The relationship between Young's modulus Y, B | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The relationship between the elastic moduli is derived from the Poisson's ratio $\sigma$. We have the following standard relations: $Y = 3K(1 - 2\

Vedclass · Accepted Answer

$Y = \frac{9\eta K}{\eta + 3K}$

Vedclass · Answer

(A) The relationship between the elastic moduli is derived from the Poisson's ratio $\sigma$. We have the following standard relations: $Y = 3K(1 - 2\sigma)$ $Y = 2\eta(1 + \sigma)$ From the first equation,$1 - 2\sigma = \frac{Y}{3K} \implies 2\sigma = 1 - \frac{Y}{3K} \implies \sigma = \frac{1}{2} - \frac{Y}{6K}$. From the second equation,$1 + \sigma = \frac{Y}{2\eta} \implies \sigma = \frac{Y}{2\eta} - 1$. Equating the two expressions for $\sigma$: $\frac{Y}{2\eta} - 1 = \frac{1}{2} - \frac{Y}{6K}$ $\frac{Y}{2\eta} + \frac{Y}{6K} = \frac{3}{2}$ Multiplying by $6\eta K$: $3YK + Y\eta = 9\eta K$ $Y(3K + \eta) = 9\eta K$ $Y = \frac{9\eta K}{3K + \eta}$ Thus,the correct option is $A$.

The relationship between Young's modulus $Y,$ Bulk modulus $K,$ and modulus of rigidity $\eta$ is:

Similar Questions

$A$ copper wire of cross-sectional area $0.01 \ cm^2$ is under a tension of $22 \ N$. Find the percentage change in the cross-sectional area (Young's modulus of copper $= 1.1 \times 10^{11} \ N \ m^{-2}$ and Poisson ratio $= 0.32$).

Explain Poisson's ratio and show that its value is less than $0.5$.

$A$ material has a Poisson's ratio of $0.50$. If a uniform rod made of this material suffers a longitudinal strain of $2 \times 10^{-3}$,what is the percentage change in its volume (in $\%$)?

Liquids have no Poisson's ratio,because

If there is no change in the volume of a wire upon stretching, then the Poisson's ratio for the material of the wire is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module