The speed of DC shunt motor can be increased above its rated speed by

Correct option is C

​$$\text{The speed of a DC shunt motor is given by the formula:}N = \frac{V - I_a R_a}{k \phi}\text{where:}\bullet \text{ N is the speed of the motor,} \\\bullet \text{ V is the supply voltage,} \\\bullet \text{ I}_a \text{ is the armature current,} \\\bullet \text{ R}_a \text{ is the armature resistance,} \\\bullet \text{ k is a constant,} \\\bullet \text{ } \phi \text{ is the flux per pole.}$$​​
From this formula, we can see that the speed N is inversely proportional to the flux ϕ. Therefore, to increase the speed of the motor above its rated speed, the flux ϕ must be decreased.
From this formula, we can see that the speed N is inversely proportional to the flux ϕ. Therefore, to increase the speed of the motor above its rated speed, the flux ϕ must be decreased.
 Increasing the supply voltage within its rated value (a): This would increase the speed, but only within the rated limits and not above the rated speed in a safe manner.
 Increasing the flux per pole (b): This would decrease the speed because the speed is inversely proportional to the flux.
 Connecting additional resistance in series with the armature (d): This is typically used for speed control below the rated speed because it causes a voltage drop in the armature circuit, thus reducing the effective voltage and speed.
Therefore, the correct answer is (c) decreasing the flux per pole