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(B) Let $l$ be the original length of the spring and $k$ be the spring constant.When the body is whirled in a horizontal plane,the centripetal force i

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$15$

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(B) Let $l$ be the original length of the spring and $k$ be the spring constant.When the body is whirled in a horizontal plane,the centripetal force is provided by the spring force.For the first case,the angular velocity is $\omega$ and elongation $x_1 = 1 \, cm$:$m \omega^2 (l + x_1) = k x_1$  --- $(1)$For the second case,the angular velocity is $2\omega$ and elongation $x_2 = 5 \, cm$:$m (2\omega)^2 (l + x_2) = k x_2$  --- $(2)$Dividing equation $(2)$ by equation $(1)$:$\frac{4 \omega^2 (l + x_2)}{\omega^2 (l + x_1)} = \frac{k x_2}{k x_1}$$4 \frac{l + 5}{l + 1} = \frac{5}{1}$$4(l + 5) = 5(l + 1)$$4l + 20 = 5l + 5$$l = 15 \, cm$.Thus,the original length of the spring is $15 \, cm$.

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