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(B) Let $L$ be the original length of the string and $k$ be the force constant of the string. The final length $L'$ is given by the sum of the origina

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$5 b-4 a$

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(B) Let $L$ be the original length of the string and $k$ be the force constant of the string. The final length $L'$ is given by the sum of the original length and the elongation: $L' = L + \frac{F}{k}$.For the first condition: $a = L + \frac{4}{k}$ $(i)$For the second condition: $b = L + \frac{5}{k}$ (ii)Subtracting equation $(i)$ from (ii): $b - a = \frac{5}{k} - \frac{4}{k} = \frac{1}{k}$,which implies $k = \frac{1}{b-a}$.Substituting $k$ into equation $(i)$: $L = a - 4(b-a) = a - 4b + 4a = 5a - 4b$.Now,for a longitudinal tension of $9 \ N$,the length $L_{new}$ is:$L_{new} = L + \frac{9}{k} = (5a - 4b) + 9(b-a) = 5a - 4b + 9b - 9a = 5b - 4a$.

The length of an elastic string is $a$ metre when the longitudinal tension is $4 \ N$ and $b$ metre when the longitudinal tension is $5 \ N$. The length of the string in metre when longitudinal tension is $9 \ N$ is

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