(i) For block B in limiting equilibrium, the frictional force is given by:

\(\text{Frictional force} = \mu \times 0.25g\)

Since the system is in equilibrium, the tension in the string due to P and Q must balance the frictional force and the weight of P:

\(0.3g = 0.2g + \mu \times 0.25g\)

Solving for \(\mu\):

\(\mu = \frac{0.3g - 0.2g}{0.25g} = 0.4\)

(ii) When Q is removed, apply Newton's second law to P and B:

For P:

\(0.2g - T = 0.2a\)

For B:

\(T - 0.4 \times 0.25g = 0.25a\)

Solving these equations simultaneously:

From the equation for B:

\(T = 0.25a + 0.4 \times 0.25g\)

Substitute into the equation for P:

\(0.2g - (0.25a + 0.4 \times 0.25g) = 0.2a\)

Simplifying:

\(0.2g - 0.1g = 0.45a\)

\(0.1g = 0.45a\)

\(a = \frac{0.1g}{0.45} = 2.22 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 0.25 \times 2.22 + 0.1g = 1.56 \text{ N}\)