Test: Belt Construction II & V Belts


10 Questions MCQ Test Machine Design | Test: Belt Construction II & V Belts


Description
This mock test of Test: Belt Construction II & V Belts for Mechanical Engineering helps you for every Mechanical Engineering entrance exam. This contains 10 Multiple Choice Questions for Mechanical Engineering Test: Belt Construction II & V Belts (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Belt Construction II & V Belts quiz give you a good mix of easy questions and tough questions. Mechanical Engineering students definitely take this Test: Belt Construction II & V Belts exercise for a better result in the exam. You can find other Test: Belt Construction II & V Belts extra questions, long questions & short questions for Mechanical Engineering on EduRev as well by searching above.
QUESTION: 1

 Calculate the angle of wrap if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.

Solution:

Explanation: ὰ=180 – 2sin¯¹(D-d/2C).

QUESTION: 2

Calculate the arc of contact if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.

Solution:

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

QUESTION: 3

Calculate the belt length if diameter of the two pulleys are 550mm and 300mm. Also the centre distance is 2800mm.

Solution:

Explanation: L=2C + π(D+d)/2 + (D-d)²/4C.

QUESTION: 4

 Crowns are never mounted on the pulley.

Solution:

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

QUESTION: 5

In a cast iron pulley minor axis is generally kept in the plane of rotation.

Solution:

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

QUESTION: 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.

Solution:

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

QUESTION: 7

The pitch diameter of bigger pulley D in terms of small diameter d is given by

Solution:

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

QUESTION: 8

 If maximum tension in the belt is 900N and allowable belt load is 500N. Calculate the number of belts required to transmit power.

Solution:

Explanation: No of belts=900/500.

QUESTION: 9

 The belt tension is maximum when velocity of belt is 0.

Solution:

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

QUESTION: 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.

Solution:

Explanation: Power=(P₁-P₂)xv.

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