Answer 1

Given:
Force (F) = m₂g/2
Area of cross-section of the string = A
 Young's modulus = Y
Let a be the acceleration produced in block m₂ in the downward direction and T be the tension in the string.
From the free body diagram:

\[\text{m}_2 \text{g - T = m}_2 \text{a} . . . \left( \text{i} \right)\]
\[\text{ T - F = m}_1 \text{a . . . (ii)}\]

From equations (i) and (ii), we get:

\[a = \frac{\text{m}_2 \text{g - F}}{\text{m}_1 + \text{m}_2}\]
\[\text{ Applying F }= \frac{\text{m}_2 \text{g}}{2}\]
\[ \Rightarrow a = \frac{\text{m}_2 \text{g}}{2\left( \text{m}_1 + \text{m}_2 \right)}\]

Again, T = F + m₁a

On applying the values of F and a, we get: 

\[\Rightarrow T = \frac{\text{m}_2 \text{g}}{2} + \text{m}_1 \frac{\text{m}_2 \text{g}}{2\left( \text{m}_1 + \text{m}_2 \right)}\]

We know that:

\[\text{Y} = \frac{\text{FL}}{\text{A ∆ L}}\]
\[ \Rightarrow \text{ Strain }= \frac{∆ \text{L}}{\text{L}} = \frac{\text{F}}{\text{AY}}\]
\[ \Rightarrow \text{ Strain  } = \frac{\left( \text{m}_2^2 + 2 \text{m}_1 \text{m}_2 \right)\text{g}}{2\left( \text{m}_1 + \text{m}_2 \right) \text{AY}}\]
\[ = \frac{\text{m}_2 g\left( 2 \text{m}_1 + \text{m}_2 \right)}{2\text{AY} \left( \text{m}_1 + \text{m}_2 \right)}\]

∴ Required strain developed in the string \[ = \frac{\text{m}_2 g\left( 2 \text{m}_1 + \text{m}_2 \right)}{2\text{AY} \left( \text{m}_1 + \text{m}_2 \right)}\] .