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In a flat belt drive the maximum tension which the belt can be subjected to is T and the mass of the belt per unit length is m kg. The velocity of the belt for maximum power transmission is
1. \(\sqrt {\frac {T}{3m}}\)
2. \(\sqrt {\frac {T}{m}}\)
3. \({\frac {T}{3m}}\)
4. \( {\frac {T}{m}}\)

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Correct Answer - Option 1 : \(\sqrt {\frac {T}{3m}}\)

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:

\(v = \sqrt {\frac{T}{{3m}}} \)

where m is mass of belt per unit length.

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

Centrifugal Tension (Tc):

Since the belt continuously runs over the pulleys, therefore some centrifugal force is caused, whose effect is to increase the tension on both the tight as well as the slack sides.

The tension caused by this centrifugal force is called centrifugal tension (0.5 mv2).

Condition for Maximum power transmitted by the belt is T = 3TC 

i.e. power transmitted will be maximum when tension is equal to three-time centrifugal tension or it shows that when the power transmitted is maximum, 1/3rd of the maximum tension is absorbed as centrifugal tension.

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