1. When large gear reductions are needed _________ gears are used.

a) helical

b) spur

c) worm

d) bevel

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2. The driven gear in the worm gear is a helical gear. True or false?

a) True

b) False

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3. 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

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4. 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

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5. 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

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6. 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

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7. What is the velocity ratio of worm gears?

a) (lπ)/d_{2}

b) (πd_{2})/l

c) l/(πd_{2})

d) d_{2}/(lπ)

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_{2})/(2π) = l/(πd

_{2}).

8. What is the centre distance for the worm gear?

a) (m_{n}/2)(T_{1} cotλ_{1} – T_{2})

b) (m_{n}/2)(T_{2} cotλ_{1} + T_{1})

c) (m_{n}/2)(T_{2} cotλ_{1} – T_{1})

d) (m_{n}/2)(T_{1} cotλ_{1} + T_{2})

### View Answer

_{n}/2)(T

_{1}cotλ

_{1}+ T

_{2}) This equation can be derived by using the formula for centre distance of a helical gear which is given as C = (m

_{n}/2) ((T

_{1}/cos Ψ

_{1}) + (T

_{2}/cos Ψ

_{2}) As, Ψ

_{2}= λ

_{1}, Ψ

_{1}= 90° – λ

_{1}.

9. What is the formula to calculate maximum efficiency of a worm gear?

a) (1+sinø)/(1-sinø)

b) (1-sinø)/(1+sinø)

c) (tan(λ_{1}-ø))/tan λ_{1}

d) (tan(λ_{1}+ø))/tan λ_{1}

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_{1}-ø))/tan λ

_{1}

10. 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°

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_{1}= Lead / Pitch circumference = 2p/πd1 = 24/50π = 0.1528 λ

_{1}= 8.687°.

11. 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

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_{1}/N

_{2}= T

_{2}/T

_{1}500/N

_{2}= 20/3 N

_{2}= 75 rpm Thus, the gear rotates at a speed of 75 rpm.

12. 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

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_{n}/2)(T

_{1}cotλ

_{1}+ T

_{2}) Therefore, C = (p

_{n}/2π)(T

_{1}cotλ

_{1}+ T

_{2}) = 109.53 mm.

13. 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°

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^{-1}(0.05) = 2.862°; θ = 90° For maximum efficiency, Ψ

_{1}= (θ+ ø)/2 = 92.862/2 = 46.43° Ψ

_{1}= 90° – λ

_{1}= 46.43° λ

_{1}= 90° – 46.43° = 43.57°.

14. 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%

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_{1}= 10°, ø = tan

^{-1}(0.07) = 4° Efficiency = tan(λ

_{1})/ tan(λ

_{1}+ ø) = 0.7072 = 70.72%.

15. 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%

### View Answer

^{-1}(0.06) = 3.43° Maximum efficiency = (1-sin ø)/(1+sin ø) = 0.8871 = 88.71%.

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