Correct Answer - Option 1 : 1250 N-m

__Concept: __

Speed of the motor is given as

\({ω} = \frac{{{N} \times 2\pi }}{{60}}\)

Torque developed by the motor is

T = P / ω

where,

Power developed P = E Ia

\(E = \frac{ϕ ZNp}{60A}\)

∴ \(T= \frac{\frac{ϕ ZNp}{60A}I_a}{\frac{{{N_s} \times 2\pi }}{{60}}}=\frac{ϕ ZpI_a}{2\pi A}\)

__Calculation:__

Given,

Armature current Ia = 130 A

Number of poles P = 6

Lap wound A = P = 6

Number of conductors Z = 864

The flux per pole ϕ = 0.07 Wb

\(T=\frac{ϕ ZpI_a}{2\pi A}=\frac{0.07\times 864\times 6 \times130}{2\pi\times 6 }=1251.97\;Nm\)

T ≈ 1250 N-m