As the tension on the wires is the same, the extension of each wire will also be the same. Now, as the length of the wires is the same, the strain on them will also be equal.

Now, we know :

Y = \( \dfrac{Stress}{Strain}\)

= \(\dfrac{F}{A},\dfrac{1}{Strain}\)

= \( \dfrac{4F}{πd^2}.\dfrac{1}{Strain}\) . . . . . . . . . . (1)

Where,

A = Area of cross-section

F = Tension force

d = Diameter of the wire

We can conclude from equation ( 1 ) that Y ∝ (\( \dfrac{1}{d^2}\))

We know that Young’s modulus for iron, Y_{1} = 190 × 10^{9} Pa

Let the diameter of the iron wire = d_{1}

Also, Young’s modulus for copper, Y_{2 }= 120 × 10^{9} Pa

let the diameter of the copper wire = d_{2}

Thus, the ratio of their diameters can be given as :

\( \dfrac{d_{1}}{d_{2}} = \sqrt{\dfrac{Y_{1}}{Y_{2}}}\)

= \( \sqrt{\dfrac{190×10^{9}}{120×10^{9}}} =1 : 25 : 1\)

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