JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A body of mass \(200\,g\) is tied to a spring of spring constant \(12.5\,N / m\), while the other end of spring is fixed at point \(O\). If the body moves about \(O\) in a circular path on a smooth horizontal surface with constant angular speed \(5\,rad / s\), 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
\(kx = m \left( L _0+ x \right) \omega^2\) \(\Rightarrow 12.5 x =\frac{1}{5}\left( L _0+ x \right) 25 \Rightarrow 1.5 x = L _0\) \(\Rightarrow \frac{ x }{ L _0}=\frac{2}{3}\)
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