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

Refer the figure, let x be the depression at the mid point i.e., CD=x
In fig. AC=CB=l=0.5m
`m=100g=0.100kg`
`AD=BD=(l^(2)+x^(2))^(1//2)`
Increase in length, `Deltal=AD+DB-AB`
`=2AD-AB`
`=2(l^(2)+x^(2))^(1//2)-2l`
`=2l(1+(x^(2))/(l^(2)))^(1//2)-2l`
`=2l(1+(x^(2))/(2l^(2)))-2l=(x^(2))/(l)`
Strain`=(Deltal)/(2l)=(x^(2))/(2l^(2))`
If T is the tension in the wire, then 2T `costheta`
`=mg or T=(mg)/(2costheta)`

Here, `costheta=(x)/((l^(2)+x^(2))^(1//2)`
`=(x)/((1+(x^(2))/(l^(2)))^(1//2))=(x)/(l(1+(1)/(2)(x^(2))/(l^(2))))`
As x`ltlt1`, so `1gtgt (1)/(2)(x^(2))/(l^(2)) and 1+(1)/(2)(x^(2))/(l^(2))=1`
`therefore costheta=(x)/(l)`
Hence, `T=(mg)/(2(n//l))=(mgl)/(2x)`
Stress=`(T)/(A)=(mgl)/(2Ax)`
`Y=("Stress")/("Strain")=(mgl)/(2Ax)xx(2l^(2))/(x^(2))=(mgl^(3))/(Ax^(3))`
`therefore x=l((mg)/(YA))^(1//3)`
`=0.5((0.1xx10)/(2xx10^(11)xx0.5xx10^(-6)))^(1//3)`
`=1.074xx10^(-2)m=1.074m`.