Correct Answer - Option 4 :

\(V\; = \;\sqrt {\dfrac{{\left( {{T_1} + \;{T_c}} \right)}}{{3m}}} \)
__Explanation:__

**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}).

When the centrifugal tension in the flat belt is considered. Then maximum tension in the belt,

⇒ T = T_{1} + T_{c}

Power transmitted by a belt is given by:

P = (T_{1} – T_{2}) × v …(1)

For** Flat Belt:**

⇒ T_{1} = T_{2}e^{μθ}

Also, T_{c} = mV^{2} …(2)

where, T_{1} = friction tension on the tight side, T_{2} = friction tension on the slack side

Using equation (1) and (2),

\(P = {T_1}\left( {1 - \dfrac{1}{{{e^{\mu \theta }}}}} \right)V = \left( {T - {T_c}} \right)VK\)

Where,

\(K = \left( {1 - \dfrac{1}{{{e^{\mu \theta }}}}} \right)\;,\)

⇒ P = (T – mV^{2}) VK

**For Maximum power transmitted by the belt**,

\( \Rightarrow \dfrac{{dP}}{{dV}} = 0\)

**⇒**** T = 3T**_{C} ..(3)

So, the **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.

For maximum power transmission,

Using equation (2) and (3),

\(\Rightarrow V = \sqrt {\dfrac{T}{{3m}}} = \sqrt {\dfrac{{\left( {{T_1} + {T_c}} \right)}}{{3m}}} \)