A steel wire of length 3 . 2m( Y _ S = 2 . 0 × 10 ^ 11 N m ^ - 2 )...

When two wires are connected end to end, the tension in them will be same.

Change in length using Hooke's law can be found as,

$Y = \dfrac{Fl}{A ∆ l} ⇒ \Deltal = \dfrac{Fl}{AY}$

Total change in length can be written as,

$\Deltal = \Delta{l}_{1} + \Delta{l}_{2}$

$⇒ \Deltal = \dfrac{F{l}_{1}}{{A}_{1}{Y}_{1}} + \dfrac{F{l}_{2}}{{A}_{2}{Y}_{2}}$

$⇒ F = \dfrac{\Deltal}{\dfrac{{l}_{1}}{{A}_{1}{Y}_{1}} + \dfrac{{l}_{2}}{{A}_{2}{Y}_{2}}} = \dfrac{1 . 4 \times {10}^{- 3}}{\dfrac{3 . 2}{\pi{\left(1 . 4 \times {10}^{- 3}\right)}^{2} \times 2 . 0 \times {10}^{11}} + \dfrac{4 . 4}{\pi{\left(1 . 4 \times {10}^{- 3}\right)}^{2} \times 1 . 1 \times {10}^{11}}} ⇒ F = 1 . 54 \times {10}^{2} = 154N$