MHT CET · Physics · Laws of Motion
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

- A \(1: 2\)
- B \(1:1\)
- C \(2:3\)
- D \(2:5\)
Answer & Solution
Correct Answer
(C) \(2:3\)
Step-by-step Solution
Detailed explanation
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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