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1. Identify the given gear. a) Spur gear
b) Helical gear
c) Worm and worm gear
d) Bevel gear

Answer: b [Reason:] The given diagram is of a helical gear. In helical gears, the teeth are inclined to the axis of the gear. The gears can be either left handed or can be right handed depending on the direction in which the helix slopes when viewed. Here, 1 is left handed gear and 2 is a right handed gear.

2. The angle at which the teeth of the gear are inclined to the axis of a gear is called as __________
a) pitch angle
b) normal angle
c) helix angle
d) gear angle

Answer: c [Reason:] Helix angle is the angle at which the teeth are inclined to the axis of a gear. In the diagram above, Ψ is called as helix angle. Depending upon the direction of this angle, the gear can be either left handed or right handed.

3. The distance between the corresponding points on adjacent teeth measured on the pitch circle is called ______________
a) helical pitch
b) normal pitch
c) gear pitch
d) circular pitch

Answer: d [Reason:] Circular pitch is the distance between the corresponding points on adjacent teeth measured on the pitch circle. Here, the distance p is called as circular pitch. p = πm, where m is the module.

4. The shortest distance measured along the normal to the helix between corresponding points on the adjacent teeth is called ____________
a) gear pitch
b) helical pitch
c) circular pitch
d) normal circular pitch

Answer: d [Reason:] Normal circular pitch or normal pitch is the shortest distance measured along the normal to the helix between corresponding points on the adjacent teeth. In the given figure, pn is the normal circular pitch. pn = p cos Ψ Therefore, mn = m cos Ψ.

5. Two spiral gears have a normal module of 10 mm and the angle between the skew shaft axes is 50°. The driver has 20 teeth and the helix angle of it is 30°. If the velocity ratio is 1/3 and the driver as well as the follower are both left handed, find the centre distance between the shafts.
a) 434.72 mm
b) 452.83 mm
c) 582.19 mm
d) 523.39 mm

Answer: a [Reason:] Ψ1 = 30°, mn = 10 mm and Ψ2 = 50° – 30° = 20°, T1 = 20. VR = 1/3 = T1/T2 T2 = 3 x T1 = 3 x 20 = 60 Centre distance = C = (mn/2) ((T1/cos Ψ1) + (T2/cos Ψ2)) = 434.72 mm.

6. The centre distance between two meshing gears is 250 mm and the angle between the shafts is 60°. The normal circular pitch is 10 mm and the gear ratio is 2. Ψ1 = 35°. Find the value of number of teeth on both the wheels.
a) 46,92
b) 45, 90
c) 54,108
d) 62, 124

Answer: a [Reason:] Given: Ψ1 = 35°, Ψ2 = 60° – 35° = 25°, G = T2/T1 = 2, pn = 10 mm and C = 250 mm Centre distance = C = 250 = (mn/2) ((T1/cos Ψ1) + (T2/cos Ψ2)) = (pn/2π) ((T1/cos Ψ1) + (T2/cos Ψ2)) = 1.59 x (1.22T1 + 2.207T1) = 5.45 T1 T1 = 250/5.45 = 45.88 ~ 46 T2 = 2 x T1 = 92.

7. Find the exact centre distance for the above problem.
a) 248.992 mm
b) 250.934 mm
c) 251.831 mm
d) 251.029 mm

Answer: b [Reason:] Ψ1 = 35°, Ψ2 = 60° – 35° = 25°, pn = 10 mm We found out that T1 = 46 and T2 = 92. Centre distance = (pn/2π) ((T1/cos Ψ1) + (T2/cos Ψ2)) = 250.934 mm.

8. Find the efficiency in the above problem if the friction angle is 6°.
a) 89.23 %
b) 91.02 %
c) 88.58 %
d) 85.69 %

Answer: c [Reason:] Given: φ = 6°, Ψ1 = 35°, Ψ2 = 60° – 35° = 25° Efficiency = η = (cos (Ψ2+ φ) x cos Ψ1¬)/ (cos (Ψ1 – φ) x cos Ψ2 ¬) = 0.8858 = 88.58 %.

9. The shaft angle of two helical gears is 60°. The normal module is 6 mm. The larger gear has 50 teeth and the gear ratio is 2. The centre distance is 300 mm. Find the helix angles of the two gears.
a) Ψ1 = 11.68°, Ψ2 = 48.32°
b) Ψ1 = 13.02°, Ψ2 = 46.98°
c) Ψ1 = 10.11°, Ψ2 = 49.89°
d) Ψ1 = 12.21°, Ψ2 = 47.79°

Answer: d [Reason:] Given: Ψ1 + Ψ2 = 60°; Ψ1 = 60° – Ψ2; mn = 6 mm; T2 = 50 and T1 = T2/G = 25. C = 300 mm C = 300 = (mn/2) ((T1/cos Ψ1) + (T2/cos Ψ2)) = 3((25/cos(60° – Ψ2))+(50/cos Ψ2) 100 = ((25/cos(60° – Ψ2))+(50/cos Ψ2) By trial and error method, we get Ψ2 = 47.79°, Ψ1 = 12.21°.

10. Find the spiral angles of the two helical gears for maximum efficiency if the shaft angle is 100° and the friction angle is 10°.
a) 55°,55°
b) 50°,50°
c) 55°,45°
d) 65°,45°

Answer: c [Reason:] For maximum efficiency of the helical gears, Ψ1 = (θ + φ)/2 = (100° + 10°)/2 = 55° Ψ2 = 100° – 55° = 45°.

11. Find the normal circular pitch as well as the normal module if the larger gear has 40 teeth and the gear ratio is 4. The helix angle is 20° and the shaft angle is 50°. Assume the centre distance between the gears to be 200 mm.
a) 22.12 mm, 7.04 mm
b) 21.31 mm, 6.78 mm
c) 20.37 mm, 6.48 mm
d) 21.98 mm, 6.996 mm

Answer: a [Reason:] Given: T2 = 40 and T1 = T2/G = 40/4 = 10, Ψ1 + Ψ2 = 50°; Ψ1 = 20°; Therefore, Ψ2 = 30°; C = 200 mm C = 200 = (mn/2) ((T1/cos Ψ1) + (T2/cos Ψ2)) 400 = mn x (56.83) mn = 7.04 mm and thus pn = πmn = 22.12 mm.

12. The normal circular pitch of two helical gears in a mesh is 7 mm. If the number of teeth on the smaller gear is 22 and the gear ratio is 2, find the pitch diameters of the gears. Ψ1 = 40° and Ψ2 = 25°.
a) 83.01 mm, 109.93 mm
b) 63.99 mm, 108.17 mm
c) 85.34 mm, 96.32 mm
d) 79.12 mm, 100.13 mm

Answer: b [Reason:] T1 = 22 and T2 = T1 x G = 22 x 2 = 44, pn = 7 mm, Ψ1 = 40° and Ψ2 = 25° d1 = P1T1/π = pnT1/(cos Ψ1 x π) = 63.99 mm d2 = P2T2/π = pnT2/(cos Ψ2 x π) = 108.17 mm.

13. For two gears having shaft angle 90° and friction angle 6°, find the maximum efficiency of the helical gears.
a) 89.23%
b) 85.64%
c) 79.02%
d) 81.07%