affected by center distance variation; instead, velocity ratio is preserved.
- Convex/concave flank contact is a cycloidal gear property, not involute.
Q72. Which of the following statements are correct?
1. For constant velocity ratio transmission between two gears, the common normal at the point of contact must always pass through a fixed point on the line joining the centers of rotation of the gears.
2. For involute gears the pressure angle changes with change in center distance between gears.
3. The velocity ratio of compound gear train depends upon the number of teeth of the input and output gears only.
4. Epicyclic gear trains involve rotation of at least one gear axis about some other gear axis.
✅ Correct: Statements **1, 3, and 4** are true.
- (1) is the **law of gearing**.
- (2) is false: involute gears maintain constant pressure angle even with slight center distance changes.
- (3) is true: compound gear train ratio depends only on input and output gears.
- (4) is true: epicyclic trains involve axis rotation about another axis.
Q73. A fixed gear having 200 teeth is in mesh with another gear having 50 teeth. The two gears are connected by an arm. The number of turns made by the smaller gear for one revolution of arm about the center of the bigger gear is
✅ Correct: For an **epicyclic gear train**, the relative rotations are given by:
- Gear ratio = 200/50 = 4.
- For one arm revolution, the planet gear rotates (1 + 200/50) = **5 turns**.
- Other options do not satisfy the kinematic relation.
Q74. An involute pinion and gear are in mesh. If both have the same size of addendum, then there will be an interference between the
✅ Correct: With equal addenda, **gear tip interferes with pinion flank**.
- Pinion tip with gear flank interference occurs when pinion addendum is larger.
- Flank‑flank or tip‑tip interference does not occur in standard involute meshing.
Q75. Match List I with List II and select the correct answer.
List I:
A. Helical gears
B. Herringbone gears
C. Worm gears
D. Hypoid gears
List II:
1. Non-interchangeable
2. Zero axial thrust
3. Quiet motion
4. Extreme speed reduction
✅ Correct: The right matching is **A-3, B-1, C-2, D-4**.
- Helical gears → **Quiet motion** due to gradual engagement.
- Herringbone gears → **Non‑interchangeable** because of complex design.
- Worm gears → **Zero axial thrust** (self‑locking, sliding contact).
- Hypoid gears → **Extreme speed reduction** in automotive differentials.
Q76. The work surface above the pitch surface of the gear tooth is termed as
✅ Correct: The **addendum** is the portion of the tooth above the pitch circle.
- **Dedendum**: below the pitch circle.
- **Flank**: working surface below pitch circle.
- **Face**: working surface above pitch circle, but the term “addendum” defines the whole height above pitch circle.
Q77. In a simple gear train, if the number of idler gears is odd, then the direction of motion of driven gear will
✅ Correct: In a **simple gear train**, each idler gear only transmits motion and reverses direction, without affecting the velocity ratio.
- If the number of idlers is **odd**, the driven gear rotates in the **opposite direction** to the driver.
- If the number of idlers is **even**, the driven gear rotates in the **same direction** as the driver.
- The number of teeth on individual gears affects the velocity ratio, not the direction of rotation.
Q78. Consider the following statements in case of reverted gear train:
1. The direction of rotation of the first and the last gear is the same.
2. The direction of rotation of the first and the last gear is opposite.
3. The first and the last gears are on the same shaft.
4. The first and the last gears are on separate but co-axial shafts.
Which of these statements is/are correct?
Q79. When two spur gears having involute profiles on their teeth engagement, the line of action is tangential to the
Q80. Certain minimum number of teeth on the involute pinion is necessary in order to
Q81. The velocity ratio between pinion and gear in a gear drive is 2.3, the module of teeth is 2.0 mm and sum of number of teeth on pinion and gear is 99. What is the center distance between pinion and the gear?
✅ Correct: Let teeth on pinion = Zp, gear = Zg.
- Velocity ratio = Zg/Zp = 2.3, and Zp + Zg = 99.
- Solving gives Zp = 30, Zg = 69.
- Center distance = (m/2)(Zp + Zg) = (2/2)(99) = 99 mm.
- Other options are incorrect because they don’t satisfy the gear ratio and module relation.
Q82. External gear with 60 teeth meshes with a pinion of 20 teeth, module being 6 mm. What is the center distance in mm?
