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Answer 1

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Explanation:

The situation is as shown in the figure.

Let T be tension in each wire.AS the bar is supported symmetrically by the three wires, therefore extension in each wire is same

as Y= F/A / ΔL/L

If D is the diameter of the wire

Then Y= F/π(D/2)² / ΔL/L

= 4FL/πD²ΔL

As per the conditions of the problem, F (tension), length L, and extension ΔL is same for each wire

Y∝ 1/D ². or D∝ 1/Y

∴

Dcopper / Diron

= √Yiron / Ycopper

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