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

From `y=(sigma)/(epsi)`
we have stress=strain`xx`Young.s modulus.
`W//A=Y_(c)xx(DeltaL_(c)//L_(c))=Y_(s)xx(DeltaL_(s)//L_(s))`
where the subscripts c and s refer to copper and strainless steel respectively,
`DeltaL_(c)//DeltaL_(s)=(Y_(s)//Y_(c))()(L_(c)//L_(s))`
Given `L_(c)=2.2m,L_(s)=1.6m`,
`Y_(c)=1.1xx10^(11)N.m^(-2) and Y_(s)=2.0xx10^(11)N.m^(-2)`
`DeltaL_(c)//DeltaL_(s)=(2.0xx10^(11))/(1.1xx10^(11))=(2.2)/(1.6)=2.5`
`DeltaL_(c)+DeltaL_(s)=7.0xx10^(-4)m`
Solving the above equations,
`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)xx[((5.0xx10^(-4)xx1.1xx10^(11)))/(2.2)]=1.8xx10^(2)N`.