Two blocks of masses m₁ and m₂ are joined by a wire of Young’s modulus Y via a massless pulley. The area of cross-section of the wire is S and its length is L. When the system is released, increase in length of the wire is

The correct option is 2m₁m₂gL/YS(m₁+m₂).

Let's break down the step-by-step derivation to find the increase in the length of the wire when the system is released:

Step 1: Understand the tension in the wire
When two blocks of masses m₁ and m₂ are suspended over a massless and frictionless pulley by a wire, the system accelerates under gravity. The tension T in the wire connecting the two masses in this Atwood machine setup is given by the standard formula:

$T=\frac{2m_1{m}_{2}g}{m_1+m_2}$

Step 2: Relate tension to Young's Modulus
Young's modulus Y of a material is defined as the ratio of tensile stress to tensile strain:

$Y=\frac{\text{Stress}}{\text{Strain}}$

Here, stress is the restoring force per unit cross-sectional area:

$\text{Stress}=\frac{T}{S}$

And strain is the fractional change in length:

$\text{Strain}=\frac{\Delta L}{L}$

where ΔL is the increase in length of the wire, S is the cross-sectional area, and L is the original length of the wire.

Step 3: Solve for the change in length (ΔL)
Substituting stress and strain into the Young's modulus formula, we get:

$Y=\frac{T/S}{\Delta L/L}=\frac{TL}{S\Delta L}$

Rearranging this equation to solve for the increase in length ΔL:

$\Delta L=\frac{TL}{YS}$

Step 4: Substitute the tension (T) into the equation
Substitute the value of T from Step 1 into the expression for ΔL:

$\Delta L=\frac{\left(\frac{2m_1{m}_{2}g}{m_1+m_2}\right)L}{YS}$

Simplifying the expression, we obtain:

$\Delta L=\frac{2m_1{m}_{2}gL}{YS(m_1+m_2)}$