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Spring Force

The lengths o...

Question

The lengths of spring are $l_{1}$ and $l_{2}$ when stretched with forces of $4 N$ and $5 N$ respectively. Its natural length is

l_{2} + l_{1}

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2 (l_{1} - l_{2})

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5 (l_{2} - 4 l_{1})

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(5 l_{1} - 4 l_{2})

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Solution

The correct option is D $(5 l_{1} - 4 l_{2})$
Let $l$ be natural length.

$l_{1}$ is length when $F = 4 N$ and
$l_{2}$ is length when $F = 5 N$ ;

From $F = k x;$ Where $x$ is deformation length
$\Rightarrow 4 = k (l_{1} - l) \Rightarrow 5 = k (l_{2} - l);$
$\Rightarrow \frac{4}{5} = \frac{l_{1} - l}{l_{2} - l}$
$\Rightarrow 4 l_{2} - 4 l = 5 l_{1} - 5 l$
$\Rightarrow l = 5 l_{1} - 4 l_{2}$

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