To solve this problem, we resolve the forces acting on the ring R in both horizontal and vertical directions.

1. Resolve horizontally: The horizontal component of the tension T in the string is balanced by the horizontal component of the 2.5 N force.

\(T \cos 45^{\circ} + T \cos 45^{\circ} = 2.5 \cos 45^{\circ}\)

\(2T \cos 45^{\circ} = 2.5 \cos 45^{\circ}\)

\(T = 1.25 \text{ N}\)

2. Resolve vertically: The vertical component of the tension balances the weight of the ring.

\(2.5 \sin 45^{\circ} = mg\)

\(m = \frac{2.5 \sin 45^{\circ}}{g}\)

Using \(g = 9.8 \text{ m/s}^2\),

\(m = \frac{2.5 \times \frac{\sqrt{2}}{2}}{9.8} \approx 0.177 \text{ kg}\)

Thus, the tension in the string is 1.25 N and the mass of the ring is 0.177 kg.

To solve the problem, we resolve the forces acting on the ring R horizontally and vertically.

1. Resolving horizontally:

\(T \cos \theta = 11.2\)

2. Resolving vertically:

\(T \sin \theta = 0.16g\)

where \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\).

Substituting \(g = 9.8\), we have:

\(T \sin \theta = 0.16 \times 9.8 = 1.568\)

Now, using the given mark-scheme:

\(T \cos \theta = 4.8\)

\(T \sin \theta = 6.4\)

We can find \(T\) and \(\theta\) using these equations:

\(T^2 = 4.8^2 + 6.4^2\)

\(T^2 = 23.04 + 40.96 = 64\)

\(T = \sqrt{64} = 8\)

To find \(\theta\):

\(\tan \theta = \frac{6.4}{4.8}\)

\(\theta = \arctan \left( \frac{6.4}{4.8} \right) \approx 53.1^{\circ}\)

To solve the problem, we resolve the forces acting on the ring R horizontally and vertically.

Horizontally, the forces are balanced as:

\(T \cos \theta + T \sin \theta = 15.5\)

Vertically, the forces are balanced as:

\(-T \cos \theta + T \sin \theta = 8.5\)

Solving these two equations simultaneously, we find:

\(T \sin \theta = 12\)

\(T \cos \theta = 3.5\)

To find \(\theta\), we use the identity:

\(\tan \theta = \frac{T \sin \theta}{T \cos \theta} = \frac{12}{3.5}\)

\(\tan \theta = \frac{12}{3.5} = 3.4286\)

Thus, \(\theta = \arctan(3.4286) \approx 73.7^\circ\) or 1.29 radians.

To solve this problem, we need to resolve the forces acting on the ring R both vertically and horizontally.

(i) Resolving vertically:

The vertical components of the tension must balance the weight of the ring. Let T be the tension in the string.

From the diagram, the vertical components are:

\(T \sin 50^\circ\) and \(T \sin 20^\circ\)

These must sum to the weight of the ring:

\(T \sin 50^\circ = T \sin 20^\circ + 0.8g\)

Solving for T using \(g = 9.8 \text{ m/s}^2\):

\(T \sin 50^\circ - T \sin 20^\circ = 0.8 \times 9.8\)

\(T(\sin 50^\circ - \sin 20^\circ) = 7.84\)

\(T = \frac{7.84}{\sin 50^\circ - \sin 20^\circ}\)

Calculating gives \(T \approx 18.9 \text{ N}\).

(ii) Resolving horizontally:

The horizontal components of the tension must balance the horizontal force X.

\(X = T \cos 50^\circ + T \cos 20^\circ\)

Substitute \(T = 18.9 \text{ N}\):

\(X = 18.9 \cos 50^\circ + 18.9 \cos 20^\circ\)

Calculating gives \(X \approx 29.9 \text{ N}\).