A small ball of mass 0.30 kg is tied to a vertical rod by two light strings, each 0.50 m long.

Step 1. Geometry. The ball is level with the midpoint of PQ, $0.40\ \text{m}$ from each point. So each string makes an angle $\varphi$ with the rod, where $\cos\varphi = \dfrac{0.40\ \text{m}}{0.50\ \text{m}} = 0.8$ and $\sin\varphi = 0.6$. The radius of the circle is $r = L\sin\varphi = 0.30\ \text{m}$.

Step 2. Forces. Let $T_1$ be the upper tension and $T_2$ the lower one.

Vertically: $T_1\cos\varphi - T_2\cos\varphi = mg$.

Along the radius: $(T_1 + T_2)\sin\varphi = m\omega^2 r = m\omega^2 L\sin\varphi$, so $T_1 + T_2 = m\omega^2 L$.

Together:

Step 3. At $8.0\ \text{rad/s}$. $\omega^2 L = 32\ \text{m/s}^2$ and $g/\cos\varphi = 12.5\ \text{m/s}^2$. So $T_1 = 0.15\ \text{kg} \times 44.5\ \text{m/s}^2 = 6.68\ \text{N}$ and $T_2 = 0.15\ \text{kg} \times 19.5\ \text{m/s}^2 = 2.93\ \text{N}$.

Step 4. The lower string goes slack when $T_2 = 0$, that is when $\omega^2 = \dfrac{g}{L\cos\varphi}$. Here $L\cos\varphi = 0.40\ \text{m}$, half of PQ, so

Notice that this depends only on the distance PQ, not on the length of the strings.

Step 5. Breaking. $T_1$ is always larger than $T_2$, so the upper string breaks first. Set $T_1 = 12\ \text{N}$: $\omega^2 L = \dfrac{2T_1}{m} - \dfrac{g}{\cos\varphi} = 80\ \text{m/s}^2 - 12.5\ \text{m/s}^2 = 67.5\ \text{m/s}^2$. So $\omega^2 = 135\ \text{s}^{-2}$ and $\omega = 11.6\ \text{rad/s}$.