Two wires of same length and radius are joined end to end and...

In series combination, $\Deltal = {ℓ}_{1} + {ℓ}_{2}$

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

$⇒ \Deltaℓ ∝ \dfrac{ℓ}{Y}$

Equivalent length of rod after joining is $= 2ℓ$

As, lengths are same and force is also same in series

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

$\dfrac{{ℓ}_{eq}}{{Y}_{eq}} = \dfrac{ℓ}{{Y}_{1}} + \dfrac{ℓ}{{Y}_{2}} ⇒ \dfrac{2ℓ}{Y} = \dfrac{ℓ}{{Y}_{1}} + \dfrac{ℓ}{{Y}_{2}}$

$∴ Y = \dfrac{2{Y}_{1}{Y}_{2}}{{Y}_{1} + {Y}_{2}}$