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 ...........

Answer 1

$T_{initial}=20g$

$T\alpha k\alpha \frac{YA}{L}$

$T\alpha \frac{1}{L}$

$T_{initial}\alpha \frac{1}{L}$

$T_{final}\alpha \frac{1}{L/2}$

$\frac{T_{final}}{T_{initial}}=2T_{initial}$

$(T_{final}{)}_{max}=40g$

$Maximumweight=40kg$