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Poisson ratio - class-XI

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A material has Poisson's ratio $0.5$. if a uniform rod of it suffers a longitudinal strain of $2\times {10}^{3}$, then the percentage change in volume is

  1. $0.6$

  2. $0.4$

  3. $0.2$

  4. zero


Correct Option: D
Explanation:

Poisson's ratio is a material property. Therefore it doesn't change with dimensions.  

Option D

Which of the following statements is correct regarding Poisson's ratio?

  1. It is the ratio of the longitudinal strain to the lateral strain

  2. Its value is independent of the nature of the material

  3. It is unitless and dimensionless quantity

  4. The practical value of Poisson's ratio lies between $0$ and $1$


Correct Option: C
Explanation:

The ratio of the lateral strain to longitudinal strain is called Poisson's ratio.
Hence, option (a) is an incorrect statement.
Its value depends only on the nature of the material. 
Hence, option (b) is an incorrect statement.
It is the ratio of two like physical quantities.
Therefore, it is unitless and dimensionless quantity.
Hence option (c) is a correct statement
The practical value of Poisson's ratio lies between $0$ and $0.5$
hence option (d) is an incorrect statement.

If the volume of a wire remains constant when subjected to tensile stress, the value of Poisson's ratio of the material of the wire is:

  1. $0.1$

  2. $0.2$

  3. $0.4$

  4. $0.5$


Correct Option: D
Explanation:

Let $L$ be the length, $r$ be the radius of the wire.
Volume of the wire is
$V=\pi { r }^{ 2 }L$
Differentiating both sides, we get
$\Delta V=\pi (2r\Delta r)L+\pi { r }^{ 2 }\Delta L\quad $
As the volume of the wire remains unchanged when it gets stretched, so $\Delta V=0$. Hence
$0=2\pi rL\Delta r+\pi { r }^{ 2 }\Delta L$
$\therefore \cfrac { \Delta r/r }{ \Delta L/L } =-\cfrac { 1 }{ 2 } $
$Poisson's\quad ratio=\cfrac { Lateral\quad strain }{ Longitudinal\quad strain } =-\cfrac { \Delta r/r }{ \Delta L/L } =\cfrac { 1 }{ 2 } =0.5\quad $

A material has Poisson's ratio $0.2$. If a uniform rod of its suffers longitudinal strain $4.0\times {10}^{-3}$, calculate the percentage change in its volume.

  1. $0.15$%

  2. $0.02$%

  3. $0.24$%

  4. $0.48$%


Correct Option: C
Explanation:

Given:
 Poisson's ratio $0.2$. 
 longitudinal strain $4.0\times 10^{−3}$
As $\sigma =-\cfrac { \Delta R/R }{ \Delta l/l } $
$\therefore \cfrac { \Delta R }{ R } =-\sigma \cfrac { \Delta l }{ l } =-0.2\times 4.0\times { 10 }^{ -3 }=-0.8\times { 10 }^{ -3 }\quad $
$V=\pi { R }^{ 2 }l$
$\therefore \cfrac { \Delta V }{ V } \times 100=\left( 2\cfrac { \Delta R }{ R } +\cfrac { \Delta l }{ l }  \right) \times 100=\left[ 2\times \left( -0.8\times { 10 }^{ -3 } \right) +4.0\times { 10 }^{ -3 } \right] \times 100=2.4\times { 10 }^{ -3 }\times 100=0.24$%

One end of a nylon rope of length $4.5m$ and diameter $6mm$ is fixed to a stem of a tree. A monkey weighting $100N$ jumps to catch the free end and stays there. what will be the change in the diameter of the rope. (Given Young's modulus of nylon $=4.8\times { 10 }^{ 11 }N{ m }^{ -2 }\quad $ and Poisson's ratio of nylon $=0.2$)

  1. $8.8\times { 10 }^{ -9 }m$

  2. $7.4\times { 10 }^{ -9 }m$

  3. $6.4\times { 10 }^{ -8 }m$

  4. $5.6\times { 10 }^{ -9 }m$


Correct Option: A
Explanation:

