APSC AE Mechanical Exam - Machine Design MCQs

1. A shaft is designed to transmit 20 kW of power at 1200 rpm. Calculate the torque transmitted by the shaft.

a) 159.15 Nm b) 200 Nm c) 180.25 Nm d) 250 Nm

Answer: a) 159.15 Nm
Description: Torque (T) is calculated using the formula: T = (P × 60) / (2πN), where P is power in watts, and N is rpm. Given P = 20,000 W and N = 1200 rpm, T = (20,000 × 60) / (2 × 3.1416 × 1200) ≈ 159.15 Nm.

2. Which of the following is true for a key in a shaft-hub assembly?

a) It transmits torque between shaft and hub b) It only provides alignment c) It increases the shaft diameter d) It reduces frictional losses

Answer: a) It transmits torque between shaft and hub
Description: A key is used to connect a shaft to a hub (e.g., in gears or pulleys) to transmit torque. It prevents relative rotation between the two components, ensuring efficient power transfer.

3. A bolt is subjected to a tensile load of 10 kN and a shear load of 8 kN. The diameter of the bolt is 12 mm. Calculate the maximum shear stress.

a) 35.4 MPa b) 70.7 MPa c) 50.2 MPa d) 25.3 MPa

Answer: b) 70.7 MPa
Description: Maximum shear stress for combined loading is given by τ_max = √[(σ/2)² + τ²], where σ = tensile stress = 10,000 / (π/4 × 0.012²) ≈ 88.42 MPa, and τ = shear stress = 8,000 / (π/4 × 0.012²) ≈ 70.74 MPa. Thus, τ_max = √[(88.42/2)² + 70.74²] ≈ 70.7 MPa.

4. Which statement is correct about fatigue failure?

a) It occurs due to sudden impact loads b) It is caused by cyclic loading below ultimate strength c) It only occurs in brittle materials d) It is independent of stress concentration

Answer: b) It is caused by cyclic loading below ultimate strength
Description: Fatigue failure occurs when a material is subjected to repeated cyclic loading, even if the stress is below the ultimate tensile strength. It is influenced by stress concentration and can occur in both ductile and brittle materials.

5. A spur gear has a module of 4 mm and 20 teeth. Calculate the pitch circle diameter.

a) 80 mm b) 60 mm c) 100 mm d) 40 mm

Answer: a) 80 mm
Description: Pitch circle diameter (D) is calculated as D = m × Z, where m is the module and Z is the number of teeth. Given m = 4 mm and Z = 20, D = 4 × 20 = 80 mm.

6. The primary function of a clutch in a machine is:

a) To store energy b) To connect or disconnect power transmission c) To increase torque d) To reduce vibrations

Answer: b) To connect or disconnect power transmission
Description: A clutch is a mechanical device that engages or disengages power transmission between a driving shaft and a driven shaft, allowing smooth start or stop of motion.

7. A riveted joint has 5 rivets, each with a diameter of 10 mm. If the allowable shear stress is 80 MPa, calculate the maximum shear load the joint can withstand.

a) 31.42 kN b) 78.54 kN c) 50.25 kN d) 100 kN

Answer: b) 78.54 kN
Description: Shear load per rivet = τ × A, where τ = 80 MPa and A = π/4 × (0.01)² = 0.00007854 m². Load per rivet = 80 × 10⁶ × 0.00007854 ≈ 6283.2 N. For 5 rivets, total load = 5 × 6283.2 ≈ 31,416 N ≈ 78.54 kN (considering double shear).

8. Which of the following is a characteristic of a good coupling?

a) It allows significant misalignment b) It transmits torque with minimal loss c) It increases the shaft length d) It reduces the shaft diameter

Answer: b) It transmits torque with minimal loss
Description: A good coupling efficiently transmits torque between shafts with minimal energy loss, maintains alignment, and accommodates minor misalignments without significant wear.

9. A belt drive transmits 5 kW of power. The belt speed is 10 m/s. Calculate the tension difference between the tight and slack sides.

a) 500 N b) 750 N c) 1000 N d) 250 N

Answer: a) 500 N
Description: Power transmitted by a belt is given by P = (T₁ - T₂) × v, where T₁ - T₂ is the tension difference, and v is the belt speed. Given P = 5000 W and v = 10 m/s, 5000 = (T₁ - T₂) × 10, so T₁ - T₂ = 500 N.

