Tag: properties of material substances

Questions Related to properties of material substances

A material has poisson's ratio 0.5. If a uniform rod of it suffers a longitudinal strain of $3\times { 10 }^{ -3 }$, what will be percentage increase in volume?

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

  1. 2%

  2. 3%

  3. 5%

  4. 0%

Show answer
Correct Option: D
Explanation:

Here, $E=3k(1-2\mu)$

where, $E=$Modulus of elasticity
$\mu=$Poisson's ratio
$k=$Modulus of elasticity
Here, $\mu=0.5$ then $k\longrightarrow \infty $
$k=\cfrac { \Delta P }{ \left( \cfrac { \Delta V }{ V }  \right)  } $
If $k\longrightarrow \infty $, then $\Delta V\longrightarrow 0$.
Hence the percentage change in volume$=0\%$

Which of the following is not dimension less

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

  1. Poission ratio

  2. Sharing strain

  3. Longitudinal strain

  4. Volume stress

Show answer
Correct Option: D
Explanation:

Strains are dimensionless, while stresses are not.

When a body undergoes a linear tensile strain if experience a lateral contraction also. The ratio of lateral contraction to longitudinal strain is known as

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

  1. Young's modulus

  2. Bulk modulus

  3. Poisson's law

  4. Hooke's law

Show answer
Correct Option: C
Explanation:

Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force

A compressive force is applied to a uniform rod of rectangular cross-section so that its length decreases by $1\%$. If the Poisson’s ratio for the material of the rod be $0.2$, which of the following statements is correct ? The volume approximately .....”

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

  1. decreases by $1\%$

  2. decreases by $0.8\%$

  3. decreases by $0.6\%$

  4. increases by $0.2\%$

Show answer
Correct Option: C
Explanation:

$V=Al=abl; \dfrac{\triangle a}{a}=\dfrac{\triangle b}{b}\left[\because \sigma=\dfrac{\dfrac{-\triangle a}{a}}{\dfrac{\triangle l}{l}}=\dfrac{\dfrac{\triangle b}{b}}{\dfrac{\triangle l}{l}}\right]$
$\Rightarrow \dfrac{\triangle V}{V}=2\dfrac{\triangle a}{a}+\dfrac{\triangle I}{l}=-2\sigma \dfrac{\triangle I}{I}+\dfrac{\triangle I}{I}\Rightarrow \dfrac{\triangle V}{V}=\dfrac{\triangle I}{I}(1-2\sigma)-1(1-2\times 0.2)=-1(1-0.4)=-0.6$
$\because$ The volume approximately decreases by $0.6\%$.

When a rubber cord is stretched, the change in volume is negligible compared to the change in its linear dimension. Then poisson's ratio for rubber is

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

  1. infinite

  2. zero

  3. 0.5

  4. -1

Show answer
Correct Option: C
Explanation:

By Lame's relation, $\ \nu = \dfrac { 1 }{ 2 } -\dfrac { E  }{ 6B} ,$ where  $B$ is bulk modulus.
Given, volume change is negligible, thus B tends to infinity. $(B=-V\dfrac { dP }{ dV } )$
 Thus, $\nu=\dfrac { 1 }{ 2 } $

The Poisson's ratio $\sigma$ should satisfy the relation :

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

  1. -1< $\sigma $ < 0.5

  2. -0.5 < $\sigma $ < 1.0

  3. 0.5 < $\sigma $ < 1.0

  4. -1.0 < $\sigma $ < -0.5

Show answer
Correct Option: A
Explanation:

Poisson's ratio is the ratio of transverse contradiction strain to longitudinal extension strain in the direction of stretching force.

The Poisson's ratio $\sigma$ should satisfy the relation,
$-1<\sigma <0.5$

A metallic wire of young's modulus Y and poisson's ratio $\sigma$, length L and area of cross section A is stretched by a load of W kg. The increase in volume of the wire is:

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

  1. $\sigma (W^2 L/2AY^2)$

  2. $\sigma (W^2 L/AY^2)$

  3. $\sigma (W^2 L/4AY^2)$

  4. $\sigma (2W^2 L/AY^2)$

Show answer
Correct Option: B
Explanation:

We know that $\sigma =(\Delta A/A) / (\Delta L/L)=(\Delta V/V)/(\Delta L/L)^2 \implies \Delta V=\sigma (\Delta L/L)^2V$

We also know $Y=(W/A)/(\Delta L/L) \implies (\Delta L/L)=W/AY$

Substituting this value in the previous expression, we get, $\Delta V=\sigma (W/AY)^2V=\sigma (W^2 L/AY^2)$

The correct option is (b)

Poisson' ratio is defined as the ratio of 

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

  1. longitudinal stress and longitudinal strain

  2. longitudinal stress and lateral stress

  3. lateral stress and longitudinal stress

  4. lateral stress and lateral strain

Show answer
Correct Option: C
Explanation:

Poisson' ratio is defined as the ratio of lateral stress and longitudinal stress

The correct option is (c)

For which material the poisson's ratio is greater than 1

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

  1. Steel

  2. Copper

  3. Aluminium

  4. None of the above

Show answer
Correct Option: D
Explanation:

Poisson's ratio can lie only between 0 to 1. It cannot be greater than 1 for any material

The correct option is (d)

A metal wire of length L is loaded and an elongation of $\Delta L$ is produced. If the area of cross section of the wire is A, then the change in volume of the wire, when elongated is . Take Poisson's ratio as 0.25

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

  1. $\Delta V=(\Delta L)^2A/L$

  2. $\Delta V=(\Delta L)^2A/4L$

  3. $\Delta V=(\Delta L)^2A/2L$

  4. $\Delta V=(\Delta L)^2A/3L$

Show answer
Correct Option: B
Explanation:

We know that $\sigma =(\Delta A/A) / (\Delta L/L)=(\Delta V/V)/(\Delta L/L)^2 \implies \Delta V=\sigma (\Delta L/L)^2V=\sigma (\Delta L/L)^2(LA)=\sigma (\Delta L)^2A/L$

Substituting $\sigma=0.25$, we get, $\Delta V=(\Delta L)^2A/4L$

The correct option is (b)