Poisson's ratio $\sigma =\cfrac { \Delta D/D }{ \Delta l/l } =\cfrac { \Delta D }{ D } .\cfrac { l }{ \Delta l } $
$\therefore \Delta D=\cfrac { \sigma D\Delta l }{ l } =\cfrac { 0.2\times 6\times { 10 }^{ -3 }\times 3.32\times { 10 }^{ -5 }m }{ 4.5 } =8.8\times { 10 }^{ -9 }m\quad $

For a given material, the Young's modulus is $2.4$ times that of the modulus of rigidity. Its Poisson's ratio is

  1. $2.4$

  2. $1.2$

  3. $0.4$

  4. $0.2$


Correct Option: D
Explanation:

As  $Y=2\eta \left( 1+\sigma  \right) $
where the symbols have their usual meanings
Given: $Y=2.4\eta $
$\therefore 2.4\eta =2\eta \left( 1+\sigma  \right) $
$1.2=1+\sigma \quad or\quad \sigma 1.2-1=0.2$

One end of a nylon rope of length $4.5m$ and diameter $6mm$ is fixed to a free limb. A monkey weighting $100N$ jumps to catch the free end and stays there. Find the elongation of the rope, (Given Young's modulus of nylon $=4.8\times { 10 }^{ 11 }N{ m }^{ -2 }$ and Poisson's ratio of nylon $=0.2$)

  1. $0.332\mu m$

  2. $0.151\mu m$

  3. $0.625\mu m$

  4. $0.425\mu m$


Correct Option: A
Explanation:

Here, $=4.5m,D=6mm=6\times { 10 }^{ -3 }m,F=100N,Y=4.8\times { 10 }^{ 11 }N\quad { m }^{ -2 },\sigma =0.2$
As $Y=\cfrac { F }{ A } \cfrac { l }{ \Delta l } $
$\therefore \Delta l=\cfrac { F }{ A } \cfrac { l }{ Y } =\cfrac { 100\times 4.5 }{ 3.14\times { \left( 3\times { 10 }^{ -3 } \right)  }^{ 2 }\times 4.8\times { 10 }^{ 11 } } =3.32\times { 10 }^{ -5 }m$

The increase in length of a wire on stretching is 0.025%. If its Poisson's ratio is 0.4, then the percentage decrease in diameter is

  1. 0.01%

  2. 0.02%

  3. 0.03%

  4. 0.04%


Correct Option: A
Explanation:

Given,

% increase in length of the wire on stretching, $\dfrac{\Delta L}{L}=0.025$%
Poisson's ratio, $v=0.4$
To find: 
% decrease in diameter $= ?$

Poisson's ratio can be given by the formula: 
 $v=-\dfrac{\Delta D/D}{\Delta L/L}$

$\dfrac{\Delta D}{D}=-v\dfrac{\Delta L}{L}$

$\dfrac{\Delta D}{D}=-0.4\times 0.025=-0.01$%
The percentage decreases in diameter is $0.01$%.
The correct option is A.

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

  1. 0.01%

  2. 0.02%

  3. 0.03%

  4. 0.04%


Correct Option: A
Explanation:

Increase in length, $\dfrac{\Delta 1}{1}  = 0.025 $%

${\dfrac{\Delta d}{d}} = ?$

Poisson’s ratio is $0.4.$

${Poisson's \ ratio} = \dfrac {\dfrac{\Delta d}{d}} {\dfrac{\Delta l}{l}}\\$

$0.4=\dfrac {\dfrac{\Delta d}{d}} {\dfrac{0.025}{100}}\\$

$\dfrac{\Delta d}{d} = \dfrac{0.4\times 0.025}{100}\\$

$\dfrac{\Delta d}{d} = 0.01$ %

Then the percentage decrease  $= 0.01$ %.

Option A is correct. 

