The starting torque of an induction motor is a crucial factor for its performance. A squirrel cage induction motor does not develop maximum starting torque due to the following reasons:

1. Option: Slip is less than unity at starting

   Slip, denoted by s, is defined as s = \frac{N_s - N_r}{N_s}, where N_s is the synchronous speed and N_r is the rotor speed. At starting, the rotor is stationary, so N_r = 0. Thus, slip is 1 (unity) at starting, not less than unity. Hence, this option is incorrect.

2. Option: Rotor resistance is fixed and low

   This is the correct answer. A squirrel cage induction motor has a rotor design with uniformly spaced bars short-circuited by end rings. This results in low and fixed resistance. For maximum starting torque, the motor requires higher rotor resistance; however, due to its inherently low rotor resistance, the starting torque is less than maximum.

3. Option: Rotor reactance is zero at starting

   The reactance of the rotor is not zero at starting. In fact, the inductive reactance is maximum at starting because the slip is one, making this option incorrect.

4. Option: Rotor resistance can be externally varied at starting

   This option is incorrect because in a squirrel cage rotor, the rotor resistance is fixed and cannot be externally varied. This characteristic contrasts with wound rotor induction motors where external resistance can be added to improve starting torque.

Thus, the inability of a squirrel cage induction motor to develop maximum starting torque is primarily due to its low and fixed rotor resistance.