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Q.

The elastic potential energy of a stretched spring is given by E = 50x2 . Where x is the displacement in meter and E is in joule, then the force constant of the spring is

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a

50 Nm

b

100 Nm-1

c

100 N/m2

d

100 Nm

answer is B.

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Detailed Solution

U=12Kx2--(1), U=50x2--(2), compare equation (1) and (2) to find K
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