MHT CET · Physics · Laws of Motion
A body of mass 100 gram is tied to a spring of spring constant \(8 \mathrm{~N} / \mathrm{m}\), while the other end of a spring is fixed. If the body moves in a circular path on smooth horizontal surface with constant angular speed \(8 \mathrm{rad} / \mathrm{s}\) then the ratio of extension in the spring to its natured length will be
- A 1:1
- B 8:1
- C 2:1
- D \(4: 1\)
Answer & Solution
Correct Answer
(D) \(4: 1\)
Step-by-step Solution
Detailed explanation
\(kx = m(L_0+x)\omega^2\) \(\frac{x}{L_0} = \frac{m\omega^2}{k - m\omega^2}\)
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