SSC JE Mechanical Engineering 2018 Set 2 Paper

Question : 106 of 200

Marks: +1, -0

A pulley is driven by a flat belt and the maximum tension produced in the belt is of 1400 N. The belt has the density of 1000 kg/m

^{3}

, 100 mm wide and 5 mm thick. What is the speed (m/sec) of the belt for the maximum power?

Solution:

For max. power condition

T = 3T_C \or T = 3(mv^2 )

1400 = 3(mv^2 )

V=√{1400/{3×.5}}

V = 30.55 m/s
∵

ρ=\mass/\volume

mass

= 1000 × .1 × .005 = .5 kg

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