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Question

A wire can bear a weight of $$20kg$$ before it breaks. If the wire is divided into two equal parts, then each part will support a maximum weight ...........


Solution

$$T_{initial}=20 g$$
$$T\quad \alpha \quad k\quad \alpha \quad \dfrac{YA}{L}$$
$$T \quad \alpha \dfrac{1}{L}$$
$$T_{initial} \quad \alpha \dfrac{1}{L}$$
$$T_{final} \alpha \dfrac{1}{L/2}$$
$$\dfrac{T_{final}}{T_{initial}}=2T_{initial}$$
$$(T_{final})_{max}=40 g$$
$$Maximum  \quad weight=40 kg$$

Physics

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