For perfectly rigid bodies, the elastic constants Y, B and n are 

  1. Y=B=n =0

  2. Y=B=n =infinity

  3. Y=2B=3n

  4. Y=B=n =0.5


Correct Option: B
Explanation:

Perfectly rigid bodies cannot be deformed upon application of any amount of force. Thus, strain is zero or the modulus of elasticity which is inversely proportional to strain becomes infinity

Thus option (b) is the correct option

The ratio of lateral strain to the linear strain within elastic limit is known as:

  1. Young's modulus

  2. Bulk's modulus

  3. Rigidity modulus

  4. Poisson's ratio


Correct Option: D
Explanation:

The ratio of lateral strain to the linear strain within elastic limit is known as Poisson's ratio

The correct option is (d)

When a uniform metallic wire is stretched the lateral strain produced in it $ \beta.  If \sigma  $ and Y are the pisson 's' ration Young's modulus for wire,then elastic potential energy density of wire is

  1. $ \dfrac {Y\beta^2}{2} $

  2. $ \dfrac {Y\beta^2}{2\sigma^2} $

  3. $ \dfrac {Y \sigma \beta^2}{2} $

  4. $ \dfrac {Y\sigma^2}{2\beta} $


Correct Option: A

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?

  1. 2%

  2. 3%

  3. 5%

  4. 0%


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

  1. Poission ratio

  2. Sharing strain

  3. Longitudinal strain

  4. Volume stress


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

  1. Young's modulus

  2. Bulk modulus

  3. Poisson's law

  4. Hooke's law


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 .....”

  1. decreases by $1\%$

  2. decreases by $0.8\%$

  3. decreases by $0.6\%$

  4. increases by $0.2\%$


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

  1. infinite

  2. zero

  3. 0.5

  4. -1


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 :

  1. -1< $\sigma $ < 0.5

  2. -0.5 < $\sigma $ < 1.0

  3. 0.5 < $\sigma $ < 1.0

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


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:

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

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

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

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


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 

  1. longitudinal stress and longitudinal strain

  2. longitudinal stress and lateral stress

  3. lateral stress and longitudinal stress

  4. lateral stress and lateral strain


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

  1. Steel

  2. Copper

  3. Aluminium

  4. None of the above


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

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

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

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

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


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)

The change in unit volume of a material under tension with increase in its poisson's ratio will be

  1. Increase

  2. Decrease

  3. Remains same

  4. Initially increases and then decreases


Correct Option: B
Explanation:

The poisson's ratio is related to modulus of elasticity as $Y = 3B(1-2 \sigma)$. Since stress is same for Y and B, we get, $dL/L=dV/3V(1-2 \sigma) \implies dV=3V (dL/L)(1-2 \sigma)$
As $\sigma$ is increased, $dV$ decreases. 

The correct option is (b)

The formula relating youngs modulus (Y), rigidity modulus (n) and Poisson's ratio ($\sigma$) is 

  1. $Y=2n(1- \sigma)$

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

  3. $Y=n(1- 2\sigma)$

  4. $Y=n(1+2 \sigma)$


Correct Option: B
Explanation:

Young's modulus and rigidity modulus can be related to poisson's ratio as $Y=2n(1+\sigma)$

The correct option is (b)

A student measures the poisson's ratio to be greater than 1 in an experiment. The meaning of this statement would be

  1. An increase in length would also result in decrease in area of cross section of the wire

  2. An increase in length would also result in increase in area of cross section of the wire

  3. An decrease in length would also result in decrease in area of cross section of the wire

  4. An increase in length will not change the area of cross section of the wire


Correct Option: B
Explanation:

Poisson's ratio = change in area /  change in length. If poisson's ratio >1, then change in area > change in length. Thus area expands when length increases

The option (b) is the correct option

The formula that relates Bulk's modulus with poisson's ratio is 

  1. $Y=3B(1+2 \sigma)$

  2. $Y=3B(1- \sigma)$

  3. $Y=3B(1-2 \sigma)$

  4. $Y=3B(1+ \sigma)$


Correct Option: C
Explanation:

