The length of an elastic string is $a$ metres when the longitudinal tension is $4 \mathrm{~N}$ and $b$…
The length of an elastic string is $a$ metres when the longitudinal tension is $4 \mathrm{~N}$ and $b$ metres when the longitudinal tension is $5 \mathrm{~N}$. The length of the string in metres when the longitudinal tension is $9 \mathrm{~N}$ is
$a-b$
$5 b-4 a$
$2 b-\frac{1}{4} a$
$4 a-3 b$
Solution
Let the initial length of the string be " $?$.
Tension $\propto$ Elongation of the string
$\begin{aligned} \frac{T_1}{T_2} & =\frac{a-l}{b-l} \Rightarrow \frac{4}{5}=\frac{a-l}{b-l} \\ 4 b-4 l & =5 a-5 l \\ l & =5 a-4 b \\ \text { Again, } \quad \frac{T_1}{T_3} & =\frac{a-l}{x-l}\end{aligned}$
$\begin{aligned} & x=\text { length of the string at tension } T(=9 \mathrm{~N}) \\ & \qquad \begin{aligned} \frac{4}{9} & =\frac{a-(5 a-4 b)}{x-(5 a-4 b)} \\ \frac{4}{9} & =\frac{-4 a+4 b}{x-(5 a-4 b)} \Rightarrow \frac{4}{9}=\frac{4(-a+b)}{x-(5 a-4 b)} \\ \Rightarrow x-(5 a-4 b) & =9 a+9 b \\ x & =-9 a+9 b+5 a-4 b \\ & =5 b-4 a\end{aligned}\end{aligned}$