Poisson ratio - class-XI

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 $

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

The ratio of lateral strain to the linear strain within elastic limit is known as Poisson's ratio

The correct option is (d)

Here, $E=3k(1-2\mu)$

Strains are dimensionless, while stresses are not.

Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force

$V=Al=abl; \dfrac{\triangle a}{a}=\dfrac{\triangle b}{b}\left[\because
\sigma=\dfrac{\dfrac{-\triangle a}{a}}{\dfrac{\triangle
l}{l}}=\dfrac{\dfrac{\triangle b}{b}}{\dfrac{\triangle l}{l}}\right]$
$\Rightarrow
\dfrac{\triangle V}{V}=2\dfrac{\triangle a}{a}+\dfrac{\triangle
I}{l}=-2\sigma \dfrac{\triangle I}{I}+\dfrac{\triangle I}{I}\Rightarrow
\dfrac{\triangle V}{V}=\dfrac{\triangle I}{I}(1-2\sigma)-1(1-2\times
0.2)=-1(1-0.4)=-0.6$
$\because$ The volume approximately decreases by $0.6%$.

By Lame's relation, $\ \nu = \dfrac { 1 }{ 2 } -\dfrac { E  }{ 6B} ,$ where  $B$ is bulk modulus.
Given, volume change is negligible, thus B tends to infinity. $(B=-V\dfrac { dP }{ dV } )$
 Thus, $\nu=\dfrac { 1 }{ 2 } $

Poisson's ratio is the ratio of transverse contradiction strain to longitudinal extension strain in the direction of stretching force.

We know that $\sigma =(\Delta A/A) / (\Delta L/L)=(\Delta V/V)/(\Delta L/L)^2 \implies \Delta V=\sigma (\Delta L/L)^2V$

We also know $Y=(W/A)/(\Delta L/L) \implies (\Delta L/L)=W/AY$

Substituting this value in the previous expression, we get, $\Delta V=\sigma (W/AY)^2V=\sigma (W^2 L/AY^2)$

The correct option is (b)

Poisson' ratio is defined as the ratio of lateral stress and longitudinal stress

The correct option is (c)

Poisson's ratio can lie only between 0 to 1. It cannot be greater than 1 for any material

The correct option is (d)

We know that $\sigma =(\Delta A/A) / (\Delta L/L)=(\Delta V/V)/(\Delta L/L)^2 \implies \Delta V=\sigma (\Delta L/L)^2V=\sigma (\Delta L/L)^2(LA)=\sigma (\Delta L)^2A/L$

Substituting $\sigma=0.25$, we get, $\Delta V=(\Delta L)^2A/4L$

The correct option is (b)

The poisson's ratio is related to modulus of elasticity as $Y = 3B(1-2 \sigma)$. Since stress is same for Y and B, we get, $dL/L=dV/3V(1-2 \sigma) \implies dV=3V (dL/L)(1-2 \sigma)$
As $\sigma$ is increased, $dV$ decreases. 

The correct option is (b)

Young's modulus and rigidity modulus can be related to poisson's ratio as $Y=2n(1+\sigma)$

The correct option is (b)

Poisson's ratio = change in area /  change in length. If poisson's ratio >1, then change in area > change in length. Thus area expands when length increases

The option (b) is the correct option

The formula relating young's modulus and bulk's modulus with poisson's ratio is $Y=3B(1-2 \sigma)$

The option (c) is the correct option

Longitudinal strain = 0.3 cm/3 m = 0.0001

Lateral strain = poisson's ratio x longitudinal strain =$ 0.25 \times 0.0001 = 2.5 \times 10^{-4}$

The correct option is (b)

Let Y, K, n and $\sigma$ be the Young's Modulus, Bulk modulus, Modulus of Rigidity and Poisson's Ratio, respectively. 
Y = 3K (1 - 2$\sigma$) [Standard formula] 
Y = 2n (1 + $\sigma$) [Standard formula] 
Hence, 3K (1 - 2$\sigma$) = 2n (1 + $\sigma$) 
Now K and n are always positive, so 
i) If $\sigma$ be +ve, then RHS is always +ve. So LHS must also be +ve. Therefore, 2$\sigma$ < 1 or $\sigma$ <1/2 
ii) If $\sigma$ be -ve, then LHS will always be +ve. Therefore, 1+$\sigma$ > 0 or $\sigma$ > -1 
Thus the limiting values of Poisson's ratio are -1 < $\sigma$ < 1/2

The correct option is (a)

The formula that relates all three elastic constants is 9/Y = 3/n + 1/B

The correct option is (b)

We know that $Y=3B(1-\sigma)$. Substituting, Y=B, we get, $1/3=1-2 \sigma$ or poisson's ratio = 1/3

The correct option is (c)

In the formula derived by the student, in order that Y is positive, $\sigma<0.25$, else Y will be negative, which is not possible

Hence, poisson's ratio should be less than 0.25

The correct option is option(c)

