A metal wire of length $2.5m$and area of cross section $1.5 \times {10^{ - 6}}{m^2}$ is stretched through $2mm$. Calculate the work done during the stretching. ($Y = 1.25 \times {10^{11}}N{m^{ - 2}}$)
A. $0.15J$
B. $0.51J$
C. $1.5J$
D. $5.1J$

Hint:To solve this problem, we need to understand the concept of stretching the wire. In stretching a wire work is done against internal restoring forces. This work is stored as elastic potential energy or strain energy. We will first find the force and then average extension to determine the work done during the stretching.

Formula used:
$W = \dfrac{1}{2}Fx$
where, $W$ is the work done during the stretching of the wire, $F$ is the force and $x$ is the extension of the wire due to stretching
$F = \dfrac{{YAx}}{L}$
where, $F$ is the force, $Y$ is the young’s modulus, $A$ is the area of cross section of wire, $x$ is the extension of the wire due to stretching and $L$ is the length of the wire

Complete step by step answer:
We are given the total length of wire $L = 2.5m$.
Area of cross section $A = 1.5 \times {10^{ - 6}}{m^2}$.
Extension $x = 2mm$, Young’s modulus $Y = 1.25 \times {10^{11}}N{m^{ - 2}}$
We know that the work done by the stretching is given by $W = \dfrac{1}{2}Fx$ and the force is given by $F = \dfrac{{YAx}}{L}$
$
W = \dfrac{{YA{x^2}}}{{2L}} \\
\Rightarrow W = \dfrac{{\left( {1.25 \times {{10}^{11}}} \right) \times \left( {1.5 \times {{10}^{ - 6}}} \right) \times {{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}}{{2 \times 2.5}} \\
\therefore W= 0.15J \\ $
Thus, the work done during the stretching is $0.15J$.

Hence, option A is the right answer.

Note:We have used the formula $F = \dfrac{{YAx}}{L}$ to determine the force. This formula is derived from the definition of Young’s modulus which is the ratio of stress and strain. The stress is defined as force applied per unit area and strain is defined as the ratio of change in length to the original length.
$
Y = \dfrac{{Stress}}{{Strain}} = \dfrac{{\dfrac{F}{A}}}{{\dfrac{x}{L}}} \\
\Rightarrow F = \dfrac{{YAx}}{L} \\ $