Question

Two rods each of mass $m$ and length $\ell$ are joined at the center to from a cross. The moment of inertia of this center of the rods and perpendicular to the plane formed by them, is:

• A. $\frac{m{\ell }^2}{12}$
• B. $\frac{m{\ell }^2}{6}$
• C. $\frac{m{\ell }^2}{3}$
• D. $\frac{m{\ell }^2}{2}$

Answer 1

The correct option is B $\frac{m{\ell }^2}{6}$

Moment of inertia for each rod $I=\frac{m{l}^2}{12}$

About the middle point and perpendicular to the plane thus making sum total for two rods

So,

Moment inertia will become $I^{\prime }=2I=2\times \frac{m{l}^2}{12}=\frac{m{l}^2}{6}$