If the field of a DC shunt motor is opened while running, what will happen to the speed?

Correct option is C

If the field of a D.C shunt motor is opened while running then its speed becomes dangerously high.
Because in DC Shunt motor

​$$N \propto \frac{E_b}{\phi}$$​​

If field winding is opened then ϕ=0

$$N \propto \frac{E_b}{0}$$​

(N=infinite(∞) Where, N = speed of the motor Eb= back E.M.F. of the motor