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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