Mechanical Properties of Matter
When a rubber cord is stretched, the change in volume with respect to change in its linear dimensions is negligible. The Poisson's ratio for rubber is
The relationship between the impressibility $$\beta $$ and the elastic constants $$E$$ and $$\mu$$. Show that Poisson's ratio $$\mu$$ cannot exceed $$1/2$$.
Let us consider a cube under an equal compressive stress $$\sigma$$, acting on all its faces.
Then,
Volume strain $$=-\dfrac{\Delta V}{V}=\dfrac{\sigma}{k'}$$...............$$1$$
where $$k$$ is the bulk modulus of elasticity.
So
$$\dfrac{\sigma}{k}=\dfrac{3\sigma }{E}(1-2\mu)$$
or,
$$E=3k(1-2\mu)=\dfrac{3}{\beta}(1-2\mu)\left(as\ k=\dfrac{1}{\beta}\right)$$
$$\mu \le \dfrac{1}{2}$$ if $$E$$ and $$\beta$$ are both to remain positive.
When a rubber cord is stretched, the change in volume with respect to change in its linear dimensions is negligible. The Poisson's ratio for rubber is
The dimensions of Poisson's ratio is?
The Poisson's ratio of a material is 0.4. If a force is applied to a wire of this material, there is a decrease of cross-sectional area by 2%. The percentage increase in its length is :
The Poissons ratio of a material is $$0.4$$. If a force is applied to a wire of this material, there is a decrease of the cross-sectional area by $$2$$%. The percentage increase in its length is
Minimum and maximum values of Poisson’s ratio for a metal lies between
Ratio of transverse to axial strain is
A $$3 cm$$ long copper wire is stretched to increase its length by $$0.3cm.$$ If poisson's ratio for copper is $$0.26$$, the lateral strain in the wire is
A wire is subjected to a longitudinal strain of $$0.05.$$ If its material has a Poisson's ratio $$0.25$$, the lateral strain experienced by it is