MCQOPTIONS
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MCQOPTIONS
This section includes 52 Mcqs, each offering curated multiple-choice questions to sharpen your Machine Design knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Find the maximum efficiency if the lead angle is given to be 10° and the coefficient of friction is 0.07. |
| A. | 79.82% |
| B. | 72.23% |
| C. | 76.29% |
| D. | 70.72% |
| Answer» E. | |
| 2. |
Calculate the lead angle of the worm gear for maximum efficiency if θ = 90° and the coefficient of friction is 0.05. |
| A. | 48.21° |
| B. | 42.23° |
| C. | 43.57° |
| D. | 46.43° |
| Answer» D. 46.43° | |
| 3. |
For a two start worm gear having a pitch of 20 mm and a lead angle 12°, find the centre distance if the larger gear has 25 teeth. |
| A. | 148.22 mm |
| B. | 124.93 mm |
| C. | 121.19 mm |
| D. | 109.53 mm |
| Answer» E. | |
| 4. |
Find the speed of the gear if the worm is a three start worm rotating at 500 rpm. The gear has 20 teeth. |
| A. | 125 rpm |
| B. | 100 rpm |
| C. | 75 rpm |
| D. | 50 rpm |
| Answer» D. 50 rpm | |
| 5. |
Find the helix angle of the worm if the pitch of the worm gear is 12 mm and the pitch diameter is 50 mm. |
| A. | 8.687° |
| B. | 11.231° |
| C. | 9.212° |
| D. | 10.319° |
| Answer» B. 11.231° | |
| 6. |
What is the formula to calculate maximum efficiency of a worm gear? |
| A. | (1+sinø)/(1-sinø) |
| B. | (1-sinø)/(1+sinø) |
| C. | (tan(λ1-ø))/tan λ1 |
| D. | (tan(λ1+ø))/tan λ1 |
| Answer» C. (tan(λ1-ø))/tan λ1 | |
| 7. |
When large gear reductions are needed _________ gears are used. |
| A. | helical |
| B. | spur |
| C. | worm |
| D. | bevel |
| Answer» D. bevel | |
| 8. |
If tangential force on worm is 1500N, then axial force on worm wheel will be? |
| A. | 1500N |
| B. | 3000N |
| C. | 1500√2 N |
| D. | 750N |
| Answer» B. 3000N | |
| 9. |
A pair of worm gear is written as 2/40/12/6. Calculate the root diameter of the worm wheel. |
| A. | 186.22mm |
| B. | 250.4mm |
| C. | 225.6mm |
| D. | 250.44mmView Answer |
| Answer» D. 250.44mmView Answer | |
| 10. |
A pair of worm gear is written as 2/40/12/6. Calculate the throat diameter of the worm wheel. |
| A. | 220.5mm |
| B. | 246.4mm |
| C. | 190.44mm |
| D. | 251.7mm |
| Answer» E. | |
| 11. |
A pair of worm gear is written as 2/40/12/6. Calculate the pitch circle diameter of worm wheel. |
| A. | 72mm |
| B. | 240mm |
| C. | 260mm |
| D. | 320mm |
| Answer» C. 260mm | |
| 12. |
A pair of worm gear is written as 2/40/12/6. Calculate the speed reduction. |
| A. | 2 |
| B. | 20 |
| C. | 15 |
| D. | 6 |
| Answer» C. 15 | |
| 13. |
A pair of worm gear is written as 2/40/12/6. Calculate the centre distance. |
| A. | 40mm |
| B. | 156mm |
| C. | 200mm |
| D. | 80mm |
| Answer» C. 200mm | |
| 14. |
If worm helix angle is 30⁰, then worm should have at least ___ threads. |
| A. | 5 |
| B. | 6 |
| C. | 7 |
| D. | 8 |
| Answer» B. 6 | |
| 15. |
The axial thrust on the worm (WA) is given by |
| A. | is given bya) WA = WT . tan φ |
| B. | WA = WT / tan φ |
| C. | WA = WT . tan λ |
| D. | WA = WT / tan λ |
| Answer» E. | |
| 16. |
The number of starts on the worm for a velocity ratio of 40 should be |
| A. | single |
| B. | double |
| C. | triple |
| D. | quadruple |
| Answer» B. double | |
| 17. |
The normal lead, in a worm having multiple start threads, is given by |
| A. | lN = l / cos λ |
| B. | lN = l . cos λ |
| C. | lN = l |
| D. | lN = l tan |
| Answer» C. lN = l | |
| 18. |