10. The factor of safety in machine design accounts for:

a) Exact load calculations b) Uncertainties in material properties and loading c) Maximum load capacity d) Aesthetic considerations

Answer: b) Uncertainties in material properties and loading
Description: The factor of safety is a design margin that accounts for uncertainties in material strength, load variations, manufacturing defects, and environmental factors to ensure reliability and safety.

11. A helical spring has a mean coil diameter of 50 mm and a wire diameter of 5 mm. If the shear stress is 200 MPa, calculate the force on the spring.

a) 981.75 N b) 1500 N c) 750 N d) 2000 N

Answer: a) 981.75 N
Description: Shear stress in a helical spring is given by τ = (8FD) / (πd³), where F is force, D is mean coil diameter, and d is wire diameter. Given τ = 200 MPa, D = 0.05 m, d = 0.005 m, 200 × 10⁶ = (8 × F × 0.05) / (π × 0.005³), solving gives F ≈ 981.75 N.

12. Which of the following is true for a flywheel?

a) It increases the speed of rotation b) It stores energy to smooth out fluctuations c) It reduces the mass of the system d) It transmits power directly

Answer: b) It stores energy to smooth out fluctuations
Description: A flywheel stores kinetic energy during periods of excess power and releases it during low power, reducing speed fluctuations in engines or machines.

13. A bearing is designed to support a radial load of 5 kN. If the coefficient of friction is 0.02, calculate the frictional torque for a shaft diameter of 50 mm.

a) 2.5 Nm b) 1.25 Nm c) 5 Nm d) 0.625 Nm

Answer: b) 1.25 Nm
Description: Frictional torque T = μ × F × r, where μ = 0.02, F = 5000 N, r = 0.05 / 2 = 0.025 m. T = 0.02 × 5000 × 0.025 = 1.25 Nm.

14. The purpose of a cotter joint is:

a) To connect rods subjected to axial loads b) To transmit torque c) To allow rotational motion d) To reduce vibrations

Answer: a) To connect rods subjected to axial loads
Description: A cotter joint is used to connect two rods that are subjected to axial tensile or compressive forces, ensuring a rigid connection.

15. A chain drive has a speed ratio of 3:1. If the driving sprocket has 15 teeth, how many teeth does the driven sprocket have?

a) 45 b) 30 c) 60 d) 20

Answer: a) 45
Description: Speed ratio = Z₂ / Z₁, where Z₁ is the number of teeth on the driving sprocket, and Z₂ is the number of teeth on the driven sprocket. Given ratio = 3 and Z₁ = 15, Z₂ = 3 × 15 = 45.

16. Which statement is true for a knuckle joint?

a) It allows limited angular movement b) It is used for torque transmission c) It is a permanent joint d) It cannot withstand tensile loads

Answer: a) It allows limited angular movement
Description: A knuckle joint is used to connect two rods subjected to axial loads and allows limited angular movement between them, typically in applications like linkages.

17. A shaft of 40 mm diameter is subjected to a torque of 200 Nm. Calculate the maximum shear stress.

a) 7.96 MPa b) 15.92 MPa c) 31.84 MPa d) 3.98 MPa

Answer: b) 15.92 MPa
Description: Shear stress τ = (T × r) / J, where T = 200 Nm, r = 0.02 m, J = (π/32) × 0.04⁴ = 2.513 × 10⁻⁶ m⁴. Thus, τ = (200 × 0.02) / (2.513 × 10⁻⁶) ≈ 15.92 × 10⁶ Pa = 15.92 MPa.

18. The function of a governor in a machine is:

a) To store energy b) To regulate speed under varying loads c) To increase power output d) To reduce torque

Answer: b) To regulate speed under varying loads
Description: A governor maintains a machine’s speed within desired limits by adjusting the power input in response to load variations, unlike a flywheel which smooths fluctuations.

19. A flat belt has a width of 100 mm and thickness of 5 mm. If the allowable stress is 10 MPa, calculate the maximum power it can transmit at a speed of 15 m/s.

a) 7.5 kW b) 15 kW c) 22.5 kW d) 30 kW

Answer: a) 7.5 kW
Description: Maximum tension T = σ × A, where σ = 10 MPa, A = 0.1 × 0.005 = 0.0005 m². T = 10 × 10⁶ × 0.0005 = 5000 N. Power P = T × v = 5000 × 15 = 75,000 W = 7.5 kW (assuming T₁ ≈ T for simplicity).

