Tag: elastic energy

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

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. $Y=\eta(1+\sigma)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By using stress relations on unit solid element, this relation can be derived:
$\eta \quad =\quad \dfrac { Y }{ 2(1+\sigma ) } \ Thus,\quad Y=2\eta (1+\sigma )$

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

Poisson's ratio cannot exceed

  1. 0.25

  2. 1.0

  3. 0.75

  4. 0.5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Poisson's ratio = Lateral strain/Longitudinal strain

$Y=3K(1-2\mu)\Rightarrow \mu=0.5-Y/6K$
$Y$ is young's modulus.
$\mu$ is poisson ratio
$K$ is compressibility of the substance which is inverse of Bulk's modulus. Maximum value of $K$ is $\infty$
So maximum value of Poisson's ratio $\mu=0.5$

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

A wire of mass $M ,$ density $\rho$ and radius $R$ is stretched. If $r$ is the change in the radius and $l$ is the change in its length, then Poisson's ratio is given by :

  1. $\dfrac { \pi l } { \rho M r R ^ { 3 } }$

  2. $\dfrac { R M \pi } { l \rho r ^ { 3 } }$

  3. $\dfrac { r M } { \pi l \rho R ^ { 3 } }$

  4. $\dfrac { l M } { \pi l \rho R ^ { 3 } }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Poisson's ratio = - (dr/r) / (dl/l). Using the volume of the wire V = pi * R^2 * l, and assuming constant volume or relating the changes, the expression for Poisson's ratio is derived as (r * M) / (pi * l * rho * R^3).

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

A uniform bar of length 'L' and cross sectional area 'A' is subjected to a tensile load 'F'. 'Y' be the Young modulus and '$\sigma$' be the Poisson's ratio then volumetric strain is  

  1. $\frac{F}{AY}(1 - \sigma)$

  2. $\frac{F}{AY}(2 - \sigma)$

  3. $\frac{F}{AY}(1 - 2\sigma)$

  4. $\frac{F}{AY} \sigma$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volumetric strain is defined as the change in volume divided by the original volume. For a bar under uniaxial stress, the longitudinal strain is F/(AY) and the lateral strains are -sigma * F/(AY). Summing these gives the volumetric strain: F/(AY) * (1 - 2*sigma).

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

A copper rod of length $l$ is suspended from the ceiling by one of its ends. Find the relative increment of its volume $\displaystyle\frac{\Delta V}{V}$.

  1. $\displaystyle\frac{\Delta V}{V}=(1-2\mu)\frac{\Delta l}{l}$

  2. <span>$\displaystyle\frac{\Delta V}{V}=(1-3\mu)\frac{\Delta l}{l}$</span>

  3. <span>$\displaystyle\frac{\Delta V}{V}=(1-2\mu)\frac{2\Delta l}{l}$</span>

  4. <span>$\displaystyle\frac{\Delta V}{V}=(1-3\mu)\frac{3\Delta l}{l}$</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can take copper rod as cylindrical rod

$v=\pi r^2 l$

$E=\dfrac{\Delta l}{l}$ (longitudinal strain)

$E'=\dfrac{\Delta r}{r}=-\mu E$  ,where $\mu$ is Poisson ratio,$E'$ is lateral strain

$\dfrac{\Delta V}{V}=\dfrac{2\Delta r}{r}+\dfrac{\Delta l}{l}$

$\dfrac{\Delta V}{V}=(1-2\mu)\dfrac{\Delta l}{l}$

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

One end of a wire $2$ m long and diameter $2$ mm, is fixed in a ceiling. A naughty boy of mass $10$ kg jumps to catch the free end and stays there. The change in length of wire is (Take $g=10m/s^2, Y=2\times 10^{11} N/m^2$).
In above problem, if Poisson's ratio is $\sigma =0.1$, the change in diameter is?

  1. $3.184\times 10^{-5}$ m

  2. $31.84\times 10^{-5}$ m

  3. $3.184\times 10^{-8}$ m

  4. $31.84\times 10^{-8}$ m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

First, calculate longitudinal strain: delta_L/L = F/(AY) = (mg)/(pi * r^2 * Y). Then, use Poisson's ratio sigma = -(delta_d/d) / (delta_L/L) to find the change in diameter. Plugging in the values yields 3.184 * 10^-8 m.