Radius of each pulley = 50mm,

R_{1} = R_{2} = 50mm

R_{1} = Radius of the driver = 50mm

R_{2} = Radius of the driven = 50mm

θ = Angle of contact(In radian) = 1800 = p,

µ = Coefficient of friction = 0.3

T_{1} = Allowable tension = 3KN

T_{1} always greater than T_{2}

Using the relation T_{1}/T_{2 }= e^{µθ}

Putting all the value,

3/T2 = e^{(0.3)(π)}

On solving T_{2} **= 1.169KN**

Since Radius of both pulley is same;

So, Torque exerted on both pulley is same an

= (T_{1}–T_{2}).R_{1} = (T_{1}–T_{2}).R_{2}

Putting all the value we get,

**(3 – 1.169) X 50 = 91.55 KN-mm**