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((R²)/4 + (L²)/12). If such a cylinder is to be made for a given mass of a material, the ratio L/R for it to have minimum possible I is

Rotational Motion: JEE Main 2020 Physics question

JEE Main 2020 · 3 Sep, Morning shift · Q8095

JEE Main 20203 Sep, Morning shiftPhysicsHardRotational Motion140× since 2002

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( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)

. If such a cylinder is to be made for a given mass of a material, the ratio

{L \over R}

for it to have minimum possible I is

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JEE Main 2020 Rotational Motion Question with Solution (3 Sep Morning shift, Q8095) | JEE Arc