A mild steel wire of length 1.0 m and cross-sectional area `0.50xx10^(-2) cm^(-2)` is stretched, 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.

Explanation

`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`