**Given date: **

Density of belt = 1gm/cm^{3 }= 1/1000 Kg/cm^{3 }

Maximum permissible stress(f) = 250 N/cm^{2 }

Width of belt(b) = 20cm

Thickness of belt(t) = 8mm 0.8cm

Ratio of tension (T_{1}/T_{2}) = 2

Maximum Power = ?

We know that, Tmax = σ.b.t

= 250 X 20 X 1.2 = 6000N ...(i)

Since Unit mass (m) = σ.b.t.L

= 1/1000 X 20 X 1.2 X 100 = 2.4Kg ...(ii)

Since velocity(V) is not given,

So we find the velocity using formula T_{C }=1/3 T_{max} = mV^{2,} for maximum power

Using the formula, 1/3 T_{max} = mV^{2}

V = (Tmax/3m)^{1/2}

V = (6000/3 X 2.4)^{1/2}

V = 28.86 m/sec ...(iii)

Using the formula, T_{C} = mV^{2}, For finding T_{C}

= 2.4(28.86)^{2}

= 1998.96N ...(iv)

Using the formula, T_{max }= T_{1} + T_{c} , for finding T_{1 }

6000 = T_{1 }+ 1998.96

T_{1} = 4001N...(v)

Now, For T_{2,} Using relation Ratio of belt tension = T_{1}/T_{2} = e^{µθ} = 2

4001/T_{2} = 2

T_{2} = 2000.5 N ...(vi)

Now Maximum Power Transmitted = (T_{1}–T_{2}).V/1000 KW

P = (4001 - 2000.5) X 28.86/1000 KW

**P = 57.73KW**