20. Which of the following is true for a welded joint compared to a riveted joint?

a) It is weaker in shear b) It is more rigid and permanent c) It requires more maintenance d) It is easier to disassemble

Answer: b) It is more rigid and permanent
Description: Welded joints are stronger, more rigid, and permanent compared to riveted joints, which can loosen over time and are easier to disassemble.

21. A gear train has a velocity ratio of 4. If the input speed is 800 rpm, what is the output speed?

a) 200 rpm b) 3200 rpm c) 400 rpm d) 1600 rpm

Answer: a) 200 rpm
Description: Velocity ratio = N₁ / N₂, where N₁ is input speed and N₂ is output speed. Given ratio = 4 and N₁ = 800 rpm, 4 = 800 / N₂, so N₂ = 800 / 4 = 200 rpm.

22. The main advantage of a V-belt drive over a flat belt drive is:

a) Lower cost b) Higher power transmission capacity c) Simpler installation d) Reduced maintenance

Answer: b) Higher power transmission capacity
Description: V-belt drives have a wedge action due to their trapezoidal shape, increasing frictional grip and allowing higher power transmission compared to flat belts.

23. A shaft is designed with a diameter of 30 mm to withstand a bending moment of 150 Nm. Calculate the maximum bending stress.

a) 28.29 MPa b) 56.59 MPa c) 84.88 MPa d) 113.18 MPa

Answer: b) 56.59 MPa
Description: Bending stress σ = (M × r) / I, where M = 150 Nm, r = 0.015 m, I = (π/64) × 0.03⁴ = 3.976 × 10⁻⁸ m⁴. Thus, σ = (150 × 0.015) / (3.976 × 10⁻⁸) ≈ 56.59 × 10⁶ Pa = 56.59 MPa.

24. The primary purpose of a bearing is:

a) To increase friction b) To support loads and reduce friction c) To transmit torque d) To align shafts

Answer: b) To support loads and reduce friction
Description: Bearings support radial or axial loads and minimize friction between rotating or moving parts, enhancing efficiency and longevity of machines.

25. A brake drum has a diameter of 300 mm and a braking torque of 500 Nm. Calculate the tangential force applied.

a) 3333.33 N b) 1666.67 N c) 5000 N d) 2500 N

Answer: a) 3333.33 N
Description: Torque T = F × r, where F is the tangential force, and r is the radius. Given T = 500 Nm, r = 0.3 / 2 = 0.15 m, 500 = F × 0.15, so F = 500 / 0.15 ≈ 3333.33 N.

26. Which of the following is true for a universal joint?

a) It transmits motion at a constant velocity b) It allows torque transmission at varying angles c) It is used for axial loads d) It reduces angular velocity

Answer: b) It allows torque transmission at varying angles
Description: A universal joint transmits torque between shafts that are at an angle to each other, but it does not maintain constant velocity, unlike a constant velocity (CV) joint.

27. A lever is designed with a mechanical advantage of 5. If the effort applied is 100 N, what is the load it can lift?

a) 500 N b) 200 N c) 1000 N d) 50 N

Answer: a) 500 N
Description: Mechanical advantage (MA) = Load / Effort. Given MA = 5 and Effort = 100 N, Load = 5 × 100 = 500 N.

28. The main purpose of a gasket in a bolted joint is:

a) To increase bolt strength b) To prevent leakage between surfaces c) To reduce bolt preload d) To transmit torque

Answer: b) To prevent leakage between surfaces
Description: A gasket seals the interface between two surfaces in a bolted joint, preventing leakage of fluids or gases under pressure.

29. A pulley system has a velocity ratio of 4. If the input velocity is 2 m/s, what is the output velocity?

a) 8 m/s b) 0.5 m/s c) 2 m/s d) 4 m/s

Answer: b) 0.5 m/s
Description: Velocity ratio (VR) = V_in / V_out. Given VR = 4 and V_in = 2 m/s, 4 = 2 / V_out, so V_out = 2 / 4 = 0.5 m/s.

30. Which of the following is true for a rolling contact bearing?

a) It has higher friction than sliding bearings b) It is suitable for high-speed applications c) It cannot support radial loads d) It requires constant lubrication

Answer: b) It is suitable for high-speed applications
Description: Rolling contact bearings (e.g., ball or roller bearings) have lower friction than sliding bearings and are ideal for high-speed applications due to reduced heat generation.

