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

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Solution

The correct option is C
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 $= \frac{f o r c e}{A r e a} = \frac{M g}{\frac{A}{2}} = \frac{2 M g}{A}$

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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

The correct option is C 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 =forceArea=MgA2=2MgA

The correct option is C
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 $= \frac{f o r c e}{A r e a} = \frac{M g}{\frac{A}{2}} = \frac{2 M g}{A}$