When a force is applied to a spring of force constant K, elongation produced in the spring is x. When the force is slowly increased to twice is present value, work done on the spring will be
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a
12Kx2
b
Kx2
c
3Kx22
d
2Kx2
answer is C.
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Detailed Solution
F=K·x, 2F=K·x1⇒x1=2FK∴ Work done =12Kx12-12Kx2=12K2FK2-12K·FK2=3F22K=3(Kx)22K=32Kx2