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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 N/m²

c

100 Nm

d

100 Nm^-1

answer is B.

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Detailed Solution

$U = \frac{1}{2} {Kx}^{2} - - (1), U = 50 x^{2} - - (2)$ , compare equation (1) and (2) to find K

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