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 x Young.s modulus. Therefore
`W//A = Y_C xx (DeltaL_C // L_C) = Y_S xx (Delta L_S// L_S)`
`Delta L_C // Delta L_S = (Y_S // Y_C) xx (L_C // L_S)`
` = (2.0 xx 10^11 // 1.1 xx 10^11 ) xx (2.2 // 1.6) = 2.5` ....(1)
The total elongation is given to be
`Delta L_C + Delta L_S = 7.0 xx 10^(-4) m ` .... (2)
Solving the above equations (1) & (2).
`Delta L_C = 5.0 xx 10^(-4)m " and " Delta L_S = 2.0 xx 10^(-4) m`
Therefore `W = (A xx Y_C xx Delta L_C// L_C`
` = pi (1.5 xx 10^(-3))^2 xx [(5.0 xx 10^(-4) xx 1.1 xx 10^11)// 2.2]`
` = 1.8 xx 10^2 N`