A_PAIR_OF_WORM_GEAR_IS_WRITTEN_AS_2/40/12/6._CALCULATE_THE_SPEED_REDUCTION.?$ |
| A. | 2 |
| B. | 20 |
| C. | 15 |
| D. | 6 |
| Answer» C. 15 | |
| 19. |
WHAT_IS_THE_FORMULA_TO_CALCULATE_MAXIMUM_EFFICIENCY_OF_A_WORM_GEAR??$ |
| A. | (1+sin‚àö‚àè)/(1-sin‚àö‚àè) |
| B. | (1-sin‚àö‚àè)/(1+sin‚àö‚àè) |
| C. | (tan(λ<sub>1</sub>-ø))/tan λ<sub>1</sub> |
| D. | (tan(λ<sub>1</sub>+ø))/tan λ<sub>1</sub> |
| Answer» C. (tan(‚âà√≠¬¨‚Ñ¢<sub>1</sub>-‚Äö√†√∂‚Äö√†√®))/tan ‚âà√≠¬¨‚Ñ¢<sub>1</sub> | |
| 20. |
THE_NUMBER_OF_STARTS_ON_THE_WORM_FOR_A_VELOCITY_RATIO_OF_40_SHOULD_BE?$ |
| A. | single |
| B. | double |
| C. | triple |
| D. | quadruple |
| Answer» B. double | |
| 21. |
A pair of worm gear is written as 2/40/12/6. Calculate the throat diameter of the worm wheel.$ |
| A. | 220.5mm |
| B. | 246.4mm |
| C. | 190.44mm |
| D. | 251.7mm |
| Answer» E. | |
| 22. |
Find the speed of the gear if the worm is a three start worm rotating at 500 rpm. The gear has 20 teeth.$ |
| A. | 125 rpm |
| B. | 100 rpm |
| C. | 75 rpm |
| D. | 50 rpm |
| Answer» D. 50 rpm | |
| 23. |
A pair of worm gear is written as 2/40/12/6. Calculate the pitch circle diameter of worm wheel.$ |
| A. | 72mm |
| B. | 240mm |
| C. | 260mm |
| D. | 320mm |
| Answer» C. 260mm | |
| 24. |
Find_the_helix_angle_of_the_worm_if_the_pitch_of_the_worm_gear_is_12_mm_and_the_pitch_diameter_is_50_mm.$ |
| A. | 8.687° |
| B. | 11.231° |
| C. | 9.212° |
| D. | 10.319° |
| Answer» B. 11.231¬¨¬®‚Äö√†√ª | |
| 25. |
The_axial_thrust_on_the_worm_(WA)_is_given_by$ |
| A. | W<sub>A</sub> = W<sub>T</sub> . tan φ |
| B. | W<sub>A</sub> = W<sub>T</sub> / tan φ |
| C. | W<sub>A</sub> = W<sub>T</sub> . tan λ |
| D. | W<sub>A</sub> = W<sub>T</sub> / tan λ |
| Answer» E. | |
| 26. |
Calculate the maximum efficiency of the worm gears which have a friction angle of 0.06. |
| A. | 88.71% |
| B. | 83.23% |
| C. | 89.91% |
| D. | 86.49% |
| Answer» B. 83.23% | |
| 27. |
Find the maximum efficiency if the lead angle is given to be 10° and the coefficient of friction is 0.07.$ |
| A. | 79.82% |
| B. | 72.23% |
| C. | 76.29% |
| D. | 70.72% |
| Answer» E. | |
| 28. |
22mm |
| A. | 250.4mm |
| B. | 225.6mm |
| C. | 250.44mm |
| Answer» B. 225.6mm | |
| 29. |
Calculate the lead angle of the worm gear for maximum efficiency if θ = 90° and the coefficient of friction is 0.05.$ |
| A. | 48.21° |
| B. | 42.23° |
| C. | 43.57° |
| D. | 46.43° |
| Answer» D. 46.43¬¨¬®‚Äö√†√ª | |
| 30. |
For a two start worm gear having a pitch of 20 mm and a lead angle 12°, find the centre distance if the larger gear has 25 teeth.$ |
| A. | 148.22 mm |
| B. | 124.93 mm |
| C. | 121.19 mm |
| D. | 109.53 mm |
| Answer» E. | |
| 31. |
A pair of worm gear is written as 2/40/12/6. Calculate the centre distance? |
| A. | 40mm |
| B. | 156mm |
| C. | 200mm |
| D. | 80mm |
| Answer» C. 200mm | |
| 32. |
What is the centre distance for the worm gear? |
| A. | (m<sub>n</sub>/2)(T<sub>1</sub> cotλ<sub>1</sub> – T<sub>2</sub>) |
| B. | (m<sub>n</sub>/2)(T<sub>2</sub> cotλ<sub>1</sub> + T<sub>1</sub>) |
| C. | (m<sub>n</sub>/2)(T<sub>2</sub> cotλ<sub>1</sub> – T<sub>1</sub>) |
| D. | (m<sub>n</sub>/2)(T<sub>1</sub> cotλ<sub>1</sub> + T<sub>2</sub>) |
| Answer» E. | |
| 33. |
The normal lead, in a worm having multiple start threads, is given b? |
| A. | l<sub>N</sub> = l / cos λ |
| B. | l<sub>N</sub> = l . cos λ |
| C. | l<sub>N</sub> = l |
| D. | l<sub>N</sub> = l tan |
| Answer» C. l<sub>N</sub> = l | |
| 34. |
