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(B) According to Hooke's Law,the extension of an elastic string is proportional to the applied tension. Let $L_0$ be the natural length of the string

Vedclass · Accepted Answer

$5b-4a$

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(B) According to Hooke's Law,the extension of an elastic string is proportional to the applied tension. Let $L_0$ be the natural length of the string and $k$ be the force constant.The length $L$ under tension $T$ is given by $L = L_0 + \frac{T}{k}$.For $T_1 = 4 \ N$,$L_1 = a = L_0 + \frac{4}{k}$  --- $(1)$For $T_2 = 5 \ N$,$L_2 = b = L_0 + \frac{5}{k}$  --- $(2)$Subtracting $(1)$ from $(2)$: $b - a = \frac{5}{k} - \frac{4}{k} = \frac{1}{k} \Rightarrow k = \frac{1}{b-a}$.Substituting $k$ into $(1)$: $L_0 = a - 4(b-a) = a - 4b + 4a = 5a - 4b$.Now,for $T_3 = 9 \ N$,the length $x$ is:$x = L_0 + \frac{9}{k} = (5a - 4b) + 9(b - a)$$x = 5a - 4b + 9b - 9a$$x = 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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