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Q.

A 5 m long aluminum wire Y=7×1010N/m2 of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire Y=12×1010N/m2 of the same length under the same weight, the diameter should now be (in mm).

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a

1.75

b

1.5

c

2.3

d

5.0

answer is C.

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Detailed Solution

l=FLπr2Y⇒r2∝1Y (F,L and l are constant )r2r1=Y1Y21/2=7×101012×10101/2⇒r2=1.5×7121/2=1.145mm⇒diameter=2.29mm
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