The formula relating young's modulus and bulk's modulus with poisson's ratio is $Y=3B(1-2 \sigma)$

The option (c) is the correct option

A copper wire 3 m long is stretched to increase its length by 0.3 cm. Find the lateral strain produced in the wire , if poisson's ratio for copper is 0.25

  1. $5 \times 10^{-4}$

  2. $2.5 \times 10^{-4}$

  3. $5 \times 10^{-3}$

  4. $2.5 \times 10^{-3}$


Correct Option: B
Explanation:

Longitudinal strain = 0.3 cm/3 m = 0.0001

Lateral strain = poisson's ratio x longitudinal strain =$ 0.25 \times 0.0001 = 2.5 \times 10^{-4}$

The correct option is (b)

The theoretical limits of poisson's ratio lies between -1 to 0.5 because

  1. Shear modulus and bulk's modulus should be positive

  2. Bulk's modulus is negative during compression

  3. Shear modulus is negative during compression

  4. Young's modulus should be always positive


Correct Option: A
Explanation:

Let Y, K, n and $\sigma$ be the Young's Modulus, Bulk modulus, Modulus of Rigidity and Poisson's Ratio, respectively. 
Y = 3K (1 - 2$\sigma$) [Standard formula] 
Y = 2n (1 + $\sigma$) [Standard formula] 
Hence, 3K (1 - 2$\sigma$) = 2n (1 + $\sigma$) 
Now K and n are always positive, so 
i) If $\sigma$ be +ve, then RHS is always +ve. So LHS must also be +ve. Therefore, 2$\sigma$ < 1 or $\sigma$ <1/2 
ii) If $\sigma$ be -ve, then LHS will always be +ve. Therefore, 1+$\sigma$ > 0 or $\sigma$ > -1 
Thus the limiting values of Poisson's ratio are -1 < $\sigma$ < 1/2

The correct option is (a)

The formula that relates all three elastic constants is 

  1. 9/Y = 3/n - 1/B

  2. 9/Y = 3/n + 1/B

  3. 9/Y = 3/n + 2/B

  4. 9/Y = 3/n - 2/B


Correct Option: B
Explanation:

The formula that relates all three elastic constants is 9/Y = 3/n + 1/B

The correct option is (b)

What is the poisson's ratio of a wire, whose Young's modulus and Bulk's modulus are equal

  1. 1/2

  2. 2/3

  3. 1/3

  4. 1/4


Correct Option: C
Explanation:

We know that $Y=3B(1-\sigma)$. Substituting, Y=B, we get, $1/3=1-2 \sigma$ or poisson's ratio = 1/3

The correct option is (c)

The formula $Y=3B(1-2 \sigma)$ relates young's modulus and bulk's modulus with poisson's ratio. A theoretical physicist derives this formula incorrectly as $Y=3B(1-4 \sigma)$. According to this formula, what would be the theoretical limits of poisson's ratio:

  1. Poisson's ratio should be less than 1

  2. Poisson's ratio should be less than 0.5

  3. Poisson's ratio should be less than 0.25

  4. Poisson's ratio should be less than 0


Correct Option: C
Explanation:

In the formula derived by the student, in order that Y is positive, $\sigma<0.25$, else Y will be negative, which is not possible

Hence, poisson's ratio should be less than 0.25

The correct option is option(c)

The ice storm in the state of Jammu strained many wires to the breaking point. In a particular situation, the transmission towers are separated by $500\ m$ of wire. The top grounding wire $15^{o}$ from horizontal at the towers, and has a diameter of $1.5cm$. The steel wire has a density of $7860\ kg\ m^{-3}$. When ice (density $900\ kg\ m^{-3}$) built upon the wire to a diameter $10.0\ cm$, the wire snapped. What was the breaking stress (force/ unit area) in $N\ m^{-2}$ in the wire at the breaking point? You may assume the ice has no strength.

