Conservation of Energy

Apply conservation of energy to find the speed for a given $\theta$ :

$\frac{1}{2} m v_0^2 + m g \, l \,cos\theta = \frac{1}{2} m v^2 \\ m v^2 = m v_0^2 + 2 m g \, l \,cos\theta$

Total centripetal force:

$\frac{m v^2}{l} = \frac{m v_0^2}{l} + 2 m g \,cos\theta$

Represent total centripetal force as a superposition of the string tension and gravity:

$\frac{m v_0^2}{l} + 2 m g \,cos\theta = T - mg \, cos\theta \\ T = \frac{m v_0^2}{l} + 3 m g \,cos\theta$

To find the breaking angle, equate the tension expression to the maximum tension:

$T_{max} = 2mg = \frac{m v_0^2}{l} + 3 m g \,cos\theta_b$

Cancelling out the mass and plugging in numbers:

$T_{max} = 2(10) = \frac{ 5^2}{2} + 3(10) \,cos\theta_b \\ \boxed{cos\theta_b = \frac{1}{4}}$

@Md Zuhair , @Steven Chase , @Hemamalinivenkatasesha Nanduri , how did u solve this problem? Could u pls psot ur solutions for this problem ?