Two blocks of mass 2kg and 4kg are connected by a metal wire going...

For the given system, $T = \left(\dfrac{2{m}_{1}{m}_{2}}{{m}_{1} + {m}_{2}}\right)g = \dfrac{80}{3}N$

Area of the wire, $A = \pi{r}^{2} = 16\pi \times {10}^{- 10}{m}^{2}$

Strain $= \dfrac{\Deltal}{l} = \dfrac{F}{AY} = \dfrac{T}{AY}$

$= \dfrac{\dfrac{80}{3}}{16\pi \times {10}^{- 10} \times 2 \times {10}^{11}} = \dfrac{1}{12\pi}$

$⇒ \alpha = 12$