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

MHT CETMHT CET 2006Gravitation

Solution:

Let the material of length 1 and side s If a material maintains constant volume during stretching $V = 1 \times s ^{2}$
Differentiate wrt dl $d V = s ^{2} \cdot d l + 1 . 2 s \cdot d s$
dl.s $=$ 2l.ds $\frac{d s}{d l}=-\frac{1}{2} \frac{s}{1}$
$\eta=-\frac{\frac{d s}{s}}{\frac{d l}{1}}=\frac{1}{2}$