Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in fig. The unloaded length of steel wire is `1.5 m ` and that of brass wire is 1.0m. Compute the elongations of the steel and the brass wires.

Explanation

Elongation of the steel wire = `1.49xx10^(-4)"m"`
Elongation of the brass wire = `1.3xx10^(-4)"m"`
Diameter of the wires, d = 0.25 m
Hence, the radius of the wires, `r=(d)/(2)`= 0.125 cm
Length of the steel wire, `L_(1)` = 1.5 m
Length of the brass wire, `L_(2)` = 1.0 m
Total force exerted on the steel wire:
`F_(1)=(4+6)g=10xx9.8`= 98 N
Young's modulus for steel:
`Y_(1)=(((F_(1))/(A_(1))))/(((DeltaL_(1))/(L_(1))))`
Where,
`DeltaL_(1)` = Change in the length of the steel wire
`A_(1)` = Area of cross-section of the steel wire = `pir_(1)^(2)`
Young’s modulus of steel, `Y_(1)=2.0xx10^(11)` pa
`thereforeDeltaL_(1)=(F_(1)xxL_(1))/(A_(1)xxY_(1))=(F_(1)xxL_(1))/(pir_(1)^(2)xxY_(1))`
`" "=(98xx1.5)/(pi(0.125xx10^(-2))^(2)xx2xx10^(11))=1.49xx10^(-4)"m"`
Total force on the brass wire: `F_(2)=6xx9.8=58.8"N"`
Young’s modulus for brass:
`Y_(2)=(((F_(2))/(A_(2))))/(((DeltaL_(2))/(L_(2))))`
Where,
`DeltaL_(2)` = Change in length
`A_(2)` = Area of cross-section of the bass wire
`thereforeDeltaL_(2)=(F_(2)xxL_(2))/(A_(2)xxY_(2))=(F_(2)xxL_(2))/(pir_(1)^(2)xxY_(2))`
`" "(58.8xx1.0)/(pixx(0.125xx10^(-2))^(2)xx(0.91xx10^(11)))=1.3xx10^(-4)"m"`
Elongation of the steel wire = 1.49`xx``10^(-4)`m
Elongation of the brass wire = 1.3`xx``10^(-4)`m