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Standard XII

Physics

Buoyant Force

Two springs A...

Question

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 stored in A is E, then in B is (assume equilibrium):

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 $2 E$
Force is same, thus $0.5 k_{a} x_{a} = 0.5 (k_{a} / 2) (x_{b})$

Thus $x_{b} = 2 x_{a}$

Energy in $A = U = 0.5 k_{A} x^{2} = E$

Energy in $B = 0.5 k_{b} x^{2} = 0.5 (k_{A} / 2) (2 x)^{2} = 2 E$

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Two springs A and B (kA=2kB) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, then in B is (assume equilibrium):

The correct option is B 2EForce is same, thus 0.5kaxa=0.5(ka/2)(xb)Thus xb=2xaEnergy in A=U=0.5kAx2=EEnergy in B=0.5kbx2=0.5(kA/2)(2x)2=2E

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 stored in A is E, then in B is (assume equilibrium):

The correct option is B $2 E$
Force is same, thus $0.5 k_{a} x_{a} = 0.5 (k_{a} / 2) (x_{b})$

Thus $x_{b} = 2 x_{a}$

Energy in $A = U = 0.5 k_{A} x^{2} = E$

Energy in $B = 0.5 k_{b} x^{2} = 0.5 (k_{A} / 2) (2 x)^{2} = 2 E$