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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

  1. A \(\sqrt{\frac{2}{3}}\)
  2. B \(\frac{3}{2}\)
  3. C \(\sqrt{\frac{3}{2}}\)
  4. D \(\frac{2}{3}\)
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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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