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Question

Energy stored in spring A=20 J, then energy stored in spring B is (under same extension of the spring)

A
20 J
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B
30 J
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C
40 J
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D
50 J
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Solution

The correct option is C 40 J
Let, EA = Energy stored in spring A
EB = Energy stored in spring B
K1 = spring constant for spring A
K2 = spring constant for spring B

We know that spring potential energy E=12kx2=Fx2

Hence, EB=2Fx2=Fx ...(i) As, for the same extension, xA=xB=x , (FB=2F)

Given that energy stored in the spring A is 20 J
EA=Fx2=20 J ...(ii) (FA=F)

Dividing (ii) by (i), we get
EAEB=F.x2F.x

EAEB=12

20EB=12
EB=40 J

Hence option C is the correct answer

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