JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2022
- Medium
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\frac{ m \omega^{2} I_{0}}{ k - m \omega^{2}}\)
Step-by-step Solution
Detailed explanation
\(K \Delta x = m \left(\ell_{0}+\Delta x \right) w ^{2}\)
\(K \Delta x = m \ell_{0} w ^{2}+ mw ^{2} \Delta x\)
\(\Delta x =\frac{ m \ell_{0} w ^{2}}{ k - mw ^{2}}\)
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