__Question 9.10__:

__Solution__:The relation for Young’s modulus is given as:

Y = Stress / Strain

= (F/A) / Strain = (4F/πd

^{2}) / Strain ....(i)

Where,

*F*= Tension force

*A*= Area of cross-section

*d*= Diameter of the wire

It can be inferred from equation (

*i*) that Y ∝ (1/d

^{2})

Young’s modulus for iron,

*Y*

_{1}= 190 × 10

^{9}Pa

Diameter of the iron wire =

*d*

_{1}

Young’s modulus for copper,

*Y*

_{2 }= 120 × 10

^{9}Pa

Diameter of the copper wire =

*d*

_{2}

Therefore, the ratio of their diameters is given as:

d

_{2}/ d

_{1}= (Y

_{1}/ Y

_{2})

^{1/2}

= (190 X 10

^{9}/ 120 X 10

^{9})

^{1/2}

= 1.25 : 1

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