If worm helix angle is 30‚Äö√Ö‚àû, then worm should have at least ___ threads.$ |
| A. | 5 |
| B. | 6 |
| C. | 7 |
| D. | 8 |
| Answer» B. 6 | |
| 35. |
What is the velocity ratio of worm gears? |
| A. | (lπ)/d<sub>2</sub> |
| B. | (πd<sub>2</sub>)/l |
| C. | l/(πd<sub>2</sub>) |
| D. | d<sub>2</sub>/(lπ) |
| Answer» D. d<sub>2</sub>/(l‚âà√¨‚àö√ë) | |
| 36. |
In worm gears, the angle between the tangent to the thread helix on the pitch cylinder and the plane normal to the axis of worm is called |
| A. | pressure angle |
| B. | lead angle |
| C. | helix angle |
| D. | friction angle |
| Answer» C. helix angle | |
| 37. |
The worm helix angle is the _____ of the worm lead angle. |
| A. | Complement |
| B. | Half |
| C. | Double |
| D. | Supplement |
| Answer» B. Half | |
| 38. |
The angle at which the teeth are inclined to the normal of the axis of rotation is called _______________ |
| A. | pitch angle |
| B. | lead angle |
| C. | normal angle |
| D. | joint angle |
| Answer» C. normal angle | |
| 39. |
The worm gears are widely used for transmitting power at ______________ velocity ratios between non-intersecting shafts. |
| A. | high |
| B. | low |
| C. | medium |
| D. | none of the mentioned |
| Answer» B. low | |
| 40. |
Can worm gears be used in steering mechanism? |
| A. | True |
| B. | False |
| Answer» B. False | |
| 41. |
The distance by which a helix advances along the axis of the gear for one turn around is called _____________ |
| A. | joint line |
| B. | normal link |
| C. | axial pitch |
| D. | lead |
| Answer» E. | |
| 42. |
For a bevel gear having the pitch angle θ, the ratio of formative number of teeth (TE) to actual number of teeth (T) is$ |
| A. | 1/sin θ |
| B. | 1/cos θ |
| C. | 1/tan θ |
| D. | sin θ cos θ |
| Answer» C. 1/tan ‚âà√≠‚Äö√†√® | |
| 43. |
The power transmitting capacity of worm gears is high although efficiency is low. |
| A. | True |
| B. | False |
| Answer» C. | |
| 44. |
The distance between corresponding points on adjacent teeth measured along the direction of the axis is called ____________ |
| A. | joint line |
| B. | normal link |
| C. | axial pitch |
| D. | lead |
| Answer» D. lead | |
| 45. |
If b denotes the face width and L denotes the cone distance, then the bevel factor is written as |
| A. | b / L |
| B. | b / 2L |
| C. | 1 – 2 b.L |
| D. | 1 – b / L |
| Answer» E. | |
| 46. |
Is it possible to use worm gears in cranes for lifting purpose? |
| A. | True |
| B. | No self-locking and hence not possible |
| C. | Possible up to a threshold load |
| D. | None of the listed |
| Answer» B. No self-locking and hence not possible | |
| 47. |
Which is of these is an advantage of worm gear? |
| A. | It is expensive |
| B. | Has high power losses and low transmission efficiency |
| C. | Produce a lot of heat |
| D. | Used for reducing speed and increasing torque |
| Answer» E. | |
| 48. |
Which of the following is not true about worm gears? |
| A. | Compact |
| B. | Smooth and silent operation |
| C. | Low speed reduction |
| D. | All the mentioned are true |
| Answer» D. All the mentioned are true | |
| 49. |
The driven gear in the worm gear is a helical gear. True or false? |
| A. | True |
| B. | False |
| Answer» B. False | |
| 50. |
The face angle of a bevel gear is equal to |
| A. | pitch angle – addendum angle |
| B. | pitch angle + addendum angle |
| C. | pitch angle – dedendum angle |
| D. | pitch angle + dedendum angle |
| Answer» C. pitch angle ‚Äö√Ñ√∂‚àö√ë‚àö¬® dedendum angle | |