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.

A copper wire of length 2.2 m and a steel wire of length 1.6 m, both o

1.	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.
Answer» Solution :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`