Two blocks of mass 2 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is $$4.0 \times 10^{-5}$$ m and Young's modulus of the metal is $$2.0 \times 10^{11}$$ N m$$^{-2}$$. The longitudinal strain developed in the wire is $$\frac{1}{\alpha\pi}$$. The value of $$\alpha$$ is [Use $$g = 10$$ m s$$^{-2}$$]

Masses$$:m₁=2kg,m₂=4kg$$

For Atwood machine:

Acceleration:
$$a=\frac{(m₂−m₁)g}{(m₁+m₂)}$$
= $$\frac{\left((4−2)\times10\right)}{(2+4)}$$

$$=\frac{20}{6}=\frac{10}{3}$$

Tension:
$$T=m₁(g+a)$$
$$=2(10+10/3)$$
=$$2\times(40/3)$$
$$=\frac{80}{3}N$$

Now strain:

$$strain=stress/Y$$

$$Stress=T/A$$

Area of wire:
$$A=\pi r^2=\pi(4\times10⁻⁵)^2$$
$$=\pi\times16\times10⁻^1⁰$$
$$=16\pi\times10⁻^1⁰$$

So:

$$strain=T/(AY)$$

$$=(80/3)/[(16\pi\times10⁻^1⁰)(2\times10^{11})]$$

Simplify denominator:

$$16\times2=32$$
$$10⁻^1⁰\times10^{11}=10$$

$$Sodeno\min ator=32\pi\times10=320\pi$$

Now:

$$strain=(80/3)/(320\pi)$$
$$=80/(960\pi)$$
$$=1/(12\pi)$$

So given:

$$strain=1/(α\pi)$$

$$⇒α=12$$

Final answer:

$$12$$