Tag: properties of material substances

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Questions Related to properties of material substances

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