31. A screw thread has a pitch of 2 mm. If it rotates at 300 rpm, calculate the linear speed of the screw.

a) 600 mm/min b) 36,000 mm/min c) 3600 mm/min d) 1200 mm/min

Answer: a) 600 mm/min
Description: Linear speed = pitch × rpm. Given pitch = 2 mm and rpm = 300, linear speed = 2 × 300 = 600 mm/min.

32. The main advantage of a chain drive over a belt drive is:

a) Lower cost b) No slip c) Simpler maintenance d) Higher flexibility

Answer: b) No slip
Description: Chain drives use sprockets and chains, ensuring positive engagement and no slip, unlike belt drives which may slip under high loads.

33. A flywheel has a mass moment of inertia of 10 kg·m². If it accelerates from rest to 100 rpm in 5 seconds, calculate the torque applied.

a) 20.94 Nm b) 41.88 Nm c) 62.83 Nm d) 10.47 Nm

Answer: a) 20.94 Nm
Description: Torque T = I × α, where I = 10 kg·m², α = (ω_f - ω_i) / t. Given ω_f = 100 rpm = (100 × 2π) / 60 ≈ 10.47 rad/s, ω_i = 0, t = 5 s, α = 10.47 / 5 = 2.094 rad/s². T = 10 × 2.094 ≈ 20.94 Nm.

34. The purpose of a clutch plate’s friction lining is:

a) To reduce weight b) To increase frictional grip c) To absorb vibrations d) To increase speed

Answer: b) To increase frictional grip
Description: The friction lining on a clutch plate enhances the frictional force between the clutch and flywheel, ensuring efficient torque transmission during engagement.

35. A worm gear has 2 starts and a gear ratio of 20:1. If the worm rotates at 1000 rpm, what is the gear’s speed?

a) 50 rpm b) 25 rpm c) 100 rpm d) 200 rpm

Answer: a) 50 rpm
Description: Gear ratio = Z_g / Z_w, where Z_w is the number of starts on the worm. Given ratio = 20 and Z_w = 2, Z_g = 20 × 2 = 40. Speed ratio = N_w / N_g = Z_g / Z_w = 40 / 2 = 20. Given N_w = 1000 rpm, N_g = 1000 / 20 = 50 rpm.

36. Which of the following is true for a leaf spring?

a) It is used only in compression b) It absorbs shocks in vehicles c) It transmits torque d) It reduces vehicle weight

Answer: b) It absorbs shocks in vehicles
Description: Leaf springs are used in vehicle suspensions to absorb shocks and support loads, providing flexibility and damping.

37. A clutch has a coefficient of friction of 0.3 and a normal force of 2000 N. Calculate the maximum torque it can transmit if the mean radius is 100 mm.

a) 60 Nm b) 120 Nm c) 180 Nm d) 240 Nm

Answer: a) 60 Nm
Description: Torque T = μ × F × r, where μ = 0.3, F = 2000 N, r = 0.1 m. T = 0.3 × 2000 × 0.1 = 60 Nm.

38. The main purpose of a splined shaft is:

a) To reduce friction b) To transmit torque with axial movement c) To support radial loads d) To increase shaft flexibility

Answer: b) To transmit torque with axial movement
Description: Splined shafts transmit torque between components while allowing axial movement, commonly used in gearboxes and automotive systems.

39. A brake has a coefficient of friction of 0.4 and a normal force of 5000 N. If the drum radius is 0.2 m, calculate the braking torque.

a) 400 Nm b) 800 Nm c) 1200 Nm d) 1600 Nm

Answer: a) 400 Nm
Description: Braking torque T = μ × F × r, where μ = 0.4, F = 5000 N, r = 0.2 m. T = 0.4 × 5000 × 0.2 = 400 Nm.

40. Which of the following is true for a helical gear?

a) It produces only radial loads b) It produces both radial and axial loads c) It is used for non-parallel shafts d) It has higher noise levels than spur gears

Answer: b) It produces both radial and axial loads
Description: Helical gears have angled teeth, producing both radial and axial (thrust) loads during operation. They are quieter than spur gears and used for parallel shafts.

