A

Given:
Mass of the disc (m) = 1.8 kg
Radius of the disc (r) = 0.2 m
Force exerted by the string (F) = 3 N
Step 1: Calculate the angular acceleration (\alpha):
Using Newton's second law for rotational motion:
\[ \tau = I \alpha \]
Where \( \tau \) is the torque, \( I \) is the moment of inertia.
The moment of inertia for a disc about its center is given by:
\[ I = \frac{1}{2} m r^2 \]
Substituting the values:
\[ I = \frac{1}{2} \times 1.8 \times (0.2)^2 = 0.036 \, \text{kg m}^2 \]

Now, the torque (\( \tau \)) caused by the force is:
\[ \tau = r \times F = 0.2 \times 3 = 0.6 \, \text{N m} \]

Substituting \( \tau \) and \( I \) into the torque equation gives us:
\[ 0.6 = 0.036 \alpha \]
Solving for \( \alpha \):
\[ \alpha = \frac{0.6}{0.036} \approx 16.67 \, \text{rad/s}^2 \]

Step 2: Calculate the linear acceleration (a):
Using the relationship between linear acceleration and angular acceleration:
\[ a = r \alpha \]
Substituting the values:
\[ a = 0.2 \times 16.67 \approx 3.33 \, \text{m/s}^2 \]
Step 3: Direction of acceleration:
The force applied is in the northerly direction, hence the acceleration of the center of mass (a) is also directed north.
Conclusion:
Thus, the linear acceleration \( a \) is approximately 3.33 m/s² to the north, and \( \alpha \) is approximately 16.67 rad/s².
Yes, it is valid that \( a = r \alpha \).