A 5 ~m long aluminium wire (Y=7 10¹⁰ Nm ⁻² . ) of diameter 3 ~mm supports a 40 ~kg mass. In order to have the same elongation in the copper wire (Y=12 10¹⁰ Nm ⁻² ) of the same length under the same weight, the diameter should now be (in mm

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A 5 ~m long aluminium wire (Y=7 10¹⁰ Nm ⁻² . ) of diameter 3 ~mm supports a 40 ~kg mass. In order to have the same elongation in the copper wire (Y=12 10¹⁰ Nm ⁻² ) of the same length under the same weight, the diameter should now be (in mm )

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Correct answer: C. 2.5

l= F L r² Y r² 1 Y (F, L and l are constant) r₂ r₁ = [ Y₁ Y₂ ]^ 1 / 2 = [ 7 10¹⁰ 12 10¹⁰ ]^ 1 / 2 r₂=1.5 ( 7 12 )^ 1 / 2 =1.145 ~mm diameter =2.29 ~mm .

A 5 ~m long aluminium wire (Y=7 10¹⁰ Nm ⁻² . ) of diameter 3 ~mm supports a 40 ~kg mass. In order to have the same elongation in the copper wire (Y=12 10¹⁰ Nm ⁻² ) of the same length under the same weight, the diameter should now be (in mm )

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