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Q.

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

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a

E

b

2E

c

$E/4$

d

$E/2$

answer is B.

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Detailed Solution

$E = \frac{1}{2} k x^{2} = \frac{F^{2}}{2k}$

$\frac{E_{A}}{E_{B}} = \frac{k_{B}}{k_{A}} = \frac{1}{2}$

$\frac{E_{A}}{E_{B}} = \frac{1}{2}, 2 E_{A} = E_{B}$

$E_{B} = 2E$

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