Two springs A and B have force constants $k_{1} a n d k_{2}$ respectively. The ratio of the work done on A to that done on B in increasing their length by the same amount is

\frac{k_{1}}{k_{2}}

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\sqrt{\frac{k_{1}}{k_{2}}}

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\frac{k_{2}}{k_{1}}

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\sqrt{\frac{k_{2}}{k_{1}}}

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Solution

The correct option is A $\frac{k_{1}}{k_{2}}$
$F_{1} = k_{1} x, F_{2} = k_{2} x$
Work done $w_{1} = \frac{1}{2} k_{1} x^{2} a n d w_{2} = \frac{1}{2} k_{2} x^{2}$
$∴ \frac{w_{1}}{w_{2}} = \frac{k_{1}}{k_{2}}$
Hence the correct choice is (a).

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The correct option is A $\frac{k_{1}}{k_{2}}$
$F_{1} = k_{1} x, F_{2} = k_{2} x$
Work done $w_{1} = \frac{1}{2} k_{1} x^{2} a n d w_{2} = \frac{1}{2} k_{2} x^{2}$
$∴ \frac{w_{1}}{w_{2}} = \frac{k_{1}}{k_{2}}$
Hence the correct choice is (a).