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A body of mass 200 gram is tied to a spring of spring constant $12.5 \mathrm{~N} / \mathrm{m}$, while other end of spring is fixed at point ' $\mathrm{O}$ '. If the body moves about ' $\mathrm{O}$ ' in a circular path on a smooth horizontal surface with constant angular speed $5 \mathrm{rad} / \mathrm{s}$ then the ratio of extension in the spring to its natural length will be
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The correct answer is:
$2:3$
Let the normal length be $\mathrm{L}$ and the extension be $\mathrm{x}$.
$\therefore \quad$ Restoring Force $=$ Centripetal Force
$\begin{aligned}
& \mathrm{kx}=\mathrm{m}(\mathrm{L}+\mathrm{x}) \omega^2 \\
& 12.5 \mathrm{x}=0.2(\mathrm{~L}+\mathrm{x}) 25 \quad \ldots(\because \omega=5 \mathrm{rad} / \mathrm{s})
\end{aligned}$
$\begin{aligned} & 12.5 x=5(L+x) \\ & 7.5 x=5 L \\ \therefore \quad & \frac{x}{L}=\frac{5}{7.5}=\frac{2}{3}=2: 3\end{aligned}$
$\therefore \quad$ Restoring Force $=$ Centripetal Force
$\begin{aligned}
& \mathrm{kx}=\mathrm{m}(\mathrm{L}+\mathrm{x}) \omega^2 \\
& 12.5 \mathrm{x}=0.2(\mathrm{~L}+\mathrm{x}) 25 \quad \ldots(\because \omega=5 \mathrm{rad} / \mathrm{s})
\end{aligned}$
$\begin{aligned} & 12.5 x=5(L+x) \\ & 7.5 x=5 L \\ \therefore \quad & \frac{x}{L}=\frac{5}{7.5}=\frac{2}{3}=2: 3\end{aligned}$
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