A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load , the net elongation is found to be `0.70` mm. Obtain the load applied.

Explanation

The copper and steel wires are under same tensile stress because they have the same tension (equal to the load W) and the same area of cross- section A. We have stress = strain `xx` Young.s modulus. Therefore
`W//A=Y_(C )xx(DeltaL_(C )//L_(C ))=Y_(S)xx(DeltaL_(S)//L_(S))`
`DeltaL_(c )//DeltaL_(s)=(Y_(s)//Y_(c ))xx(L_(c )//L_(s))`
`=(2.0xx10^(11)//1.1xx10^(11))xx(2.2//1.6)=2.5" ".....(1)`
The total elongation is given to be
`DeltaL_(c )+DeltaL_(s)=7.0xx10^(-4)m" "....(2)`
Solving the above equations (1) & (2).
`DeltaL_(c )=5.0xx10^(-4)m and DeltaL_(s)=2.0xx10^(-4)m`.
Therefore `W=(AxxY_(c )xxDeltaL_(c ))//L_(c )`
`=pi(1.5xx10^(-3))^(2)x[(5.0xx10^(-4)xx1.1xx10^(11))//2.2]`
`=1.8xx10^(2)N`