(i) To find the tension in the string, consider the forces acting on the pulley. The resultant force exerted by the string is \(3\sqrt{3} \text{ N}\). Using the cosine of the angle \(30^\circ\), we have:

\(2T \cos 30^\circ = 3\sqrt{3}\)

\(2T \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\)

\(T \sqrt{3} = 3\sqrt{3}\)

\(T = 3 \text{ N}\)

(ii) For particle Q in limiting equilibrium, resolve forces parallel and perpendicular to AC. The tension \(T = 3 \text{ N}\), and the coefficient of friction \(\mu = 0.75\).

Resolve forces parallel to AC:

\(T = F + mg \sin 30^\circ\)

Resolve forces perpendicular to AC:

\(R = mg \cos 30^\circ\)

Using \(F = \mu R\):

\(3 = 0.75 (10 \cos 30^\circ)m + 10m \sin 30^\circ\)

\(3 = 0.75 \times \frac{\sqrt{3}}{2} \times 10m + 10m \times \frac{1}{2}\)

\(3 = 3.75m \sqrt{3} + 5m\)

Solving for \(m\):

\(m = 0.261 \text{ kg}\)

(i) The tension \(T\) in the string is given by the horizontal component of the force: \(T = 4\sqrt{2} \cos 45^\circ = 4\) N. The mass of Q is found using \(T = mg\), so \(4 = 0.4g\), giving the mass of Q as 0.4 kg.

(ii) For P on the point of slipping, the frictional force \(F\) equals the tension \(T\). Using \(F = \mu R\) and \(R = m_P g\), we have \(4 = 0.8 \times m_P \times 10\), giving \(m_P = 0.5\) kg.

(iii) When a particle of mass 0.1 kg is attached to Q, the system starts to move. Applying Newton's second law to P and Q, we have: \(T - 0.8 \times 0.5g = 0.5a\) and \(0.5g - T = 0.5a\). Solving these equations gives the tension \(T = 4.5\) N.

(a) Resolve forces for the 5 kg particle: The normal reaction is given by the weight of the 5 kg particle, so \(R = 5g\). The frictional force \(F\) is the difference between the weight of the 6 kg particle and the 4 kg particle, so \(F = 6g - 4g = 2g\). The coefficient of friction \(\mu\) is given by \(\mu = \frac{F}{R} = \frac{2g}{5g} = 0.4\).

(b) Apply Newton's second law to the 4 kg and 8 kg particles: For the 4 kg particle, \(T_1 - 4g = 4a\). For the 8 kg particle, \(8g - T_2 = 8a\). The tension difference is \(T_2 - T_1 = F = 0.4 \times 5g\). Adding the equations gives \(8g - 4g - 2g = 17a\), leading to \(a = 1.18 \text{ m/s}^2\). Solving for tensions, \(T_1 = 44.7 \text{ N}\) and \(T_2 = 70.6 \text{ N}\).

(c) For the 8 kg particle, \(T - 4g = 4a\) and \(-T - F = 5a\), or \(-4g - 2g = 9a\). Solving gives \(a = \frac{60}{9}\). Use \(v^2 = u^2 + 2as\) to find \(v\) when the 8 kg particle reaches the ground: \(v^2 = 2 \times \frac{20}{17} \times 0.5 = \frac{20}{17}\), leading to \(v = 1.0846 \ldots\). Use \(v = u + at\) to find \(t\): \(0 = \frac{20}{17} - \frac{60}{9}t\), giving \(t = 0.163 \text{ s}\).

(i) Using Newton's second law for both particles:

For particle P: \(7g - T = 7a\)

For particle Q: \(T - 3g = 3a\)

Adding these equations gives: \(7g - 3g = 10a\)

Solving for acceleration, \(a = \frac{4g}{10} = 4 \text{ m/s}^2\)

Using the equation of motion for speed of P: \(v^2 = u^2 + 2as\)

Where initial speed \(u = 0\), \(a = 4 \text{ m/s}^2\), and \(s = 0.4 \text{ m}\):

\(v^2 = 0 + 2 \times 4 \times 0.4\)

\(v^2 = 3.2\)

\(v = \sqrt{3.2} = 1.79 \text{ m/s}\)

(ii) To determine if Q comes to rest, use:

\(0 = 3.2 + 2 \times (-g) \times s\)

Solving for \(s\):

\(s = 0.16 \text{ m}\)

Since \(0.16 + 0.4 = 0.56 \text{ m}\), particle Q does not come to rest before it reaches the pulley.

(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 \(T\):

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