A mild steel wire of length `1.0m` and cross sectional area `5.0xx10^(-2)cm^(2)` is stretched, within its elastic limit horizontally between two pillars. A mass of `100g` is suspended form the midpont of the wire. Calculate the depression at the midpoint `(Y_("steel) = 200GPa)`

Explanation

`x = l [(Mg)/(YA)]^(1//3)`, Here , `2l = 1m, l = 0.5m`,
`Mass = 100g = 0.1kg`
`Y = 200Gpa = 200xx10^(9)N//m^(2) = 2xx10^(11) N//m^(2)`
`x = 0.5 [(0.1xx10)/((2xx10^(11))(0.50xx10^(-6)))]^(1//3) = 1.074xx10^(-2)m`