Correct Answer - Option 2 : 202.5 N-m

**Concept:**

\(T = \frac{ZPϕ(I_a)}{A60}\)

Where,

T is the torque of the DC motor

Z is the number of conductors

A is the number of parallel paths

P is the number of poles of the motor

ϕ is the field flux of the motor

I_{a} is armature current

For any DC machine, \((\frac{ZP}{A60})\) remains constant

∴ T ∝ ϕ Ia

**Calculation:**

Given,

Torque T_{1} = 150 N-m

There is 10% reduction in field flux

⇒ ϕ_{2} = 90% ϕ_{1} = 0.9 ϕ_{1}

Also armature current increased by 50%

⇒ I_{a2} = 150% I_{a1} = 1.5 I_{a1}

As T ∝ ϕ I_{a}

\(\frac{{{T_2}}}{{{T_1}}} = \frac{{{\phi _2}}}{{{\phi _1}}} \times \frac{{{I_{a2}}}}{{{I_{a1}}}}\)

\(\frac{{{T_2}}}{{150}} = \frac{{0.9{\phi _2}}}{{{\phi _1}}} \times \frac{{1.5{I_{a2}}}}{{{I_{a1}}}} = 1.35\)

T_{2} = 202.5 N-m

**Therefore the new value of torque is 202.5 N-m**