Calculate the clongations of (a) a copper wire of 1.4 mm diameter and (b) an aluminium wire of 1 mm iameter as shown in figure, if Young's modulus for copper and aluminium are `11 xx 10^(10)` N/`m^(3)`' and `7 xx 10^(10) N//m^(2)` ? espectively.

Explanation

By definition of Young.s modulus `DeltaL = (FL)/(YA)`
(A) As tension in copper wire will be due to load of 7 kg.
i.e., `T_(Cu) = M_(1)g = 7xx gN`
So `(DeltaL)_(Cu) = (7xx 9.8xx0.5) /((22//7)(0.7xx 10^(-3)) xx 11xx10^(10))`
i.e., `(DeltaL)_(cu) = (9.8xx5)/(22xx11)xx 10^(-3) = 0.2xx10^(-3)`
However, the tension in the aluminimum wiere will be
`T_(AI)=M_(2)g+T_(Cu) = (M_(2)+M_(1))g`
So, `(DeltaL)_(AI) = ((4+7)xx9.8xx1)/((22//7)(0.5xx10^(-3))^(2) xx 11 xx 10^(10))`