Q.

Two springs A and B having spring constant K A and KB where KA = 2KB. These are stretched by applying force of equal magnitude. If energy stored in spring A is E then energy stored in B will be :

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a

2E

b

E4

c

E2

d

4E

answer is A.

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Detailed Solution

Energy stored =12Kx2=(Kx)22K⇒E∝1K
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Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
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