A

Step 1: Define the parameters. The disk has a radius \( r_0 \) and rotates with an angular velocity \( \omega \). The viscosity of the fluid is \( \mu \).
Step 2: The linear velocity \( v \) at a radius \( r \) from the center of the disk is given by \( v = r \omega \).
Step 3: Consider a differential element of the disk at radius \( r \) with thickness \( dr \) and width \( h \) (the height of the cylinder). The area of the differential element is \( dA = h \, dr \).
Step 4: The shear stress \( \tau \) acting on this element due to viscosity is given by \( \tau = \mu \frac{du}{dy} \), where \( u \) is the velocity and \( y \) is the distance. The velocity gradient, in this case, is \( \frac{du}{dy} \approx \frac{v}{h} = \frac{r \omega}{h} \).
Step 5: Therefore, the shear stress can be expressed as \( \tau = \mu \frac{r \omega}{h} \).
Step 6: The differential torque \( dT \) contributed by this element is given by \( dT = \tau \, dA \, r = \left( \mu \frac{r \omega}{h} \right) (h \, dr) \cdot r = \mu \frac{r^2 \omega}{h} dr\).
Step 7: To obtain the total torque \( T \), integrate \( dT \) from 0 to \( r_0 \):
\[ T = \int_0^{r_0} \mu \frac{r^2 \omega}{h} dr = \mu \frac{\omega}{h} \int_0^{r_0} r^2 dr = \mu \frac{\omega}{h} \left( \frac{r_0^3}{3} \right)\].
Therefore, the total viscous torque on the disk is \( T = \frac{\mu \omega r_0^3}{3h} \).
This expression represents the viscous torque on the disk due to the fluid's viscosity.