A heavy mass is fastened to a thin wire and whirled in a vertical...

Let $m$ be the mass, $v$ be the velocity, and $r$ be the radius of the vertical circle.

At an angle $\theta$ from the lowest point, the tension $T$ in the wire is given by the equation:

$T - mg \cos \theta = \dfrac{mv^2}{r}$

$T = \dfrac{mv^2}{r} + mg \cos \theta$

By conservation of mechanical energy, the velocity $v$ of the mass is maximum at the lowest point of the vertical circle. Also, the value of $\cos \theta$ is maximum at the lowest point since $\theta = 0$ and $\cos 0 = 1$.

Therefore, both terms $\dfrac{mv^2}{r}$ and $mg \cos \theta$ are maximum at the lowest point, making the tension $T$ maximum at this position.

Since the wire is most likely to break when the tension is maximum, it will break when the mass reaches the lowest point.