D

Step 1: Understanding the torque in a concentric cylinder system
In a system like this, the torque (D) required to maintain the angular velocity is related to the viscosity of the fluid between the cylinders, the dimensions of the cylinders, and the angular velocity.

Step 2: Calculating the shear stress
The shear stress (A) in a fluid is defined as:
A =  . \tau = \mu \frac{du}{dy}
where \u is the velocity gradient and \y is the distance between the cylinder surfaces.

This cylindrical setup can be expressed as:
- Inner cylinder radius, R1 = 0.040 ft
- Outer cylinder radius, R2 = 0.42 ft

Step 3: Finding the velocity gradient
The velocity gradient (du/dy) can be simplified in this case by noting that the fluid flows under a shear due to rotation. The angular velocity, \omega, is given as:
\omega = \frac{60 \text{ rpm}}{60} \times 2\pi = 2\pi \, ext{ radians/second}.

The corresponding radius for the inner cylinder:
R1 = 0.040 ft.

The velocity of the inner cylinder (v) is given by:
v = R1 * \omega = 0.040 \times 2\pi = 0.2513 \, \text{ft/s}.

The distance (dy) between the annulus is:
dy = R2 - R1 = 0.42 - 0.040 = 0.38 ft.
Hence, the velocity gradient is:
\frac{du}{dy} = \frac{0.2513}{0.38}.

Step 4: Relating torque, viscosity, and dimensions
The torque (T) can be related to the shear stress and the dimensions of the cylinders:
\[ T = 2 \pi L (R_1^2 \tau_1 - R_2^2 \tau_2) \]
where L = 1.00 ft (length of the cylinder).

Since the outer cylinder is stationary, \tau_2 = 0, and we have:
\[ T = 2 \pi L R_1^2 \tau \]\[ \tau = \mu \frac{du}{dy} \] thus:
\[ T = 2 \pi L R_1^2 \mu \frac{du}{dy} \].

Step 5: Substituting values and solving for viscosity
Rearranging for viscosity gives:
\[ \mu = \frac{T}{2 \pi L R_1^2 \frac{du}{dy}} \].
Substitute known values:
- T = 0.650 lb.ft
- L = 1.00 ft
- R1 = 0.040 ft
- \frac{du}{dy} = \frac{0.2513}{0.38}\approx 0.660f/s.
\[ \mu = \frac{0.650}{2 \pi (1.00)(0.040^2)(0.660)} \approx 0.0687 \, ext{ lb.ft/s} imes ext{s}. ext{ ft}^{2} \].
Finally, this can be converted to conventional units of viscosity, yielding the viscosity of the liquid in the gap as approximately equal to 0.0687 lb-ft/s.
Therefore the answer is D.