**Given information:**
- Source voltage (V_s) = 200 V
- Source current (I_s) = 10.5 A
- Motor speed (N) = 200 rpm
- Armature resistance (R_a) = 0.5 ohm
- Field resistance (R_f) = 0.5 ohm

**To find:**
- Motor speed (N) when the source voltage drops to 175 V

**Assumptions:**
- The load torque is constant at the rated motor torque.
- The motor is operating under steady-state conditions.

**Solution:**

1. **Determine the rated motor torque (T_r):**
The rated motor torque can be calculated using the formula:
T_r = (V_s - V_a) / (I_a)
Where V_a is the armature voltage and I_a is the armature current.
Given that V_s = 200 V, V_a = I_a * R_a, and R_a = 0.5 ohm, we can calculate V_a and I_a as follows:
V_a = I_a * R_a
I_a = I_s - I_f
Where I_f is the field current.
Given that R_f = 0.5 ohm, we can calculate I_f as follows:
I_f = V_s / (R_a + R_f)
Substituting the values, we get:
I_f = 200 V / (0.5 ohm + 0.5 ohm) = 100 A
Now we can calculate V_a and I_a:
V_a = (10.5 A - 100 A) * 0.5 ohm = -44.75 V
I_a = 10.5 A - 100 A = -89.5 A (negative sign indicates the direction of current flow)

Substituting these values into the formula for T_r, we get:
T_r = (200 V - (-44.75 V)) / (-89.5 A) = 2.25 Nm

2. **Calculate the motor speed (N) at the reduced voltage (V'_s):**
To calculate the motor speed at the reduced voltage, we need to determine the new armature voltage (V'_a) using the voltage drop formula:
V'_a = V'_s - I_a * R_a
Given that V'_s = 175 V and I_a = -89.5 A, we can calculate V'_a as follows:
V'_a = 175 V - (-89.5 A * 0.5 ohm) = 219.75 V

Now we can calculate the new motor speed (N') using the formula:
N' = (V'_a / V_a) * N
Substituting the values, we get:
N' = (219.75 V / (-44.75 V)) * 200 rpm = -982.2 rpm

Therefore, the motor speed at the reduced voltage of 175 V is approximately -982.2 rpm. The negative sign indicates that the motor is running in the opposite direction.