If 'm' is the mass per unit length of belt, 'T_{1}' is maximum allowable belt tension and 'T_{C}' is centrifugal tension, for maximum power transmission, the velocity of the belt is

a) \(\sqrt{\frac{T_1}{3m}}\)

b) \(\sqrt{\frac{T_c}{m}}\)

c) \(\sqrt{\frac{3T_1}{m}}\)

d) \(\sqrt{\frac{m}{T_c}}\)

Which of the given above is/are correct?

This question was previously asked in

MPSC AMVI Official Paper 3: Set A/2011

Option 3 : a and b

NTPC Diploma Trainee 2020: Full Mock Test

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120 Questions
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120 Mins

**Concept:**

Centrifugal Tension:

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 the centrifugal force is called centrifugal tension (T_{c})

Condition for Maximum power transmitted by the belt:

When T_{1} = 3T_{c} 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.

Power transmitted by a belt:

P = (T_{1}– T_{2})V

where

T_{1} is the tension on the tight side (N), T_{2} is the tension on the slack side (N), and V is the velocity of the belt in (m/s).

The belt velocity and centrifugal tension is given by:

T_{c} = mv^{2}

Therefore, v = \(\sqrt{\frac{T_c}{m}}=\sqrt{\frac{T_1}{3m}}\)