As given in the question,

The acceleration \( 0.3 ms^{-2} \) is caused by the net horizontal force.

\( 2 – f_{1} = 2 * 0.3 \) \( f_{1} = 1.4 N \) \( f_{1} and f_{2} \) produces torques which are opposite in direction about the center of the ring.

The angular acceleration is

From torque equation,\( \tau = I\alpha = mr^{2}\alpha \) \( \tau = force * r \) \( (f_{1} – f_{2}) 0.5 = mr^{2}\alpha \)

=\( 2 * (0.5)^{2} \frac{0.3}{0.5} \) \( (1.4 + f_{2}) = 2 * (0.5)^{2}\frac{0.3}{0.5} \) \( (1.4 + f_{2}) = 2(0.3) \) \( f_{2} = 0.8N \)

Coefficient of friction,

\( \mu = \frac{0.8}{N} \) \( \Rightarrow \frac{0.8}{2} \) \( \frac{P}{10} = 0.4 \)

P = 4

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