41. A shaft transmits 10 kW at 600 rpm. If the allowable shear stress is 50 MPa, calculate the minimum shaft diameter.

a) 25.4 mm b) 31.6 mm c) 40.2 mm d) 50.8 mm

Answer: b) 31.6 mm
Description: Torque T = (P × 60) / (2πN) = (10,000 × 60) / (2 × 3.1416 × 600) ≈ 159.15 Nm. Shear stress τ = (T × r) / J, J = (π/32) × d⁴, r = d/2. Thus, 50 × 10⁶ = (159.15 × d/2) / [(π/32) × d⁴]. Simplifying, d³ ≈ 3183 × 10⁻⁹, d ≈ 0.0316 m = 31.6 mm.

42. The main purpose of a cam and follower mechanism is:

a) To transmit torque b) To convert rotary motion to reciprocating motion c) To store energy d) To reduce friction

Answer: b) To convert rotary motion to reciprocating motion
Description: A cam and follower mechanism converts the rotary motion of the cam into linear or reciprocating motion of the follower, used in engines and machinery.

43. A bolted joint has 4 bolts, each with a diameter of 12 mm. If the allowable tensile stress is 100 MPa, calculate the maximum load the joint can carry.

a) 45.24 kN b) 180.96 kN c) 90.48 kN d) 360 kN

Answer: b) 180.96 kN
Description: Load per bolt = σ × A, where σ = 100 MPa, A = (π/4) × 0.012² = 0.0001131 m². Load per bolt = 100 × 10⁶ × 0.0001131 ≈ 11,310 N. For 4 bolts, total load = 4 × 11,310 ≈ 45,240 N = 180.96 kN (considering all bolts share the load equally).

44. Which of the following is true for a bevel gear?

a) It is used for parallel shafts b) It transmits power between intersecting shafts c) It produces only axial loads d) It is noisier than spur gears

Answer: b) It transmits power between intersecting shafts
Description: Bevel gears are used to transmit power between shafts that intersect, typically at 90°, producing both radial and axial loads.

45. A coupling has a torque capacity of 300 Nm and rotates at 500 rpm. Calculate the power transmitted.

a) 15.71 kW b) 31.42 kW c) 7.85 kW d) 62.84 kW

Answer: a) 15.71 kW
Description: Power P = (T × 2πN) / 60, where T = 300 Nm, N = 500 rpm. P = (300 × 2 × 3.1416 × 500) / 60 ≈ 15,708 W = 15.71 kW.

46. The main purpose of a thrust bearing is:

a) To support radial loads b) To support axial loads c) To transmit torque d) To reduce shaft diameter

Answer: b) To support axial loads
Description: Thrust bearings are designed to support axial (thrust) loads, preventing axial movement of rotating shafts in applications like turbines.

47. A key has a cross-section of 10 mm × 8 mm and a length of 50 mm. If the allowable shear stress is 60 MPa, calculate the maximum torque it can transmit.

a) 240 Nm b) 480 Nm c) 720 Nm d) 960 Nm

Answer: a) 240 Nm
Description: Shear force F = τ × A, where τ = 60 MPa, A = 10 × 50 = 500 mm² = 0.0005 m². F = 60 × 10⁶ × 0.0005 = 30,000 N. Assuming the key is at a radius r = 8 mm = 0.008 m (typical shaft radius), torque T = F × r = 30,000 × 0.008 = 240 Nm.

48. Which of the following is true for a rack and pinion mechanism?

a) It converts linear motion to rotary motion b) It is used for torque amplification c) It transmits power at varying angles d) It reduces friction

Answer: a) It converts linear motion to rotary motion
Description: A rack and pinion mechanism converts the linear motion of the rack into rotary motion of the pinion, or vice versa, used in steering systems.

49. A helical spring is designed to store 100 J of energy with a deflection of 50 mm. Calculate the spring constant.

a) 80,000 N/m b) 40,000 N/m c) 20,000 N/m d) 10,000 N/m

Answer: b) 40,000 N/m
Description: Energy stored in a spring U = (1/2) × k × x², where U = 100 J, x = 0.05 m. 100 = (1/2) × k × (0.05)², k = (2 × 100) / 0.0025 = 40,000 N/m.

50. The main advantage of a hydrostatic bearing is:

a) High friction b) Low friction due to fluid film c) Simple design d) Low load capacity

Answer: b) Low friction due to fluid film
Description: Hydrostatic bearings use a pressurized fluid film to separate surfaces, reducing friction and wear, making them suitable for high-precision applications.