JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2020
- Hard
- MCQ
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
Pick an option to check your answer.
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 \((1)\)
\(\frac{ dI }{ clR }= M \left(\frac{2 R }{4}+\frac{2 L }{12} \times \frac{ dI }{ dr }\right)=0\)
\(\frac{ R }{2}+\frac{ L }{6} \frac{ dI }{ dR }=0\)
Substituting value of \(\frac{ dI }{ d R }\) from eqution \((2)\)
\(\frac{ R }{2}+\frac{ L }{6}\left(\frac{-2 L }{ R }\right)=0\)
\(\frac{ R }{2}=\frac{ L^{2}}{3 R } \Rightarrow \frac{ L }{ R }=\sqrt{\frac{3}{2}}\)
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