Tag: physics

Questions Related to physics

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