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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