Tag: physics

Questions Related to physics

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. 

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

  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

Show answer
Correct Option: C
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.

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

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

  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

Show answer
Correct Option: D
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. 

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

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

  1. 0.8

  2. 0.35

  3. 0.7

  4. 0.4

Show answer
Correct Option: D
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$

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                   

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

  1. 0.00625

  2. 0.125

  3. 0.0125

  4. 0.0625

Show answer
Correct Option: C
Explanation:

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

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

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

  1. 0.26

  2. 2.6

  3. 0.026

  4. 0.0026

Show answer
Correct Option: C
Explanation:

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

$\sigma =0.26$ (given)

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

$-0.026=\epsilon y$

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 :

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

  1. $+0.50$

  2. $-0.50$

  3. $0.25$

  4. $-0.25$

Show answer
Correct Option: A
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 $dl$
$dV = s^2.dl+ l .2s.ds$
$dl .s = 2l .ds$
$\dfrac{ds}{dl} = -\dfrac{1}{2}\dfrac{s}{l}$
$\eta = -\dfrac{\dfrac{ds}{s}}{\dfrac{dl}{l}} = \dfrac{1}{2}$

For a given material, the Young's modulus is $2.4$ times that of rigidity modulus. Its poisson's ratio is.

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

  1. $2.4$

  2. $1.2$

  3. $0.4$

  4. $0.2$

Show answer
Correct Option: D
Explanation:

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

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:

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

  1. 0.50

    • 0.50

  3. 0.25

    • 0.25
Show answer
Correct Option: A
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
$\sigma=\cfrac{1}{2}=0.5$

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 :

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

  1. 1%

  2. 2%

  3. 2.5%

  4. 4%

Show answer
Correct Option: D
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 $=4$%

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?

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

  1. $0.2$

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

  3. Zero

  4. $1.25 \times 10 ^ { - 6 }$

Show answer
Correct Option: B
Explanation:
$\sigma =.25$
$\Rightarrow \ \dfrac {-\Delta R1R}{\Delta l 10}$
$\Rightarrow \ \dfrac {\Delta R}{R}=-.25\dfrac {\Delta l}{l}$
$=-.25\times 5\times 16^4$
$v=\pi R^2 l$
$\Rightarrow \ \dfrac {\Delta v}{v}\times 100 \left (2\dfrac {\Delta R}{R} + \dfrac {\Delta l}{l}\right)\times 100$
$=(2\times (-.25\times 5\times 10^{-4})+5\times 10^{-4})\times 150$
$=2.5\times 10^{-2}=.025\%$