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

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. $0.99995 mm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the relation between Young's modulus, Poisson's ratio, and the change in dimensions, the change in radius is calculated. The result is 0.99997 mm.

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

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. $0\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volumetric strain = longitudinal_strain * (1 - 2*sigma). Since sigma = 0.5, (1 - 2*0.5) = 0, so the volume change is 0%.

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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) $
$\therefore$    $L=10cm$.
Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

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. $\frac{1}{2}\left( {1 + \mu } \right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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 its modulus of rigidity. Its Poisson's ratio is

  1. $0.2$

  2. $0.4$

  3. $1.2$

  4. $2.4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.