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Work Energy and Power question

2001 · Q158
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Work Energy and Power question

2001 · Q158

NEETPhysicsWork Energy and PowerMCQ+4 / −1
Two springs A and B having spring constant KA and KB (KA = 2KB) are stretched by applying force of equal magnitude. If energy stored in spring A is EA then energy stored in B will be
  1. A
    2EA
  2. B
    EA/4
  3. C
    EA/2
  4. D
    4EA
View written solutionFree

Correct answer: A

Energy = 12Kx2=12F2K{1 \over 2}K{x^2} = {1 \over 2}{{{F^2}} \over K}21​Kx2=21​KF2​

KAKB=2{{{K_A}} \over {{K_B}}} = 2KB​KA​​=2

∴\therefore∴ EAEB=12⇒EB=2EA{{{E_A}} \over {{E_B}}} = {1 \over 2} \Rightarrow {E_B} = 2{E_A}EB​EA​​=21​⇒EB​=2EA​

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