JEE Main 2020 (3 Sep, Shift 1) Physics Q4: Moment of Inertia

Question

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 a material, the ratio L/R for it to have minimum possible I is :

(1) $\sqrt{\frac{3}{2}}$ Official answer
(2) $\frac{2}{3}$
(3) $\sqrt{\frac{2}{3}}$
(4) $\frac{3}{2}$

Official answer

Option 1

NTA final key (Sep 2020).

Accepted Answer

Option 1: $\sqrt{\frac{3}{2}}$

Answer

$\sqrt{\frac{3}{2}}$

Answer

$\frac{2}{3}$

Answer

$\sqrt{\frac{2}{3}}$

Answer

$\frac{3}{2}$

JEE Main 3 September 2020, Shift 1, Physics Q4

Same idea in other shifts