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Two springs of spring constants 1000 Nm^-1 and 2000 Nm^-1 are stretched with same force. They will have potential energy in the ratio of

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2 : 1

2² : 1²

1 : 2

1² : 2²

answer is A.

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Detailed Solution

$\begin{array}{l} U = \frac{1}{2} {kx}^{2} = \frac{1}{2} k {(\frac{F}{k})}^{2} = \frac{F^{2}}{2 k} \\ U_{1} \times k_{1} = U_{2} \times k_{2} or \frac{U_{1}}{U_{2}} = \frac{k_{2}}{k_{1}} = \frac{2000}{1000} = \frac{2}{1} \end{array}$

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