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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A circular plate is rotating in horizontal plane, about an axis passing through its center and perpendicular to the plate, with an angular velocity \(\omega\). A person sits at the center having two dumbbells in his hands. When he stretches out his hands, the moment of inertia of the system becomes triple. If \(E\) be the initial Kinetic energy of the system, then final Kinetic energy will be \(\frac{E}{x}\).The value of \(x\) is \(....\)
- A \(3\)
- B \(6\)
- C \(9\)
- D \(12\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\(KE =\frac{ L ^2}{2 I } \Rightarrow \frac{ KE _{\text {fimpl }}}{ KE _{\text {initial }}}=\frac{ I _{\text {intitial }}}{ I _{\text {final }}} \Rightarrow \frac{ KE _{\text {fimpl }}}{ E }=\frac{1}{3}\) \(\Rightarrow KE _{\text {fimel }}=\frac{E}{3}\)
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