Given,

Poisson's ratio $=-\cfrac{\epsilon _{lateral}}{\epsilon _{Longtudinal}}$

Given that,

Young’s modulus $Y=6.6\times {{10}^{10}}\,N/{{m}^{2}}$

Bulk modulus $B=11\times {{10}^{10}}\,N/{{m}^{2}}$

We know that,

  $ Y=3K\left( 1-2\mu  \right) $

 $ 6.6\times {{10}^{10}}=3\times 11\times {{10}^{10}}-66\times {{10}^{10}}\mu  $

 $ -\mu =\dfrac{\left( 6.6-33 \right)\times {{10}^{10}}}{66\times {{10}^{10}}} $

 $ \mu =0.4 $

Hence, the poisson’s ratio is $0.4$

Suppose, D be the diameter of the wire Poissons ratio,  

$σ=\frac { lateral strain }{ longitudinal strain } $

 $σ=\frac { \frac { ΔD }{ D }  }{ \frac { ΔL }{ L }  } $

 $\frac { ΔL }{ L } =0.025$

 $σ=0.004$

 $σ=\frac { \frac { ΔD }{ D }  }{ \frac { 1 }{ 40 }  } $

 $\frac { ΔD }{ D } =\frac { 1 }{ 40 } \times 0.4=0.01$

Volumetric strain = longitudinal_strain * (1 - 2sigma). If sigma = 0.5, then (1 - 20.5) = 0, so the volume change is 0%.

$v=\dfrac { \pi { d }^{ 2 }l }{ 4 } $ 

Relation between young’s modulus and transverse strain is as follows:

$Y=2\eta \left( 1+\sigma  \right)$

Where, \[$\eta ]$is modulus of rigidity

And $\sigma $is transverse strain, $\sigma =0$

So,

$ Y=2\eta  $

$ \eta =\dfrac{Y}{2}=\dfrac{6\times {{10}^{12}}}{2}=3\times {{10}^{12}}\,N/{{m}^{2}} $

Using the relation between Young's modulus, Poisson's ratio, and the change in dimensions, the change in radius is calculated. The result is 0.99997 mm.

Volumetric strain = longitudinal_strain * (1 - 2sigma). Since sigma = 0.5, (1 - 20.5) = 0, so the volume change is 0%.

As we know$:-$

$Y = 2\eta \left( {1 + \sigma } \right)$

$V = \pi {r^2}I$

This is a factual question. The ratio is labelled $\beta$.

stress is $\dfrac{F}{A}$ hence unit $N/m^2$

Volumetric strain = longitudinal_strain * (1 - 2sigma). For volume to not change, (1 - 2sigma) must be 0, which means sigma = 0.5.

By using stress relations on unit solid element, this relation can be derived:
$\eta \quad =\quad \dfrac { Y }{ 2(1+\sigma ) } \ Thus,\quad Y=2\eta (1+\sigma )$

Poisson's ratio = Lateral strain/Longitudinal strain

Poisson's ratio = - (dr/r) / (dl/l). Using the volume of the wire V = pi * R^2 * l, and assuming constant volume or relating the changes, the expression for Poisson's ratio is derived as (r * M) / (pi * l * rho * R^3).

Volumetric strain is defined as the change in volume divided by the original volume. For a bar under uniaxial stress, the longitudinal strain is F/(AY) and the lateral strains are -sigma * F/(AY). Summing these gives the volumetric strain: F/(AY) * (1 - 2*sigma).

We can take copper rod as cylindrical rod

First, calculate longitudinal strain: delta_L/L = F/(AY) = (mg)/(pi * r^2 * Y). Then, use Poisson's ratio sigma = -(delta_d/d) / (delta_L/L) to find the change in diameter. Plugging in the values yields 3.184 * 10^-8 m.

$Y=2\eta(1+\sigma)\Rightarrow \sigma=\displaystyle\frac{0.5Y-\eta}{\eta}$

Hookes law states that stress is proportional to strain up to elastic limit. If p is the stress induced in material and e the corresponding strain, then according to Hooke's law, 
$\dfrac{P}{E}$ = E, a constant.

The Poisson's ratio of a stable, isotropic, linear elastic material cannot be less than $1.0$ nor greater than $0.5$. So only possible value among the options is $0.4.$

Experimental value of poisson's ratio is always between $0$ to $1/2$ .

1/ Young's modulis is property of a metal independent of its dimensions
2/Definition of Poisson's ratio 
Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force. Tensile deformation is considered positive and compressive deformation is considered negative. The definition of Poisson's ratio contains a minus sign so that normal materials have a positive ratio.Virtually all common materials, such as the blue rubber band on the right, become narrower in cross section when they are stretched. 

Relation  between  Young's  modulus,  bulk  modulus  and  poisson's  ratio  is  given  below :
$Y = 3B (1-2\sigma)$
So,  according  to  problem
$ 6.6 \times  10^{10} = 3 \times 11 \times 10^{10} (1-2\sigma)$
$ \sigma = 0.4$

$\epsilon x=0.05$  (given)
$\sigma =0.25$
$\dfrac{\epsilon y}{0.05}=-0.25$ (standard result)
$=-0.0125$

$\epsilon x=\dfrac{0.3}{3}$$=0.1$ (standard result)

Let the material of length $l$ and side $s$ 

$Y=2\eta(1+\sigma)$
$\Rightarrow 2.4\eta =2\eta(1+\sigma)$
$\Rightarrow 1.2=1+\sigma$
$\Rightarrow \sigma=0.2$

Volume of a wire of radius $r$ and length $l$ is,

Poisson ratio $=0.5$
Since, density is constant therefore change in volume is zero, we have
    $V=A\times l=$ constant
$\Rightarrow \log { V } =\log { A } +\log { l } $
or $\dfrac { dA }{ A } +\dfrac { dl }{ l } =0$
$\Rightarrow \dfrac { dl }{ l } =-\dfrac { dA }{ A } $
$\therefore $ Percentage increase in length $=4$%