A mild steel wire of length 1.0 m and cross-sectional are `0.5 xx 10^(-20)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=2 xx 10^(11)Nm^(-2`.

Explanation

Length of the steel wire = 1.0 m
Area of cross-section, A = 0.50 `xx 10^(-2)"cm"^(2)=0.50xx10^(-6)"m"^(2)`
A mass 100 g is suspended from its midpoint.
m = 100 g = 0.1 kg
Hence, the wire dips, as shown in the given figure.

Original length = XZ
Depression = l
The length after mass m, is attached to the wire = XO + OZ
Increase in the length of the wire:
`Deltal=(XO+OZ)-XZ`
Where, `XO=OZ=[(0.5)^(2)+l^(2)]^((1)/(2))`
`thereforeDeltal =2[(0.5)^(2)+(l)^(2)]^((1)/(2))-1.0`
`" "2xx0.5[1+((l)/(0.5))^(2)]^((1)/(2))-1.0`
Expanding and neglecting higher terms, we get:
`Deltal=(l^(2))/(0.5)`
Strain = `("Increase in length")/("Original length")`
Let T be the tension in the wire.
`thereforemg=2Tcostheta`
Using the figure, it can be written as:
`costheta=(l)/(((0.5)^(2)+l^(2))^((1)/(2)))`
`" "=(l)/((0.5)(1+((l)/(0.5))^(2))^((1)/(2)))`
Expanding the expression and eliminating the higher terms:
`costheta=(l)/((0.5)(1+
(l^(2))/(2(0.5)^(2))))`
`(1+(l^(2))/(0.5))cong1` for small `l`
`therefore costheta=(l)/(0.5)`
`thereforeT=(mg)/(2((l)/(0.5)))=(mgxx0.5)/(2l)=(mg)/(4l)`
Stress = `("Tension")/("Area")=(mg)/(4lxxA)`
Young's modulus = `=("Stress")/("Strain")`
`Y=(mgxx0.5)/(4lxxAxxl^(2))`
`l=3sqrt((mgxx0.5)/(4YA))`
Young’s modulus of steel, Y= `2xx10^(11)` Pa
`thereforel=sqrt((0.1xx9.8xx0.5)/(4xx2xx10^(11)xx0.50xx10^(-6)))`
= 0.0106 m
Hence, the depression at the midpoint is 0.0106 m.