ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion

One end of a massless spring of spring constant \(k\) and natural length \(l_{0}\) is fixed while the other end is connected to a small object of mass \(m\) lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity \(\omega\) about an axis passing through fixed end, then the elongation of the spring will be

  1. A \(\frac{ k - m \omega^{2} l_{0}}{ m \omega^{2}}\)
  2. B \(\frac{ m \omega^{2} l_{0}}{ k + m \omega^{2}}\)
  3. C \(\frac{ m \omega^{2} I_{0}}{ k - m \omega^{2}}\)
  4. D \(\frac{ k + m \omega^{2} l_{0}}{m \omega^{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{ m \omega^{2} I_{0}}{ k - m \omega^{2}}\)

Step-by-step Solution

Detailed explanation

\(K \Delta x = m \left(\ell_{0}+\Delta x \right) w ^{2}\) \(K \Delta x = m \ell_{0} w ^{2}+ mw ^{2} \Delta x\) \(\Delta x =\frac{ m \ell_{0} w ^{2}}{ k - mw ^{2}}\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app