Tag: physics

Questions Related to physics

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 } $