Physics - Elasticity Question with Solution | TestHub

$T_1=K\left(l-l_1\right)$

$T_2=K(l-l_2)$

So,$\frac{T_1}{T_2}=\frac{l_1-l}{l_2-l}$

$\therefore T_{1}l-T_1{l}_2=T_{2}l-T_2{l}_1$

$\left(T_1-T_2\right)l=T_1{l}_2-T_2{l}_1$

$l= \frac{T_1{l}_2-T_2{l}_1}{(T_1-T_2)}$

$l=\left(5a-4b\right)$...(i)

$k=\frac{1}{b-a }$...(ii)

So, length of wire when tension is 9 N

$9=k{l}^{\prime } (l^{\prime }=$change in length)

$9=\frac{1}{\left(b-a\right)} \times l^{\text{'}}\Rightarrow l^{\text{'}}= 9b-9a$

Hence, final length$=l+l^{\text{'}}$

$=5a-4a+9a-9a$

$l_0=5b-4a$