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Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

Consider the statements A and B, identify the correct answer given below :
(A) : If the volume of a body remains unchanged when subjected to tensile strain, the value of poisson's ratio is 1/2.
(B) : Phosper bronze has low Young's modulus and high rigidity modulus. 

  1. A and B are correct

  2. A and B are wrong

  3. A is correct and B is wrong

  4. A is wrong and B is right
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

As Phosper bronze is solid so, value of young's modulus is also high.

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

Consider the following two statements A and B and identify the correct answer.
A) When the length of a wire is doubled, the Young's modulus of the wire is also doubled
B) For elastic bodies Poisson's ratio is + Ve and for inelastic bodies Poissons ratio is -Ve

  1. Both A & B are true

  2. A is true but B is false

  3. A is true but B is true

  4. Both A & B are false
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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. 

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

For a material Y $=$ 6.6x10$^{10}$ N/m$^{2}$ and bulk modulus K $=$ 11x10$^{10}$ N/m$^{2}$, then its Poissons's ratio is

  1. 0.8

  2. 0.35

  3. 0.7

  4. 0.4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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$

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

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                   

  1. 0.00625

  2. 0.125

  3. 0.0125

  4. 0.0625
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

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

  1. 0.26

  2. 2.6

  3. 0.026

  4. 0.0026
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

$\sigma =0.26$ (given)

$0.26=\dfrac{-\epsilon y}{0.1}$

$-0.026=\epsilon y$

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

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 :

  1. $+0.50$

  2. $-0.50$

  3. $0.25$

  4. $-0.25$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

If a material maintains constant volume during stretching
$V = l \times s^2$
Differentiate wrt $dl$
$dV = s^2.dl+ l .2s.ds$
$dl .s = 2l .ds$
$\dfrac{ds}{dl} = -\dfrac{1}{2}\dfrac{s}{l}$
$\eta = -\dfrac{\dfrac{ds}{s}}{\dfrac{dl}{l}} = \dfrac{1}{2}$

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

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

  1. 0.50
    • 0.50

  3. 0.25
    • 0.25
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

$V=\pi{r}^{2}l$
$\therefore dV=2\pi rldr + \pi{r}^{2}dl$
$0=2\pi rldr + \pi{r}^{2}dl$
$(dv=0,$ as volume is unchanged$)$
$\therefore 2rldr = -{r}^{2}dl$
$\cfrac{dr}{r}= -\cfrac{1}{2} \cfrac{dl}{l}$
$\therefore \cfrac{\cfrac{dr}{r}}{\cfrac{dl}{l}}=-\cfrac{1}{2}$
$\therefore \sigma= -\cfrac{1}{2}$
As, here $-ve$ sign implies that if, length increased, radius decreased, so, we can write
$\sigma=\cfrac{1}{2}=0.5$

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

The Poisson's ratio of a material is $0.5$. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by 4%. The percentage increase in the length is :

  1. 1%

  2. 2%

  3. 2.5%

  4. 4%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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 $=4$%

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

The Poisson's ratio of the material of a wire is$0.25 .$ If it is stretched by a force F, the longitudinal strain produced in the wire is $5 \times 10 ^ { - 4 } .$ What is the percentage increase in its volume?

  1. $0.2$

  2. $2.5 \times 10 ^ { - 2 }$

  3. Zero

  4. $1.25 \times 10 ^ { - 6 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\sigma =.25$
$\Rightarrow \ \dfrac {-\Delta R1R}{\Delta l 10}$
$\Rightarrow \ \dfrac {\Delta R}{R}=-.25\dfrac {\Delta l}{l}$
$=-.25\times 5\times 16^4$
$v=\pi R^2 l$
$\Rightarrow \ \dfrac {\Delta v}{v}\times 100 \left (2\dfrac {\Delta R}{R} + \dfrac {\Delta l}{l}\right)\times 100$
$=(2\times (-.25\times 5\times 10^{-4})+5\times 10^{-4})\times 150$
$=2.5\times 10^{-2}=.025\%$