JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Moment of inertia of a cylinder of mass \(M,\) length \(L\) and radius \(R\) about an axis passing through its centre and perpendicular to the axis of the cylinder is \(I = M \left(\frac{ R ^{2}}{4}+\frac{ L ^{2}}{12}\right) .\) If such a cylinder is to be made for a given mass of material, the ratio \(\frac LR\) for it to have minimum possible \(I\) is
- A \(\sqrt{\frac{2}{3}}\)
- B \(\frac{3}{2}\)
- C \(\sqrt{\frac{3}{2}}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{3}{2}}\)
Step-by-step Solution
Detailed explanation
\(I=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}\right)\)\(...(1)\) as mass is constant \(\Rightarrow m =\rho V =\) constant \(V = constant\) \(\pi^{2} R l=\) constant \(\Rightarrow R ^{2} L =\) constant \(2 RL + R ^{2} \frac{ dL }{ dR }=0\)\(...(2)\) From equation \((1)\)…
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