1.

There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of lhe material of the wire is :

Answer»

`+1/2`
`-1/2`
`+1/4`
`-1/4`

Solution :Volume of cylindrical wire, V = `(pix^(2)L)/4`, where x is diameter of wire.
DIFFERENTIATING both sides,
`(DV)/(dx)=pi/4[2xL+x^(2).(DL)/(dx)]`.
Also volume remains CONSTANT `therefore (dV)/(dx)=0`
`therefore2xL+x^(2)(dL)/(dx)=0rArr2xL=-x^(2)(dL)/(dx)`
`rArr((dx)/x)/((dL)/L)=-1/2`
`rArr` Poisson.s ratio `=-1/2`
So the correct choice is (b).


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