## A steel wire of length 1.5 m and diameter 1 mm is joined to an aluminum wire of identical dimensions to make a composite wire of length 3 m.

A steel wire of length 1.5 m and diameter 1 mm is joined to an aluminum wire of identical dimensions to make a composite wire of length 3 m. Find the resulting change in the length of this composite wire if an object with a mass of 5 kg is hung vertically from one of its ends. (Neglect any effects the masses of the two wires have on the changes in their lengths.). Use 2 x 1011 N/m2 and 0.7 x 1011 N/m2 for the Young’s modulus for steel and aluminum respectively.

## Answers ( )

Answer:1.805 mmExplanation:Extension in the steel wire = WL_{steel}/AE_{steel}

Extension in the aluminium wire = WL_{Al}/AE_{Al}

Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})

we have:

W = mg

W = 5 × 9.8

W = 49 N

Area A = π/4 × (0.001)²

= 7.85398 × 10 ⁻⁷ m²

Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})

Total extension = 49/ 7.85398 × 10 ⁻⁷ ( (1.5/ 200×10⁹) + 1.5/ 70×10⁹))

Total extension = 0.0018048

Total extension = 1.805 mmThus, the total extension = the resulting change in the length of this composite wire = 1.805 mm