Consider the setup illustrated in the figure. The applied force F...

Let $a$ be the acceleration of the system and $T$ be the tension in the string.

Assuming the block of mass $m_2$ moves downwards and $m_1$ moves to the right, the equations of motion are:
For mass $m_2$:
$m_2 g - T = m_2 a$

For mass $m_1$:
$T - F = m_1 a$

Adding the two equations, we get:
$m_2 g - F = (m_1 + m_2) a$

Given that $F = \dfrac{m_2 g}{2}$, substituting this into the equation:
$m_2 g - \dfrac{m_2 g}{2} = (m_1 + m_2) a$

$\dfrac{m_2 g}{2} = (m_1 + m_2) a$

$a = \dfrac{m_2 g}{2(m_1 + m_2)}$

Now, substituting the value of $a$ in the equation for $m_2$ to find the tension $T$:
$T = m_2 g - m_2 a$

$T = m_2 g - m_2 \left( \dfrac{m_2 g}{2(m_1 + m_2)} \right)$

$T = m_2 g \left( 1 - \dfrac{m_2}{2(m_1 + m_2)} \right)$

$T = m_2 g \left( \dfrac{2m_1 + 2m_2 - m_2}{2(m_1 + m_2)} \right)$

$T = \dfrac{m_2 g (2m_1 + m_2)}{2(m_1 + m_2)}$

The strain developed in the string is given by the ratio of stress to Young's modulus:
$\text{Strain} = \dfrac{\text{Stress}}{Y} = \dfrac{T}{A Y}$

Substituting the value of $T$:
$\text{Strain} = \dfrac{m_2 g (2m_1 + m_2)}{2 A Y (m_1 + m_2)}$