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(B) According to Hooke's Law, the extension of an elastic string is proportional to the applied tension. Let $l$ be the natural length of the string a

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$5b-4a$

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(B) According to Hooke's Law, the extension of an elastic string is proportional to the applied tension. Let $l$ be the natural length of the string and $k$ be the force constant.The length of the string under tension $T$ is given by $L = l + \frac{T}{k}$.For $T_1 = 4 \ N$, $L_1 = a = l + \frac{4}{k} \implies \frac{4}{k} = a - l$  (Equation $1$).For $T_2 = 5 \ N$, $L_2 = b = l + \frac{5}{k} \implies \frac{5}{k} = b - l$  (Equation $2$).Subtracting Equation $1$ from Equation $2$: $\frac{5}{k} - \frac{4}{k} = (b - l) - (a - l) \implies \frac{1}{k} = b - a$.Substituting $\frac{1}{k}$ into Equation $1$: $a = l + 4(b - a) \implies a = l + 4b - 4a \implies l = 5a - 4b$.Now, for $T_3 = 9 \ N$, the length $x$ is $x = l + \frac{9}{k}$.Substituting $l = 5a - 4b$ and $\frac{1}{k} = b - a$:$x = (5a - 4b) + 9(b - a) = 5a - 4b + 9b - 9a = 5b - 4a$.

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

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