Poisson's Ratio

Poisson's ratio is “the ratio of transverse contraction strain to longitudinal extension strain in the direction of the _____”?

What happens when Poisson ratio is zero?
 

Why Poisson ratio has no unit?
 

What is meant Poisson's ratio What is its unit and significance?
 

Lateral strain produced in a wire depends on:

For a given material, the Young’s modulus is $2.4$ times that of the modulus of rigidity. Its Poisson’s ratio is:

A load $1 kg$ produces a certain extension in the wire of length $3 \mathrm{m}$ and radius $5\times {10}^{-4} \mathrm{m}$. How much will be the lateral strain produced in the wire?$\left(Y=7.48\times {10}^{10 }\mathrm{N}/{\mathrm{m}}^2 and \sigma =0.291\right)$ 

Explain Poisson's ratio. Discuss its limiting values.
 

A material has Poisson’s ratio $0.2$. If a uniform rod made out of it suffers longitudinal strain $4.0\times {10}^{–3}$, then calculate the percentage change in its volume.

If the volume of the wire remains constant when it is subjected to tensile stress, the value of Poisson’s ratio of the material of the wire is

The ratio of lateral strain to the longitudinal strain is called

A copper wire of cross-section A is under tension T. Find the fractional decrease in the cross-sectional area (Young's modulus is $Y$ and Poisson's ratio is $\sigma$ )

Given the following values for an elastic material Young's modulus $=\mathrm{ }7\mathrm{ }\times \mathrm{ }{10}^{10}\mathrm{ }\mathrm{N}{\mathrm{m}}^{-2}$ and Bulk modulus $=\mathrm{ }11\mathrm{ }\times \mathrm{ }{10}^{10}\mathrm{ }\mathrm{N}{\mathrm{m}}^{-2}$. The Poisson's ratio of the material is -

Given the following values for an elastic material Young's $\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{l}\mathrm{u}\mathrm{s}=7\mathrm{ }\times \mathrm{ }{10}^{10}\mathrm{ }{\mathrm{N}\mathrm{m}}^{–2}$ and Bulk $\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{l}\mathrm{u}\mathrm{s}=11\mathrm{ }\times \mathrm{ }{10}^{10}\mathrm{ }{\mathrm{N}\mathrm{m}}^{–2}$. The Poisson's ratio of the material is -

If there is no change in the volume of wire on stretching, then Poisson's ratio for the material of wire is -

Given the following values for an elastic material: Young's $\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{l}\mathrm{u}\mathrm{s}=7\times {10}^{10}\mathrm{ }{\mathrm{N}\mathrm{m}}^{-2}$ and $\mathrm{B}\mathrm{u}\mathrm{l}\mathrm{k}\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{l}\mathrm{u}\mathrm{s}=11\times {10}^{10}\mathrm{ }{\mathrm{N}\mathrm{m}}^{-2}$. The Poisson's ratio of the material is -

The increase in length on stretching a wire is 0.05%. If its Poisson's ratio is 0.4, the diameter is reduced by

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 wire is

The Poisson's ratio cannot have the value

When a wire of length $10 \mathrm{m}$ is subjected to a force of $100 \mathrm{N}$ along its length, the lateral strain produced is $0.01\times {10}^{-3} \mathrm{m}$. The Poisson's ratio was found to be $0.4$. If the area of cross-section of wire is $0.025\mathrm{ }{\mathrm{m}}^2$, its Young's modulus is