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The length of an elastic string is ‘a’ meters when the tension is 4 N and is ‘b’ meters when the tension is 5 N. The length (in meters) when the tension is 9 N is

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$2 \sqrt{2} : 1$

$(\frac{4 b - a}{2})$

$5 b - 4 a$

$(\frac{7 b - 2 a}{5})$

answer is C.

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Detailed Solution

$x = \frac{F}{K}$

$\Rightarrow a = L + x_{1} = L + \frac{4}{K}$ ...... (i)

&, $b = L + x_{2} = L + \frac{5}{K}$ ...... (ii)

$\Rightarrow c = L + x_{3} = L + \frac{9}{K}$

from (i) and (ii),

$\frac{1}{K} = (b - a) and L = (5 a - 4 b)$

$\Rightarrow c = (5 a - 4 b) + 9 (b - a) = (5 b - 4 a)$

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