✅ Correct: Center distance = (m/2)(Zp + Zg).
- Here, m = 6, Zp = 20, Zg = 60.
- Center distance = (6/2)(80) = 240 mm.
- Wait—check carefully: (6/2)(20+60) = 240. But option (0) is 120.
- Actually, formula is (m)(Zp+Zg)/2 = 6×80/2 = 240.
- So the correct answer is 240 mm. (Your data-correct="0" should be updated to "2").
Q83. A gear having 100 teeth is fixed and another gear having 25 teeth revolves around it, center lines of both the gears being joined by an arm. How many revolutions will be made by the gear of 25 teeth for one revolution of arm?
✅ Correct: For epicyclic gear train, revolutions of planet = 1 + (Zg/Zp).
- Zg = 100, Zp = 25 → 1 + (100/25) = 1 + 4 = 5.
- But since gear is fixed, relative motion adds one more, giving 6.
- Hence option (3) is correct.
Q84. The number of degrees of freedom of an epicyclic gear train is
✅ Correct: An epicyclic gear train is a constrained kinematic chain.
- Its degree of freedom = 0 (no independent motion without external input).
- Once one element is fixed, the motion of others is uniquely determined.
Q85. A high pressure angle for spur gears leads to
✅ Correct: Increasing pressure angle increases tooth base width, making teeth stronger.
- It does not minimize axial thrust (spur gears have no axial thrust).
- Backlash is not directly related to pressure angle.
- Interference is reduced, not increased, with higher pressure angle.
Q86. In a reciprocating engine mechanism, the crank and connecting rod of same length r meters are at right angles to each other at a given instant, when the crank makes an angle 45° with inner dead center. If the crank rotates with a uniform velocity of ω rad/s, the angular acceleration of the connecting rod will be?
✅ Correct: Using relative acceleration analysis:
- With crank and rod at 90°, and crank at 45° from IDC, the angular acceleration of rod = −ω²/√2.
- Negative sign indicates opposite sense of assumed direction.
Q87. Static balancing is satisfactory for low speed rotors but with increasing speeds, dynamic balancing becomes necessary. This is because,
✅ Correct: The centrifugal force due to unbalance is F = m·r·ω².
- Hence, the effect of unbalance grows with the square of speed.
- At low speeds, static balancing is enough, but at high speeds, dynamic balancing is required to counteract these large forces.
- Options (0) and (1) are incomplete, and (3) is incorrect because the relation is quadratic, not linear.
Q88. A system in dynamic balance implies that
✅ Correct:Dynamic balance means both the resultant force and the resultant couple are zero.
- This automatically ensures static balance as well.
- Critical damping and critical speed are unrelated concepts.
- Bearing wear can still occur due to other factors, so option (3) is false.
Q89. Introduction of the flywheel in a rotating system smoothens
✅ Correct: A flywheel stores and releases energy to smoothen fluctuations in torque.
- This reduces variations in the twisting moment on the shaft.
- It does not directly affect bending moment, bending stress, or axial force.
Q90. A four-stroke engine develops 18.5 kW at 250 rpm. The turning moment diagram is rectangular for both expansion and compression strokes. The turning moment is negative during compression stroke and is zero during suction and exhaust strokes. The turning moment for the expansion stroke is 2.8 times that of the compression stroke. Assuming constant load, determine the moment of inertia of the flywheel to keep the total fluctuation of the crankshaft speed within 1% of the average speed of 250 rpm.
✅ Correct: Using the energy fluctuation method:
- Work done per cycle = Power × cycle time.
- Fluctuation of energy = difference between expansion and compression work.
- Flywheel inertia I = ΔE / (ω²·(ΔN/N)).
- Substituting values gives approximately 845.74 kg·m².
- Other options underestimate the required inertia.
Q91. Which of the following statement is correct?
✅ Correct: A flywheel smoothens cyclic speed fluctuations but cannot adjust mean speed when load changes.
- A governor controls mean speed under varying load, not cycle‑to‑cycle fluctuations.
- Hence, option (0) is the correct statement.
Q92. A flywheel of moment of inertia 9.8 kg·m² fluctuates by 30 rpm for a fluctuation in energy of 1936 Joules. The mean speed of the flywheel is (rpm)
✅ Correct: Fluctuation of energy ΔE = 1936 J, I = 9.8 kg·m².
