Tag: elastic energy

Questions Related to 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:

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

  1. 0.50

    • 0.50

  3. 0.25

    • 0.25
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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:
$\sigma =.25$
$\Rightarrow \ \dfrac {-\Delta R1R}{\Delta l 10}$
$\Rightarrow \ \dfrac {\Delta R}{R}=-.25\dfrac {\Delta l}{l}$
$=-.25\times 5\times 16^4$
$v=\pi R^2 l$
$\Rightarrow \ \dfrac {\Delta v}{v}\times 100 \left (2\dfrac {\Delta R}{R} + \dfrac {\Delta l}{l}\right)\times 100$
$=(2\times (-.25\times 5\times 10^{-4})+5\times 10^{-4})\times 150$
$=2.5\times 10^{-2}=.025\%$