Tag: elastic energy

Questions Related to elastic energy

Which of the following relation is true?

  1. $3Y=K(1+\sigma)$

  2. $K=\displaystyle \frac{9\eta Y}{Y+\eta}$

  3. $\sigma=(6K+\eta)Y$

  4. $\sigma=\displaystyle\frac{0.5Y-\eta}{\eta}$


Correct Option: D
Explanation:

$Y=2\eta(1+\sigma)\Rightarrow \sigma=\displaystyle\frac{0.5Y-\eta}{\eta}$

Ratio of transverse to axial strain is 

  1. Toricelli ratio

  2. Poisson's ratio

  3. Stoke's ratio

  4. Bernoulli's ratio


Correct Option: B
Explanation:

Hookes law states that stress is proportional to strain up to elastic limit. If p is the stress induced in material and e the corresponding strain, then according to Hooke's law, 
$\dfrac{P}{E}$ = E, a constant.

Possible value of Poisson's ratio is

  1. 1

  2. 0.9

  3. 0.8

  4. 0.4


Correct Option: D
Explanation:

The Poisson's ratio of a stable, isotropic, linear elastic material cannot be less than $1.0$ nor greater than $0.5$. So only possible value among the options is $0.4.$

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


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

  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


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

  1. 0.8

  2. 0.35

  3. 0.7

  4. 0.4


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                   

  1. 0.00625

  2. 0.125

  3. 0.0125

  4. 0.0625


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

  1. 0.26

  2. 2.6

  3. 0.026

  4. 0.0026


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 :

  1. $+0.50$

  2. $-0.50$

  3. $0.25$

  4. $-0.25$


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.

  1. $2.4$

  2. $1.2$

  3. $0.4$

  4. $0.2$


Correct Option: D
Explanation:

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