Question

The increase in length on stretching a wire is 0.05%. If its poisson's ratio is 0.4, then its diameter:
(A) Reduce by 0.02%
(B) Reduce by 0.1%
(C) Reduce by 0.03%
(D) Decrease by 0.4%

Answer 1

Hint: The Poisson’s ratio for a wire is given. When the wire is stretched, the length increases by 0.05%. Now by definition, we know the Poisson’s ratio $\sigma = \dfrac{{{\text{lateral strain}}}}{{{\text{longitudinal strain}}}}$. Longitudinal strain is given, so we can find the lateral strain by substituting the given values.

Formula used:
Poisson’s ratio $\sigma = \dfrac{{{\text{lateral strain}}}}{{{\text{longitudinal strain}}}}$ $ \Rightarrow \sigma = \dfrac{{\dfrac{{\Delta D}}{D}}}{{\dfrac{{\Delta l}}{l}}}$

Complete step by step solution:
Let D and l be the diameter and length of the given wire respectively.
When the wire is stretched, the length increases by 0.05%.
We know, the Poisson’s ratio $\sigma = \dfrac{{{\text{lateral strain}}}}{{{\text{longitudinal strain}}}}$
$ \Rightarrow \sigma = \dfrac{{\dfrac{{\Delta D}}{D}}}{{\dfrac{{\Delta l}}{l}}} = 0.4$
Here, $\dfrac{{\Delta l}}{l}$= 0.05
$ \Rightarrow \sigma = \dfrac{{\dfrac{{\Delta D}}{D}}}{{0.05}} = 0.4$
$ \Rightarrow \dfrac{{\Delta D}}{D} = 0.05 \times 0.4 = 0.02$

Therefore, the correct answer is option (A), reduced by 0.02%.

Note: If the length of the wire increases on stretching, the diameter will decrease simultaneously. However, the ratio of lateral and longitudinal strain is always a constant for a material and is defined as the Poisson’s Ratio. It is denoted by the letter $\sigma $.