Tag: poisson's ratio

Poisson's ratio is a material property. Therefore it doesn't change with dimensions.  

The ratio of the lateral strain to longitudinal strain is called Poisson's ratio.
Hence, option (a) is an incorrect statement.
Its value depends only on the nature of the material. 
Hence, option (b) is an incorrect statement.
It is the ratio of two like physical quantities.
Therefore, it is unitless and dimensionless quantity.
Hence option (c) is a correct statement
The practical value of Poisson's ratio lies between $0$ and $0.5$
hence option (d) is an incorrect statement.

Let $L$ be the length, $r$ be the radius of the wire.
Volume of the wire is
$V=\pi { r }^{ 2 }L$
Differentiating both sides, we get
$\Delta V=\pi (2r\Delta r)L+\pi { r }^{ 2 }\Delta L\quad $
As the volume of the wire remains unchanged when it gets stretched, so $\Delta V=0$. Hence
$0=2\pi rL\Delta r+\pi { r }^{ 2 }\Delta L$
$\therefore \cfrac { \Delta r/r }{ \Delta L/L } =-\cfrac { 1 }{ 2 } $
$Poisson's\quad ratio=\cfrac { Lateral\quad strain }{ Longitudinal\quad strain } =-\cfrac { \Delta r/r }{ \Delta L/L } =\cfrac { 1 }{ 2 } =0.5\quad $

As $\sigma =-\cfrac { \Delta R/R }{ \Delta l/l } $
$\therefore \cfrac { \Delta R }{ R } =-\sigma \cfrac { \Delta l }{ l } =-0.2\times 4.0\times { 10 }^{ -3 }=-0.8\times { 10 }^{ -3 }\quad $
$V=\pi { R }^{ 2 }l$
$\therefore \cfrac { \Delta V }{ V } \times 100=\left( 2\cfrac { \Delta R }{ R } +\cfrac { \Delta l }{ l }  \right) \times 100=\left[ 2\times \left( -0.8\times { 10 }^{ -3 } \right) +4.0\times { 10 }^{ -3 } \right] \times 100=2.4\times { 10 }^{ -3 }\times 100=0.24$%

Poisson's ratio $\sigma =\cfrac { \Delta D/D }{ \Delta l/l } =\cfrac { \Delta D }{ D } .\cfrac { l }{ \Delta l } $
$\therefore \Delta D=\cfrac { \sigma D\Delta l }{ l } =\cfrac { 0.2\times 6\times { 10 }^{ -3 }\times 3.32\times { 10 }^{ -5 }m }{ 4.5 } =8.8\times { 10 }^{ -9 }m\quad $

As  $Y=2\eta \left( 1+\sigma  \right) $
where the symbols have their usual meanings
Given: $Y=2.4\eta $
$\therefore 2.4\eta =2\eta \left( 1+\sigma  \right) $
$1.2=1+\sigma \quad or\quad \sigma 1.2-1=0.2$

Here, $=4.5m,D=6mm=6\times { 10 }^{ -3 }m,F=100N,Y=4.8\times { 10 }^{ 11 }N\quad { m }^{ -2 },\sigma =0.2$
As $Y=\cfrac { F }{ A } \cfrac { l }{ \Delta l } $
$\therefore \Delta l=\cfrac { F }{ A } \cfrac { l }{ Y } =\cfrac { 100\times 4.5 }{ 3.14\times { \left( 3\times { 10 }^{ -3 } \right)  }^{ 2 }\times 4.8\times { 10 }^{ 11 } } =3.32\times { 10 }^{ -5 }m$

Given,

Increase in length, $\dfrac{\Delta 1}{1}  = 0.025 $%

${\dfrac{\Delta d}{d}} = ?$

Poisson’s ratio is $0.4.$

${Poisson's \ ratio} = \dfrac {\dfrac{\Delta d}{d}} {\dfrac{\Delta l}{l}}\\$

$0.4=\dfrac {\dfrac{\Delta d}{d}} {\dfrac{0.025}{100}}\\$

$\dfrac{\Delta d}{d} = \dfrac{0.4\times 0.025}{100}\\$

$\dfrac{\Delta d}{d} = 0.01$ %

Then the percentage decrease  $= 0.01$ %.

Option A is correct. 

Perfectly rigid bodies cannot be deformed upon application of any amount of force. Thus, strain is zero or the modulus of elasticity which is inversely proportional to strain becomes infinity

Thus option (b) is the correct option