Moment of inertia of a rod of mass ' M ' and length 'L' about an axis passing through its center and normal to its length is ' α '. Now the rod is cut into two equal parts and these parts are joined symmetrically to

Correct answer: B. α / 4 α= M ^2 12 aligned & α^ =2[ M 2 (( )/(2))^2 12 ] & α^ = M ^2 48 =(α)/(4) aligned Correct option is (2)

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Physics
Chapter: Rotational Motion

JEE Main 2025 · 2 April · Morning Shift · Physics · Rotational Motion

Moment of inertia of a rod of mass ' M ' and length 'L' about an...

Moment of inertia of a rod of mass ' M ' and length 'L' about an axis passing through its center and normal to its length is '

\alpha

'. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is : Moment of inertia of a rod of mass ' M ' and length 'L' about an axis passing through its center and normal to its

Moment of inertia of a rod of mass ' M ' and length 'L' about an axis passing through its center and normal to its

Options

A $\alpha$
B $\alpha / 4$
C $\alpha / 8$
D $\alpha / 2$

Correct answer

B. $\alpha / 4$

Step-by-step solution

\alpha=\frac{\mathrm{M} \ell^2}{12}

\begin{aligned} & \alpha^{\prime}=2\left[\frac{\frac{\mathrm{M}}{2}\left(\frac{\ell}{2}\right)^2}{12}\right] \\ & \alpha^{\prime}=\frac{\mathrm{M} \ell^2}{48}=\frac{\alpha}{4}\end{aligned}

Correct option is (2)

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