  1. $7.4\ \times 10^{7}\ N\ m^{-2}$

  2. $4.5\ \times 10^{8}\ N\ m^{-2}$

  3. $2.6\ \times 10^{6}\ N\ m^{-2}$

  4. $1.15\ \times 10^{7}\ N\ m^{-2}$


Correct Option: C

If Young modulus is three times of modulus of rigidity, then Poisson ratio is equal to:

  1. $0.2$

  2. $0.3$

  3. $0.4$

  4. $0.5$


Correct Option: D
Explanation:

Given,


$E=3G$

$2(1+m)=\dfrac{3G}{G}$

$m=\dfrac{3}{2}-1=\dfrac{1}{2}=0.5$

A material has Poissons ratio $0.5$. If a uniform rod made of the surface a longitudinal string of $2\times {10}^{-3}$, what is the percentage increase in its volume?

  1. $2\%$

  2. $4\%$

  3. $0\%$

  4. $5\%$


Correct Option: B

A steel wire of length $30cm$ is stretched ti increase its length by $0.2cm$. Find the lateral strain in the wire if the poisson's ratio for steel is $0.19$ :

  1. $0.0019$

  2. $0.0008$

  3. $0.019$

  4. $0.008$


Correct Option: A
Explanation:

Poisson's ratio $=-\cfrac{\epsilon _{lateral}}{\epsilon _{Longtudinal}}$

$\epsilon _{longitudinal}=\cfrac{\triangle L}{L}=\cfrac{0.2}{20}$
$\therefore \epsilon _{lateral}=-\cfrac{0.2}{20}\times 0.19$
$=-0.0019$
$|\epsilon _{lateral}|=0.0019$

For a material $Y={ 6.6\times 10 }^{ 10 }\ { N/m }^{ 2 }$ and bulk modulus $K{ 11\times 10 }^{ 10 }\ { N/m }^{ 2 }$, then its Poisson's ratio is:

  1. $0.8$

  2. $0.35$

  3. $0.7$

  4. $0.4$


Correct Option: D
Explanation:

Given that,

Young’s modulus $Y=6.6\times {{10}^{10}}\,N/{{m}^{2}}$

Bulk modulus $B=11\times {{10}^{10}}\,N/{{m}^{2}}$

We know that,

  $ Y=3K\left( 1-2\mu  \right) $

 $ 6.6\times {{10}^{10}}=3\times 11\times {{10}^{10}}-66\times {{10}^{10}}\mu  $

 $ -\mu =\dfrac{\left( 6.6-33 \right)\times {{10}^{10}}}{66\times {{10}^{10}}} $

 $ \mu =0.4 $

Hence, the poisson’s ratio is $0.4$

The increase in the length of a wire on stretching is $0.025 \%$. If its Poisson's 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: A
Explanation:

Suppose, D be the diameter of the wire Poissons ratio,  

$σ=\frac { lateral strain }{ longitudinal strain } $

 $σ=\frac { \frac { ΔD }{ D }  }{ \frac { ΔL }{ L }  } $

 $\frac { ΔL }{ L } =0.025$

 $σ=0.004$

 $σ=\frac { \frac { ΔD }{ D }  }{ \frac { 1 }{ 40 }  } $

 $\frac { ΔD }{ D } =\frac { 1 }{ 40 } \times 0.4=0.01$

When a wire is stretched, its length increases by 0.3% and the diameter decreases by 0.1%. Poisson's ratio of the material of the wire is about

  1. 0.03

  2. 0.333

  3. 0.15

  4. 0.015


Correct Option: A

A material has Poisson's ratio 0.5. If a uniform rod of it suffers a longitudinal strain of $2\times { 10 }^{ -3 }$, then the percentage increase in its volume is 

  1. 0%

  2. 10%

  3. 20%

  4. 5%


Correct Option: A

When a metal wire is stretched by a load, the fractional change in its volume $\Delta V/V$ is proportional to?

