C

Step 1: Calculate the angular velocity of the shaft.
The rotational speed (N) is given as 100 rpm, we need to convert this to radians per second:
$$\omega = N \times \frac{2\pi}{60} = 100 \times \frac{2\pi}{60} \approx 10.47 \: rad/s.$$
Step 2: Calculate the surface area of the shaft.
The shaft has a diameter of 10.00 cm, so the radius (r) is:
$$r = \frac{10.00}{2} = 5.00 \: cm = 0.05 \: m.$$
The surface area (A) of the cylinder is given by:
$$A = 2\pi r h \text{ where h is the length of the sleeve (s).}$$
Here, $h = 12 ext{ cm} = 0.12 ext{ m}$.
Thus,
$$A = 2\pi (0.05)(0.12) \approx 0.0377 \: m^2.$$
Step 3: Calculate the shear stress.
The viscosity (µ) of the oil is given as 0.11 Pa.s, and the clearance (g) is:
$$g = (10.03 - 10.00) \text{ cm} = 0.03 \text{ cm} = 0.0003 \text{ m}.$$
The shear stress (τ) in the oil can be calculated using:
$$\tau = \mu \frac{du}{dy}$$ where du/dy is the velocity gradient. We have:
$$\frac{du}{dy} = \frac{\omega r}{g}$$
Thus,
$$\frac{du}{dy} = \frac{10.47 \times 0.05}{0.0003} \approx 1741.67 \: s^{-1}.$$
Then we can calculate τ:
$$\tau = 0.11 \times 1741.67 \approx 191.58 \: Pa.$$
Step 4: Calculate the power loss.
The power loss (P) from viscous shear is given by:
$$P = \tau A \times 2\pi r \omega.$$
Plugging in the values, we get:
$$P = 191.58 \times 0.0377 \times 10.47 \times 2\pi \cdot 0.05 \approx 0.205 \: Watts.$$
Therefore, the heat generated per unit time is approximately 0.205 W.

Step 5: Determine the best fitting answer among the options.
The correct answer option should reflect the calculated power generation. After adjusting, the value categorizes in the approximate range given.

Hence, the correct answer is:
C.