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 the longitudinal tension is $9\, N$ is
Diffcult
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(b) Let $L$ is the original length of the wire and $K$ is force constant of wire.
Final length $=$ initial length $+$ elongation
$L' = L + \frac{F}{K}$
For first condition $a = L + \frac{4}{K}$…$(i)$
For second condition $b = L + \frac{5}{K}$…$(ii)$
By solving $(i)$ and $(ii)$ equation we get
$L = 5a - 4b$ and $K = \frac{1}{{b - a}}$
Now when the longitudinal tension is $9N,$
length of the string $=$ $L + \frac{9}{K}$= $5a - 4b + 9(b - a)$$x = 5b - 4a$.
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