  1. $-\dfrac{\Delta l}{l}$

  2. $\left(\dfrac{\Delta l}{l}\right)^2$

  3. $\sqrt{\Delta l/l}$

  4. None of these


Correct Option: A
Explanation:

$v=\dfrac { \pi { d }^{ 2 }l }{ 4 } $ 


$⟹\dfrac { ΔV }{ V } =\dfrac { 2Δd }{ d } +\dfrac { Δl }{ l } $

 $⟹d\frac { ΔV }{ V } =\dfrac { (1−2σ)Δl }{ l } $

$(\dfrac { Δd }{ d } =\dfrac { −σΔl }{ l } )$

where $σ$ is Poisson's ratio.

A material has poisson's ratio $0.3$. If a uniform rod of it suffers a longitudinal strain of $25\times 10^{-3}$, then the percentage increase in its volume is

  1. $1\%$

  2. $2\%$

  3. $3\%$

  4. $4\%$


Correct Option: C

The Young's modulus of the material of a wire is $6\times 10^{12}$$N/m^{2}$ and there is no transverse in it, then its modulus of rigidity will be 

  1. $3\times 10^{12}N/m^{2}$

  2. $2\times 10^{12}N/m^{2}$

  3. $ 10^{12}N/m^{2}$

  4. None of the above


Correct Option: A
Explanation:

Relation between young’s modulus and transverse strain is as follows:

$Y=2\eta \left( 1+\sigma  \right)$

Where, \[$\eta ]$is modulus of rigidity

And $\sigma $is transverse strain, $\sigma =0$

So,

$ Y=2\eta  $

$ \eta =\dfrac{Y}{2}=\dfrac{6\times {{10}^{12}}}{2}=3\times {{10}^{12}}\,N/{{m}^{2}} $

A cylinderical wire of radius $1 mm,$ length $1 m,$ Young's modulus = $2\times10^{11}N/m^2$, poisson's ratio $\mu =\pi/10$ is stretched by a force of $100N$. Its radius will become

  1. $0.99998 mm$

  2. $0.99999 mm$

  3. $0.99997 mm$

  4. $0.99995 mm$


Correct Option: C

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?

  1. $2\%$

  2. $3\%$

  3. $5\%$

  4. $0\%$


Correct Option: D

The poisson's ratio can not be

  1. $-1$

  2. $0$

  3. $0.25$

  4. $0.5$


Correct Option: C

A cube of wood supporting $200 gm$ mass just in water $(\rho =1g/cc)$. When the mass is removed, the cube rises by $2cm$. The volume of cube is

  1. $1000 cc$

  2. $800cc$

  3. $500 cc$

  4. None of these


Correct Option: D
Explanation:
Let the edge of cube be $L$ when mass is on the cube of wood 
$200g+{ L }^{ 3 }{ d } _{ wood }g={ L }^{ 3 }{ d } _{ water }g$
$\Rightarrow { L }^{ 3 }{ d } _{ wood }g={ L }^{ 3 }{ d } _{ water }g$
$\Rightarrow { L }^{ 3 }{ d } _{ wood }g={ L }^{ 3 }{ d } _{ water }g-200g$
$\Rightarrow { L }^{ 3 }{ d } _{ wood }={ L }^{ 3 }d-200$
When mass is removed
${ L }^{ 3 }{ d } _{ wood }-\left( L-2 \right) { L }^{ 2 }{ d } _{ water }\quad \longrightarrow \left( 2 \right) $
From $(1)$ and $(2)$
${ L }^{ 3 }{ d } _{ water }-200=\left( L-2 \right) { L }^{ 2 }{ d } _{ water }$
But ${ d } _{ water }=1$
$\therefore$    ${ L }^{ 3 }-200={ L }^{ 2 }\left( L-2 \right) $
$\therefore$    $L=10cm$.

what is the ratio of Youngs modulus $E$ to shear modulus $G$ in terms of poissons ratio$?$

  1. $2\left( {1 + \mu } \right)$

  2. $2\left( {1 - \mu } \right)$

  3. $\frac{1}{2}\left( {1 - \mu } \right)$

  4. $\frac{1}{2}\left( {1 + \mu } \right)$


Correct Option: A
Explanation:

As we know$:-$

$G = \frac{E}{{2\left( {1 + \mu } \right)}}$ 
so this gives the ratio of $E$ to $G = 2\left( {1 + \mu } \right)$
Hence,
option $(A)$ is correct answer.

