$A$ mass of $2 \, kg$ is suspended with a thread $AB$ as shown in the figure. $A$ thread $CD$ of the same type is attached to the other end of the $2 \, kg$ mass. The lower thread is pulled gradually,harder and harder in the downward direction,so as to apply force on $AB$. Which of the threads will break and why?

(A) Let the force applied to the lower thread $CD$ be $F$.
For the lower thread $CD$,the tension $T_{CD} = F$.
For the upper thread $AB$,the tension $T_{AB}$ must support both the applied force $F$ and the weight of the $2 \, kg$ mass $(mg)$.
Thus,$T_{AB} = F + mg$.
Since $T_{AB} = F + mg$ and $T_{CD} = F$,it is clear that $T_{AB} > T_{CD}$ for any applied force $F$.
Therefore,the tension in the upper thread $AB$ will reach its breaking point before the tension in the lower thread $CD$. Hence,thread $AB$ will break first.