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(D) Given: Mass $m = 100 \ g = 0.1 \ kg$,Spring constant $k = 8 \ N/m$,Angular speed $\omega = 8 \ rad/s$.Let the natural length of the spring be $l_0

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$4$

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(D) Given: Mass $m = 100 \ g = 0.1 \ kg$,Spring constant $k = 8 \ N/m$,Angular speed $\omega = 8 \ rad/s$.Let the natural length of the spring be $l_0$ and the extension be $x$. The total radius of the circular path is $r = l_0 + x$.The centripetal force required for circular motion is provided by the spring force: $F_c = F_s$.$m \omega^2 r = kx$.$0.1 	imes (8)^2 	imes (l_0 + x) = 8x$.$0.1 	imes 64 	imes (l_0 + x) = 8x$.$6.4(l_0 + x) = 8x$.$6.4 l_0 + 6.4 x = 8x$.$6.4 l_0 = 8x - 6.4 x$.$6.4 l_0 = 1.6 x$.$\frac{x}{l_0} = \frac{6.4}{1.6} = 4$.Thus,the ratio of extension to natural length is $4:1$.

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