Tag: properties of material substances

Quizzes (4) Questions (125)

Questions Related to properties of material substances

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

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. $0.5$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $

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

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. $0.48$%
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $
$V=\pi { R }^{ 2 }l$
$\therefore \cfrac { \Delta V }{ V } \times 100=\left( 2\cfrac { \Delta R }{ R } +\cfrac { \Delta l }{ l }  \right) \times 100=\left[ 2\times \left( -0.8\times { 10 }^{ -3 } \right) +4.0\times { 10 }^{ -3 } \right] \times 100=2.4\times { 10 }^{ -3 }\times 100=0.24$%

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

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. $5.6\times { 10 }^{ -9 }m$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $

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

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. $0.2$
Reveal answer Fill a bubble to check yourself
D Correct answer
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) $
$1.2=1+\sigma \quad or\quad \sigma 1.2-1=0.2$

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

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. $0.425\mu m$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } $
$\therefore \Delta l=\cfrac { F }{ A } \cfrac { l }{ Y } =\cfrac { 100\times 4.5 }{ 3.14\times { \left( 3\times { 10 }^{ -3 } \right)  }^{ 2 }\times 4.8\times { 10 }^{ 11 } } =3.32\times { 10 }^{ -5 }m$

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

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%
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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%
Reveal answer Fill a bubble to check yourself
A Correct answer
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. 

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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} $
Reveal answer Fill a bubble to check yourself
A Correct answer