A wire is stretched such that its volume rema | vedclass

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

Question

(B) The Poisson's ratio $\sigma$ is defined as the negative ratio of transverse strain to longitudinal strain: $\sigma = -\frac{\Delta D/D}{\Delta L/L

Vedclass · Accepted Answer

$0.50$

Vedclass · Answer

(B) The Poisson's ratio $\sigma$ is defined as the negative ratio of transverse strain to longitudinal strain: $\sigma = -\frac{\Delta D/D}{\Delta L/L}$.For a wire of length $L$ and diameter $D$,the volume $V = \frac{\pi D^2 L}{4}$.Since the volume $V$ remains constant,$\frac{\Delta V}{V} = 2\frac{\Delta D}{D} + \frac{\Delta L}{L} = 0$.This implies $2\frac{\Delta D}{D} = -\frac{\Delta L}{L}$,or $\frac{\Delta D/D}{\Delta L/L} = -0.5$.Substituting this into the definition of Poisson's ratio: $\sigma = -(-0.5) = 0.5$.Therefore,for a material with constant volume under stretching,the Poisson's ratio is $0.5$.Thus,option $(B)$ is correct.

$A$ wire is stretched such that its volume remains constant. The Poisson's ratio of the material of the wire is

Similar Questions

There is no change in the volume of a wire due to change in its length on stretching. The Poisson's ratio of the material of the wire is

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

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

Which statement is true for a metal?

For silver,Young's modulus is $7.25 \times 10^{10} \ N/m^2$ and Bulk modulus is $11 \times 10^{10} \ N/m^2$. Its Poisson's ratio will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module