For a given material, the Young's modulus is 2.4 times its modulus of rigidity. Its Poisson's ratio is

  1. $0.2$

  2. $0.4$

  3. $1.2$

  4. $2.4$


Correct Option: A
Explanation:

$Y = 2\eta \left( {1 + \sigma } \right)$

But $Y = 2.4\eta $
$\therefore 2.4\eta  = 2\eta \left( {1 + \sigma } \right)$
$\left( {1 + \sigma } \right) = 1.2$
$\sigma  = 0.2$
Hence,
option $(A)$ is correct answer.

When a wire is stretched, its length increases by $0.3$% and the diameter decreases by $0.1$%. Poisson's ratio of the material of the wire is about   

  1. $0.03$

  2. $0.333$

  3. $0.15$

  4. $0.015$


Correct Option: C

If rigidity modulus is 2.6 times of youngs modulus then the value of poission's ratio is 

  1. 0.2

  2. 0.3

  3. 0.5

  4. 0.1


Correct Option: B

When a rubber cord is stretched, the change in volume with respect to change in its linear dimensions is negligible. The Poisson's ratio for rubber is 

  1. 1

  2. 0.25

  3. 0.5

  4. 0.75


Correct Option: C
Explanation:

$V = \pi {r^2}I$

$\frac{{\Delta V}}{V} = \frac{{\Delta \left( {\pi {r^2}I} \right)}}{{\pi {r^2}I}}$
$\frac{{\Delta V}}{V} = \frac{{{r^2}\Delta I + 2rI\Delta r}}{{{r^2}I}}$
$\frac{{\Delta V}}{V} = \frac{{\Delta I}}{I} + \frac{{2\Delta r}}{r}$
But $\frac{{\Delta V}}{V} = 0$
therefore$,$ $\frac{{\Delta I}}{I} = \frac{{ - 2\Delta r}}{r}$
Now$,$ Poisson's ratio$,$ $\sigma  = \frac{{\frac{{ - \Delta r}}{r}}}{{\frac{{\Delta I}}{I}}}$-------------------$(1)$
from equation $(1),$
$\sigma  =  - \left( {\frac{{\frac{{\Delta r}}{r}}}{{\frac{{ - 2\Delta r}}{r}}}} \right) = \frac{1}{2} = 0.5$
Hence,
option $(C)$ is correct answer.

For a given material, the Youngs modulas is $2.4$ times its modulus of rigidity. What is the value of its poissons ratio ?

  1. $0.5$

  2. $0.4$

  3. $0.2$

  4. $0.3$


Correct Option: A

The ratio of change in dimension at right angles to applied force to the initial dimension is defined as

  1. $Y$

  2. $\eta$

  3. $\beta$

  4. $K$


Correct Option: C
Explanation:

This is a factual question. The ratio is labelled $\beta$.

Which of the following pairs is not correct?

  1. strain-dimensionless

  2. stress-$N/m^{2}$

  3. modulus of elasticity-$N/m^{2}$

  4. poisson's ratio-$N/m^{2}$


Correct Option: D
Explanation:

stress is $\dfrac{F}{A}$ hence unit $N/m^2$

strain is $\dfrac{\Delta l}{L}$ so unit $m/m$ therefore dimensionless
modulus of elasticity is $ \dfrac{stress}{strain}$ hence same unit  as stress as the denominator is dimensionless
poisson's ratio $\dfrac{-\epsilon _t}{\epsilon _l} $ so its also going to be dimensionless

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

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:

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

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
  2. 0.25

    • 0.25

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 :

  1. 1%

  2. 2%

  3. 2.5%

  4. 4%


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?

  1. $0.2$

  2. $2.5 \times 10 ^ { - 2 }$

  3. Zero

  4. $1.25 \times 10 ^ { - 6 }$


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


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