SSC JE Mechanical Engineering 2012 Paper

Question : 180 of 200

Marks: +1, -0

Mass per unit length of the belt of an open belt drive is 1.5 kg/m. If the linear velocity of the belt is 10 m/s and radius of driving pulley 0.5 m, centrifugal tension in the belt is

600 N
400 N
300 N
150 N

Validate

Solution:

Given:
Mass per unit length = 1.5 kg/m
Linear velocity V = 10 m/s
Radius of driving polly r = 0.5 m
Centrifugal tension

T_C = mv ^2= 1.5 × (10)^ 2

= 1.5 × 100 = 150 N

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