The velocity of belt for maximum power is

(Where m = mass of the belt in kg per metre length. T = Tension)

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UKPSC JE Mechanical 2013 Official Paper I
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Answer (Detailed Solution Below)

Option 1 :
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UKPSC JE CE Full Test 1 (Paper I)
180 Qs. 360 Marks 180 Mins

Detailed Solution

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Concept:

Power transmitted by a belt:

P = (T1 – T2)v

For maximum power:

T1 = T/3

where T1 = Tension in the tight side, T = Maximum tension to which the belt is subjected.

The velocity of the belt for maximum power (Maximum Permissible velocity)

where m is mass of belt per unit length.

m = Area × length × density = b.t.l.ρ 

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