Tag: elastic energy

Questions Related to elastic energy

For which value of Poisson's ratio the volume of a wire does not change when it is subjected to a tension?

  1. 0.5

  2. -1

  3. 0.1

  4. 0


Correct Option: A

The relationship between Y, $\eta$ and $\sigma$ is

  1. $Y=2\eta(1+\sigma)$

  2. $\eta=2Y(1+\sigma)$

  3. $\displaystyle \sigma=\frac{2Y}{(1+\eta)}$

  4. $Y=\eta(1+\sigma)$


Correct Option: A
Explanation:

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 can not have the value:

  1. 0.1

  2. 0.7

  3. 0.2

  4. 0.5


Correct Option: C

Poisson's ratio cannot exceed

  1. 0.25

  2. 1.0

  3. 0.75

  4. 0.5


Correct Option: D
Explanation:

Poisson's ratio = Lateral strain/Longitudinal strain

$Y=3K(1-2\mu)\Rightarrow \mu=0.5-Y/6K$
$Y$ is young's modulus.
$\mu$ is poisson ratio
$K$ is compressibility of the substance which is inverse of Bulk's modulus. Maximum value of $K$ is $\infty$
So maximum value of Poisson's ratio $\mu=0.5$

A wire of mass $M ,$ density $\rho$ and radius $R$ is stretched. If $r$ is the change in the radius and $l$ is the change in its length, then Poisson's ratio is given by :

  1. $\dfrac { \pi l } { \rho M r R ^ { 3 } }$

  2. $\dfrac { R M \pi } { l \rho r ^ { 3 } }$

  3. $\dfrac { r M } { \pi l \rho R ^ { 3 } }$

  4. $\dfrac { l M } { \pi l \rho R ^ { 3 } }$


Correct Option: C

The increase in length of a wire on stretching is 0.025% If its poisson ratio is 0.4, then the percentage decrease in the diameter is : 

  1. 0.01

  2. 0.02

  3. 0.03

  4. 0.04


Correct Option: C

If Poission's ratio is 0.5 for a material, then the material is

  1. Rigid

  2. Elastic fatigue

  3. Compressible

  4. None


Correct Option: A

A uniform bar of length 'L' and cross sectional area 'A' is subjected to a tensile load 'F'. 'Y' be the Young modulus and '$\sigma$' be the Poisson's ratio then volumetric strain is  

  1. $\frac{F}{AY}(1 - \sigma)$

  2. $\frac{F}{AY}(2 - \sigma)$

  3. $\frac{F}{AY}(1 - 2\sigma)$

  4. $\frac{F}{AY} \sigma$


Correct Option: C

A copper rod of length $l$ is suspended from the ceiling by one of its ends. Find the relative increment of its volume $\displaystyle\frac{\Delta V}{V}$.

  1. $\displaystyle\frac{\Delta V}{V}=(1-2\mu)\frac{\Delta l}{l}$

  2. $\displaystyle\frac{\Delta V}{V}=(1-3\mu)\frac{\Delta l}{l}$

  3. $\displaystyle\frac{\Delta V}{V}=(1-2\mu)\frac{2\Delta l}{l}$

  4. $\displaystyle\frac{\Delta V}{V}=(1-3\mu)\frac{3\Delta l}{l}$


Correct Option: A
Explanation:

We can take copper rod as cylindrical rod

$v=\pi r^2 l$

$E=\dfrac{\Delta l}{l}$ (longitudinal strain)

$E'=\dfrac{\Delta r}{r}=-\mu E$  ,where $\mu$ is Poisson ratio,$E'$ is lateral strain

$\dfrac{\Delta V}{V}=\dfrac{2\Delta r}{r}+\dfrac{\Delta l}{l}$

$\dfrac{\Delta V}{V}=(1-2\mu)\dfrac{\Delta l}{l}$

One end of a wire $2$ m long and diameter $2$ mm, is fixed in a ceiling. A naughty boy of mass $10$ kg jumps to catch the free end and stays there. The change in length of wire is (Take $g=10m/s^2, Y=2\times 10^{11} N/m^2$).
In above problem, if Poisson's ratio is $\sigma =0.1$, the change in diameter is?

  1. $3.184\times 10^{-5}$ m

  2. $31.84\times 10^{-5}$ m

  3. $3.184\times 10^{-8}$ m

  4. $31.84\times 10^{-8}$ m


Correct Option: C