The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle $\theta$ should be

Step 1: Determine the tension in the string.

The outer blocks of mass $m$ hang freely in equilibrium, meaning the tension $T$ in the continuous string balances their weight.

$$T = mg$$

Step 2: Identify the forces acting on the middle block.

The middle block of mass $\sqrt{2}m$ experiences a downward gravitational force and two tension forces $T$ directed at an angle $\theta$ relative to the vertical.

$$W = \sqrt{2}mg$$

Step 3: Apply the condition for vertical equilibrium.

The horizontal components of the tension forces cancel by symmetry, while the sum of the vertical components balances the weight of the middle block.

$$2T\cos\theta = \sqrt{2}mg$$

Step 4: Solve for the angle $\theta$.

Substitute $T = mg$ into the equilibrium equation.

$$2mg\cos\theta = \sqrt{2}mg$$

Simplify by dividing both sides by $mg$.

$$2\cos\theta = \sqrt{2}$$ $$\cos\theta = \frac{1}{\sqrt{2}}$$ $$\theta = 45^\circ$$

Answer: $\theta = 45^\circ$