Tag: elastic energy

Questions Related to 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 $=0.2$)

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

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

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

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

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

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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

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

  1. $2.4$

  2. $1.2$

  3. $0.4$

  4. $0.2$

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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 $=0.2$)

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

  1. $0.332\mu m$

  2. $0.151\mu m$

  3. $0.625\mu m$

  4. $0.425\mu m$

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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

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

  1. 0.01%

  2. 0.02%

  3. 0.03%

  4. 0.04%

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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:

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

  1. 0.01%

  2. 0.02%

  3. 0.03%

  4. 0.04%

Show answer
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 

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

  1. Y=B=n =0

  2. Y=B=n =infinity

  3. Y=2B=3n

  4. Y=B=n =0.5

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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:

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

  1. Young's modulus

  2. Bulk's modulus

  3. Rigidity modulus

  4. Poisson's ratio

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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

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

  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} $

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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?

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

  1. 2%

  2. 3%

  3. 5%

  4. 0%

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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

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Correct Option: D
Explanation:

Strains are dimensionless, while stresses are not.