Two springs $A$ and $B$ $(K_{A} = 2 K_{B})$ are stretched by applying force of equal magnitudes at the four ends.If the energy stored in $A$ is $E$ , that in $B$ is:

E / 2

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2 E

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E

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E / 4

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Solution

The correct option is B $2 E$
According to question
$E_{A} = E; F_{A} = F_{B} = F$
$k_{A} = 2 K_{B}$
$∵ E_{A} = \frac{1}{2} K_{A} x_{A}^{2}$
$∵ F_{A} = F_{B}$
$K_{A} x_{A} = K_{B} x_{B}$
$2 K_{B} x_{A} = K_{b} x_{B}$ $[∵ K_{A} = 2 K_{B}]$
$X_{B} = 2 x_{A}$
$E_{B} = \frac{1}{2} K_{B} x_{B}^{2}$
$\frac{E_{B}}{E_{A}} = \frac{K_{B} x_{B}^{2}}{K_{A} x_{A}^{2}} = \frac{1}{2} \times 4 = 2$
$E_{B} = 2 E$

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