Three blocks, each of same mass m, are connected with wires W₁ and W₂ of same cross-sectional area a and Young’s modulus Y. Neglecting friction the strain developed in wire W₂ is

The correct answer is 2mg/3aY.

Let us analyze the system to find the tension and strain in wire W₂. Under standard conditions, the setup consists of three blocks each of mass $m$ connected over a frictionless, massless pulley:
- Two blocks are suspended on one side (let's say the left side) and are connected to each other vertically by wire W₂.
- One block is suspended on the other side (the right side).
- The top block on the left side is connected to the right block by wire W₁ passing over the pulley.

Step 1: Calculate the acceleration of the system
Let $a$ be the common acceleration of the system, and $g$ be the acceleration due to gravity. Since the total mass on the left side ($2m$) is greater than the mass on the right side ($m$), the left side will accelerate downwards while the right side accelerates upwards.

The net pulling force on the system is the difference in gravitational forces on either side of the pulley:
$F_{\text{net}}=2mg-mg=mg$

The total mass being accelerated is:
$M_{\text{total}}=m+m+m=3m$

Using Newton's second law, the acceleration $a$ of the system is:
$a=\frac{F_{\text{net}}}{M_{\text{total}}}=\frac{mg}{3m}=\frac{g}{3}$

Step 2: Find the tension in wire W₂
Let $T_2$ be the tension in wire W₂. This wire supports the bottom-most block on the left side, which is accelerating downwards at $a=\frac{g}{3}$.
Writing the equation of motion for this bottom block:
$mg-T_2=ma$

Rearranging the equation to solve for $T_2$:
$T_2=mg-ma$

Substitute the value of $a=\frac{g}{3}$:
$T_2=mg-m\left(\frac{g}{3}\right)=\frac{2}{3}mg$

Step 3: Calculate the strain in wire W₂
The stress in wire W₂ is defined as the tension force per unit cross-sectional area $a$:
$\text{Stress}=\frac{T_2}{a}$

Substitute the value of $T_2$:
$\text{Stress}=\frac{2mg}{3a}$

Young's modulus $Y$ is given by the ratio of stress to strain:
$Y=\frac{\text{Stress}}{\text{Strain}}\Rightarrow \text{Strain}=\frac{\text{Stress}}{Y}$

Substitute the expression for stress to find the strain developed in wire W₂:
$\text{Strain}=\frac{2mg}{3aY}$