JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 :
- A \(\alpha\)
- B \(\alpha / 4\)
- C \(\alpha / 8\)
- D \(\alpha / 2\)
Answer & Solution
Correct Answer
(B) \(\alpha / 4\)
Step-by-step Solution
Detailed explanation
\(\alpha=\frac{\mathrm{M} \ell^2}{12}\) \(\begin{aligned} & \alpha^{\prime}=2\left[\frac{\frac{\mathrm{M}}{2}\left(\frac{\ell}{2}\right)^2}{12}\right] \\ & \alpha^{\prime}=\frac{\mathrm{M} \ell^2}{48}=\frac{\alpha}{4}\end{aligned}\) Correct option is (2)
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