Question

A steel wire of length 30 cm is stretched to increase its length by 0.2cm. Find the lateral strain in the wire if the Poisson’s ratio for steel is 0.19.
(A) 0.0012
(B) 0.0080
(C) 0.19
(D) 0.008

Answer 1

Hint:
We are given with steel wire whose length is mentioned and upon application of some force, it gets extended by 0.2 cm. We are given the value of change in length. We need to find the lateral strain in the wire given is the Poisson ratio

Complete Step by Step Solution:
We know Poisson’s ratio gives us the ratio of expansion produced to the axial compression in the wire when it experiences the force. It is a dimensionless quantity. It is denoted by Greek letter nu \[\nu \]
\[\nu =\frac{{{\varepsilon }_{t}}}{{{\varepsilon }_{l}}}\]
Here, \[{{\varepsilon }_{t}}\] is transverse strain and \[{{\varepsilon }_{l}}\] is a longitudinal strain.
Given values, \[\Delta l\]=0.2 cm and \[l\] = 30 cm
Putting them in the above mentioned formula,
\[{{\varepsilon }_{l}}=\frac{0.2}{30}=0.0067\]
Now Poisson’s ratio is \[\nu =\dfrac{{{\varepsilon }_{t}}}{{{\varepsilon }_{l}}}\]
\[
\Rightarrow 0.19=\dfrac{{{\varepsilon }_{t}}}{0.0067} \\
\Rightarrow {{\varepsilon }_{t}}=0.19\times 0.0067 \\
 \Rightarrow {{\varepsilon }_{t}}=0.0012 \\
\]
So, the correct option is (A)

Additional Information:
Stress is defined as the force acting per unit area and strain is defined as the ratio of change in length to the original length. The unit of stress is pascals and strain are a dimensionless quantity.

Note:
Poisson’s ratio range is from -1 to 0.5. Another term which is commonly used is the Poisson effect. It is when a force is applied to stretch the body in a given direction it tends to compress in a direction perpendicular to that force. For example, a rubber band tends to become thinner when stretched.