As shown in the figure, a rod PQ of length 20.0 cm and mass 200 g...

When the switch $S$ is open, no current flows through the rod $PQ$. Therefore, the magnetic force acting on the rod is zero.

The rod is in equilibrium under the action of its weight and the tensions in the two threads.

Given that the length of the rod $PQ = 20.0 \text{ cm}$ and the lengths of the threads $OP = OQ = 20.0 \text{ cm}$, $\triangle OPQ$ is an equilateral triangle.

The angle each thread makes with the horizontal rod is $60^\circ$.

Let $T$ be the tension in each thread. Balancing the forces in the vertical direction:

$2T \sin(60^\circ) = mg$

Substituting $m = 0.2 \text{ kg}$ and $g = 9.8 \text{ m/s}^2$:

$2T \left(\dfrac{\sqrt{3}}{2}\right) = 0.2 \times 9.8$

$T \sqrt{3} = 1.96$

$T = \dfrac{1.96}{\sqrt{3}} \approx 1.13 \text{ N}$