Tag: physics

Let Y, K, n and $\sigma$ be the Young's Modulus, Bulk modulus, Modulus of Rigidity and Poisson's Ratio, respectively. 
Y = 3K (1 - 2$\sigma$) [Standard formula] 
Y = 2n (1 + $\sigma$) [Standard formula] 
Hence, 3K (1 - 2$\sigma$) = 2n (1 + $\sigma$) 
Now K and n are always positive, so 
i) If $\sigma$ be +ve, then RHS is always +ve. So LHS must also be +ve. Therefore, 2$\sigma$ < 1 or $\sigma$ <1/2 
ii) If $\sigma$ be -ve, then LHS will always be +ve. Therefore, 1+$\sigma$ > 0 or $\sigma$ > -1 
Thus the limiting values of Poisson's ratio are -1 < $\sigma$ < 1/2

The correct option is (a)

The formula that relates all three elastic constants is 9/Y = 3/n + 1/B

The correct option is (b)

We know that $Y=3B(1-\sigma)$. Substituting, Y=B, we get, $1/3=1-2 \sigma$ or poisson's ratio = 1/3

The correct option is (c)

In the formula derived by the student, in order that Y is positive, $\sigma<0.25$, else Y will be negative, which is not possible

Hence, poisson's ratio should be less than 0.25

The correct option is option(c)

Given,

Poisson's ratio $=-\cfrac{\epsilon _{lateral}}{\epsilon _{Longtudinal}}$

Given that,

Young’s modulus $Y=6.6\times {{10}^{10}}\,N/{{m}^{2}}$

Bulk modulus $B=11\times {{10}^{10}}\,N/{{m}^{2}}$

We know that,

  $ Y=3K\left( 1-2\mu  \right) $

 $ 6.6\times {{10}^{10}}=3\times 11\times {{10}^{10}}-66\times {{10}^{10}}\mu  $

 $ -\mu =\dfrac{\left( 6.6-33 \right)\times {{10}^{10}}}{66\times {{10}^{10}}} $

 $ \mu =0.4 $

Hence, the poisson’s ratio is $0.4$

Suppose, D be the diameter of the wire Poissons ratio,  

$σ=\frac { lateral strain }{ longitudinal strain } $

 $σ=\frac { \frac { ΔD }{ D }  }{ \frac { ΔL }{ L }  } $

 $\frac { ΔL }{ L } =0.025$

 $σ=0.004$

 $σ=\frac { \frac { ΔD }{ D }  }{ \frac { 1 }{ 40 }  } $

 $\frac { ΔD }{ D } =\frac { 1 }{ 40 } \times 0.4=0.01$