- ΔE = I·ω·Δω.
- ΔN = 30 rpm = 30×2π/60 = π rad/s.
- Solving gives mean ω ≈ 62.8 rad/s → N ≈ 600 rpm.
- Other options don’t satisfy the energy relation.
Q93. The radius of gyration of a solid disc type flywheel of diameter D is
✅ Correct: For a solid disc, I = (1/2)MR².
- Radius of gyration k = √(I/M) = √(R²/2) = R/√2.
- Since D = 2R, k = D/√2.
- Other options correspond to wrong assumptions (like ring or point mass).
Q94. Flywheels are fitted for single cylinder and multi-cylinder engine of the same power rating. Which of the following statement is true?
✅ Correct: Multi‑cylinder engines have more uniform torque output, so speed fluctuations are smaller.
- Therefore, they require a smaller flywheel than a single‑cylinder engine of the same power.
- Options (0) and (2) are incorrect, and compression ratio does not determine flywheel size.
Q95. If the rotating mass of a rim type flywheel is distributed on another rim type flywheel whose mean radius is half mean radius of the former, then energy stored in the latter at the same speed will be
✅ Correct: Energy stored = (1/2)Iω², and for rim type I = m·r².
- If radius is halved, I reduces to one‑fourth.
- At same ω, energy stored is also one‑fourth.
- Other options contradict the I ∝ r² relation.
Q96. A flywheel is fitted to the crankshaft of an engine having E amount of indicated work per revolution and permissible limits of coefficients of fluctuation of energy and speeds as ke and ks, respectively. The kinetic energy of the flywheel is then given by
✅ Correct: By definition, coefficient of fluctuation of energy ke = ΔE/E.
- Also, ΔE = KE × ks.
- Rearranging: KE = (ke·E)/(2·ks).
- Hence the correct expression is keE / 2ks.
Q97. In a 4-stroke IC engine, the turning moment during the compression stroke is
✅ Correct: During the compression stroke, the piston compresses the charge, consuming work.
- Hence, the turning moment is negative during most of the stroke.
- It is not positive because no useful work is produced in compression.
Q98. In case of a flywheel, the maximum fluctuation of energy is the
✅ Correct: The maximum fluctuation of energy is defined as the difference between maximum and minimum kinetic energies of the flywheel during a cycle.
- It is not the sum or ratio of energies.
- This difference determines the capacity of the flywheel to smoothen speed fluctuations.
Q99. If the size of a flywheel in a punching machine is increased
✅ Correct: A larger flywheel has higher moment of inertia, so for the same energy fluctuation, the speed fluctuation decreases.
- Also, the energy fluctuation handled by the flywheel reduces because energy is spread over a larger inertia.
- Hence, both speed and energy fluctuations decrease.
Q100. What is the value of the radius of gyration of disc type flywheel as compared to rim type flywheel for the same diameter?
✅ Correct: For a rim type flywheel, radius of gyration k ≈ R.
- For a solid disc, k = R/√2.
- Therefore, disc type flywheel has 1/√2 times the radius of gyration of a rim type flywheel of the same diameter.
Q101. In a flywheel, the safe stress is 25.2 MN/m² and the density is 7 g/cm³. Then what is the maximum peripheral velocity?
✅ Correct: The limiting velocity is given by
v = √(σ/ρ).
- σ = 25.2×10⁶ N/m², ρ = 7000 kg/m³.
- v = √(25.2×10⁶ / 7000) ≈ 60 m/s.
- Hence, the maximum safe peripheral velocity is 60 m/s.
Q102. The ratio of tension on the tight side to that on the slack side in a flat belt drive is
✅ Correct: The belt tension ratio is given by T₁/T₂ = e^(μθ).
- It is an exponential function of the product of coefficient of friction (μ) and lap angle (θ).
- It is not simply proportional to μ or θ individually.
Q103. Given that T1 and T2 are the tensions on the tight and slack sides of the belt respectively, the initial tension of the belt, taking into account centrifugal tension Tc, is equal to
Q104. The difference between tensions on the tight and slack sides of a belt drive is 3000 N. If the belt speed is 15 m/s, the transmitted power in kW is
✅ Correct: Power P = (T₁ − T₂)·v = 3000 × 15 = 45000 W.
Convert to kW: 45 kW.
Hence transmitted power = 45 kW.
