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0 votes
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in Physics by (63.3k points)

On suspending a weight Mg, the length l of elastic wire and area of cross-section A its length becomes double the initial length. The instantaneous stress action on the wire is

(a)   Mg/A

(b)  Mg/2A

(c)  2Mg/A

(d)  4Mg/A

1 Answer

+1 vote
by (67.9k points)
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Best answer

Correct option (c) 2Mg/A

Explanation:

When the length of wire becomes double, its area of cross section will become half because volume of wire is constant (V = AL).

So the instantaneous stress = Force/Area = Mg/A/2 = 2Mg/A.

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