The length of a wire becomes l _ 1 and l _ 2 when 100N and 120N...

Young's modulus is given by $Y = \dfrac{\dfrac{F}{A}}{\dfrac{∆ l}{l}}$

$⇒ \dfrac{F}{A} = Y\left(\dfrac{\Deltal}{l}\right) ⇒ \Delta l = \dfrac{Fl}{AY}$

$⇒ l + \Delta l = {l}_{1} ⇒ \left(\dfrac{{F}_{1}L}{AY} + L\right) = L\left(\dfrac{{F}_{1}}{AY} + 1\right) = {l}_{1} . . . ( i )$

Similarly,

${l}_{2} = \left(\dfrac{{F}_{2}}{AY} + 1\right)L$

$⇒ 10{l}_{2} = 11{l}_{1}$

It is given that $10{l}_{1} = 11{l}_{2}$

$⇒ 10 \times L\left[\dfrac{{F}_{2}}{AY} + 1\right] = 11L\left[\dfrac{{F}_{1}}{AY} + 1\right]$

$⇒ \dfrac{10{F}_{2}}{AY} + 10 = \dfrac{11{F}_{1}}{AY} + 11$

Substituting ${F}_{1} = 100N, {F}_{2} = 120N$

$⇒ \dfrac{1200}{AY} - \dfrac{1100}{AY} = 1 ⇒ \dfrac{100}{AY} = 1$

$⇒ \dfrac{1}{AY} = \dfrac{1}{100}$

Using equation (i),

$L = \dfrac{{l}_{1}}{\left(\dfrac{{F}_{1}}{AY} + 1\right)} = \dfrac{{l}_{1}}{\dfrac{100}{100} + 1} = \dfrac{{l}_{1}}{2}$