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A mild steel wire of length 1.0 m and cr...

A mild steel wire of length 1.0 m and cross-sectional are `0.5 xx 10^(-2)cm^(2)` is streached, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid point of the wire, calculate the depression at the mid point.
`g = 10ms^(-2), Y=20 xx 10^(11)Nm^(-2`.

Text Solution

Verified by Experts

`x=l[(Mg)/(YA)]^(1//3)` Here, `m 2l=1m, l=0.5m`,
`A=0.50xx10^(-6) m^(2) , , M=0.1 kg`,
`Y=2xx10^(11) N//m^(2) and g=10m//s^(2)`
`therefore` Depression `x=1.074xx10^(-2)m`
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