# Machine Design Questions and Answers – Belt Construction II & V Belts

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This set of Machine Design Questions for campus interviews focuses on “Belt Construction II & V Belts”.

1. Calculate the angle of wrap if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.
a) 174.8⁰
b) 167.8⁰
c) 159.3⁰
d) None of the mentioned

Explanation: ὰ=180 – 2sin¯¹(D-d/2C).

2. Calculate the arc of contact if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.
a) 1.04
b) 1.03
c) 1.01
d) 1.02

Explanation: ὰ=180 – 2sin¯¹(D-d/2C). Factor=1+ (1.04-1)(180-174.8)/(180-170).

3. Calculate the belt length if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.
a) 6.5m
b) 4.66m
c) 6.94m
d) 5.26m

Explanation: L=2C + π(D+d)/2 + (D-d)²/4C.
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4. Crowns are never mounted on the pulley.
a) True
b) False

Explanation: Crowns are used to avoid slip in case of misalignment or non-parallelism.

5. In a cast iron pulley minor axis is generally kept in the plane of rotation.
a) True
b) False

Explanation: Keeping minor axis in plane of rotation increases the cross section.

6. The number of V belts required for a given application are given by (ignoring correction factor for arc of contact and belt length) Transmitted power/kW rating of single belt x Industrial Service Factor.
a) True
b) False

Explanation: It is given by Transmitted power x Industrial Service Factor /kW rating of single belt.

7. The pitch diameter of bigger pulley D in terms of small diameter d is given by
a) dx[speed of smaller pulley/speed of bigger pulley].
b) dx[speed of bigger pulley/speed of smaller pulley].
c) d
d) None of the mentioned

Explanation: Product of diameter and speed of pulley is constant.

8. If maximum tension in the belt is 900N and allowable belt load is 500N. Calculate the number of belts required to transmit power.
a) 2
b) 3
c) 4
d) 5

Explanation: No of belts=900/500.

9. The belt tension is maximum when velocity of belt is 0.
a) It is max at v=infinity
b) True
c) It is velocity independent
d) It has a constant value

Explanation: P₁-mv²/P₂-mv²=e^(fa/sinθ/2). Hence belt tension is maximum when v=0.

10. If belt tension in the two sides is 730N and 140N and belt is moving with a velocity of 10m/s, calculate the power transmitted.
a) 4.5kW
b) 5.9kW
c) 6.2kW
d) None of the mentioned

Explanation: Power=(P₁-P₂)xv.

11. If tensions in the belt are P₁ and P₂, then find P₁-mv²/P₂-mv². Contact angle for smaller pulley is 156⁰, Groove angle is 36⁰ and coefficient of friction is 0.2.
a) 6.21
b) 5.83
c) 4.66
d) 5.36

Explanation: P₁-mv²/P₂-mv²=e^(fὰ/sinθ/2).

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