JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
The moment of inertia of a uniform cylinder of length \(l\) and radius \(R\) about its perpendicular bisector is \(I\). What is the ratio \(l/R\) such that the moment of inertia is minimum?
- A \(\sqrt {\frac{3}{2}} \;\;\;\;\;\;\;\;\;\;\)
- B \(\frac{{\sqrt 3 }}{2}\)
- C \(1\)
- D \(\frac{3}{{\sqrt 2 }}\)
Answer & Solution
Correct Answer
(A) \(\sqrt {\frac{3}{2}} \;\;\;\;\;\;\;\;\;\;\)
Step-by-step Solution
Detailed explanation
As we known, moment of inertia of a solid cylinder about an axis which is perpendicular bisector…
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