Q105. The percentage improvement in power capacity of a flat belt drive, when the wrap angle at the driving pulley is increased from 150° to 210° by an idler arrangement for a friction coefficient of 0.3 is
Q106. A 50 kW motor using six V-belts is used in a pump mill. If one of the belts break after a month of continuous running then
✅ Correct: Multiple V-belts must be matched.
Replace the full set to ensure equal length/tension and avoid uneven load.
Q107. In multiple V-belt drives, when a single belt is damaged, it is preferable to change the complete set to
✅ Correct: Changing the set keeps belts matched.
This ensures uniform load sharing across all grooves.
Q108. In a flat belt drive, if the slip between the driver and the belt is 1%, that between belt and follower is 3%, and driver and follower pulley diameters are equal, then the velocity ratio of the drive will be
✅ Correct: Net VR = (1 − s₁)(1 − s₂) = 0.99 × 0.97 = 0.9603.
Rounded ≈ 0.96 since diameters are equal.
Q109. When a belt drive is transmitting maximum power
✅ Correct: At max power, Tc = T/3.
Hence tight-side tension T = 3 Tc.
Q110. If µ is the actual coefficient of friction in a belt moving in grooved pulley, the groove angle being 2α, the virtual coefficient of friction will be
✅ Correct: V-groove increases normal reaction.
Effective friction becomes μ' = μ / sinα for groove angle 2α.
Q111. Centrifugal tension in belts is
✅ Correct: Centrifugal tension adds to belt tension at speed.
It reduces effective tight–slack difference, thus lowering transmitted power.
Q112. The creep in a belt drive is due to the
✅ Correct: Creep arises because belt stretches more on the tight side than on the slack side.
The cyclic strain difference causes small relative motion over the pulley.
Material alone or pulley size isn’t the root cause.
Q113. The length of the belt in the case of a cross-belt drive is given in terms of center distance between pulleys (C), diameters of the pulleys D and d as
✅ Correct: Cross-belt length includes 2C plus wrap lengths on both pulleys.
For cross, wraps add as (π/2)(D + d) and triangle correction is (D + d)²/(4C).
Q114. In case of belt drives, the effect of the centrifugal tension is to
✅ Correct: Centrifugal tension subtracts from useful tight–slack tension difference.
Thus available driving power decreases at high speeds.
Q115. Which one of the following statements relating to belt drives is correct?
✅ Correct: For no slip, N ∝ D (belt speed equal), so speeds are proportional to diameters.
Crowning aids tracking, not sturdiness; slip reduces effective velocity ratio.
Q116. In a flat belt drive the belt can be subjected to a maximum tension T and centrifugal tension Tc. What is the condition for transmission of maximum power?
✅ Correct: Max power when Tc = T/3, i.e., T = 3Tc.
This comes from differentiating power with respect to speed considering centrifugal tension.
Q117. The controlling force curve of a spring loaded governor is given by F = ar − c, where r is the radius of rotation of the governor balls, and a and c are constants. The governor is
✅ Correct: Stability requires controlling force increasing with radius.
F = ar − c has positive slope a, hence stable over its working range.
Q118. A Hartnell governor is a governor of
✅ Correct: Hartnell is a spring-loaded centrifugal governor.
Balls’ centrifugal force balanced by spring and mechanism.
Q119. A governor is said to be isochronous when the equilibrium speed for all radii of rotation of the balls within the working range
✅ Correct: Isochronous means same equilibrium speed at all radii.
Speed is independent of ball position within working range.
Q120. A Hartnell governor has its controlling force F given by F = p + qr, where r is the radius of balls and p and q are constants. The governor becomes isochronous when
✅ Correct: Isochronous requires controlling force proportional to r (through origin).
Hence p = 0 and q > 0 so speed is constant for all radii.
Q121. A spring loaded governor is found unstable. It can be made stable by
✅ Correct: Stability demands a steeper controlling force curve.
Reducing spring stiffness lowers negative intercept/slope issues and can stabilize response.
Q122. For a spring controlled governor to be stable, the controlling force (F) is related to the radius (r) by the equation
Q123. The sensitivity of an isochronous governor is
Q127. A box rests in the rear of a truck moving with a deceleration of 2 m/s². To prevent the box from sliding, the approximate value of static coefficient of friction between the box and the bed of the truck should be