A mass of 1 kg is suspended by a string A. A | vedclass

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(A) When the string $C$ is pulled slowly,the system remains in equilibrium at each instant.Let $T_C$ be the tension in string $C$ and $T_A$ be the ten

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The portion $A$ of the string will break

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(A) When the string $C$ is pulled slowly,the system remains in equilibrium at each instant.Let $T_C$ be the tension in string $C$ and $T_A$ be the tension in string $A$.For the mass $m = 1 \ kg$,the forces acting are the tension $T_C$ downwards,the weight $mg$ downwards,and the tension $T_A$ upwards.In equilibrium,$T_A = T_C + mg$.Since $T_A = T_C + mg$,the tension in the upper string $A$ is always greater than the tension in the lower string $C$ by an amount equal to the weight of the mass.Therefore,as the tension increases,the tension in string $A$ reaches its breaking point before the tension in string $C$ does.Thus,the upper string $A$ will break first.

$A$ mass of $1 \ kg$ is suspended by a string $A$. Another string $C$ is connected to its lower end (see figure). If the string $C$ is stretched slowly,then

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