A

To find the damping torque acting on the circular disk in oil, we can use the formula for viscous torque, which can be derived from the stress due to viscosity. The damping torque \( \tau \) is given by the formula: \( \tau = -b \cdot \omega \), where \( b \) is the damping coefficient and \( \omega \) is the angular velocity. Since the disk is rotating in a fluid, the damping coefficient can be expressed based on the viscosity and geometry of the disk.
Given:
- Angular velocity, \( \omega = 0.3 \; \text{rad/s} \)
- Viscosity of oil, \( \eta = 8 \times 10^{-3} \, \text{Pa.s} \)
- Radius of the disk, \( r = 75 \, \text{mm} = 0.075 \, \text{m} \)
Step 1: Calculate the damping coefficient \( b \) for a circular disk:
\( b = 2 \pi \eta r^2 \)
Substituting the values,
\( b = 2 \pi (8 \times 10^{-3}) (0.075)^2 \)
\( b = 2 \pi (8 \times 10^{-3}) (0.005625) \)
\( b = 2 \pi (4.5 \times 10^{-5}) \)
\( b \approx 2.83 \times 10^{-4} \, \text{N.m.s} \)
Step 2: Now, we can calculate the damping torque:
\( \tau = -b \cdot \omega = - (2.83 \times 10^{-4}) (0.3) \)
\( \tau = -8.49 \times 10^{-5} \, \text{N.m} \)
Step 3: Since we are interested in the magnitude, we can say:
\( \tau = 8.49 \times 10^{-5} \, \text{N.m} \).
Therefore, the damping torque is approximately 8.49 × 10^-5 N.m.
Thus, the correct answer is A.