A small ring Z carries vertical load A of mass 27.3 kg by two strings ZX and ZY. String ZY carries a load B at its free end through a frictionless pulley. Find the value of B (rounded to nearest integer), if the string is stable at the configuration shown in fig.

Correct option is B

​$$\textbf{Let}: \\T_1 = \text{tension in ZX (angle } 45^\circ) \\T_2 = \text{tension in ZY (angle } 60^\circ) \\W = mg = 27.3 \times 9.81 = 267.513\, \text{N} \\\\\textbf{Vertical Force Balance} \\T_1 \sin(45^\circ) + T_2 \sin(60^\circ) = W \\\\\textbf{Horizontal Force Balance at Z} \\T_1 \cos(45^\circ) = T_2 \cos(60^\circ) \\\\\textbf{Step 3: Use Trig Values} \\\sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071 \\\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660, \quad \cos(60^\circ) = 0.5 \\\\\text{From Equation (2):} \\T_1 \cdot 0.7071 = T_2 \cdot 0.5 \Rightarrow T_1 = \frac{0.5}{0.7071} T_2 \Rightarrow T_1 \approx 0.7071 T_2 \tag{3} \\\\\textbf{Step 4: Plug into Equation (1)} \\0.7071 T_1 + 0.8660 T_2 = 267.513 \\0.7071 (0.7071 T_2) + 0.8660 T_2 = 267.513 \\0.5 T_2 + 0.8660 T_2 = 267.513 \\1.3660 T_2 = 267.513 \Rightarrow T_2 = \frac{267.513}{1.3660} \approx 195.85\, \text{N} \\\\\textbf{Step 5: Tension in ZY equals Weight of B} \\T_2 = W_B = m_B \cdot g \Rightarrow m_B = \frac{T_2}{9.81} = \frac{195.85}{9.81} \approx \boxed{20\, \text{kg}} \quad \text{(rounded to nearest integer)}$$​​