The spring constant of two springs are $K_{1}$ and $K_{2}$ respectively springs are stretch up to that limit when potential energy both becomes equal. The ratio of applied force $(F_{1}$ and $F_{2})$ on them will be:

K_{1} : K_{2}

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K_{2} : K_{1}

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\sqrt{K_{1}} : \sqrt{K_{2}}

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\sqrt{K_{2}} : \sqrt{K_{1}}

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Solution

The correct option is C $\sqrt{K_{1}} : \sqrt{K_{2}}$
Let for same potential energy, elongations of springs be $x_{1}$ and $x_{2}$ and there respective force constants be $K_{1}$ and $K_{2}$ .
Then we have,
$\frac{1}{2} K_{1} x_{1}^{2} = \frac{1}{2} K_{2} x_{2}^{2}$
Hence,
$\Rightarrow \frac{x_{1}^{2}}{x_{2}^{2}} = \frac{K_{2}}{K_{1}}$
$\Rightarrow \frac{f_{1}}{f_{2}} = \frac{K_{1} x_{1}}{K_{2} x_{2}} = \frac{K_{1}}{K_{2}} \times \sqrt{\frac{K_{2}}{K_{1}}} = \sqrt{\frac{K_{1}}{K_{2}}}$
$= \frac{\sqrt{K_{1}}}{\sqrt{K_{2}}}$

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