Question

Two identical rods each of mass 'M' and length 'l' are joined in crossed position as shown in figure. The moment of inertia of this system about a bisector would be

• A. $\frac{M{l}^2}{6}$
• B. $\frac{M{l}^2}{12}$
• C. $\frac{M{l}^2}{3}$
• D. $\frac{M{l}^2}{4}$

Answer 1

The correct option is B $\frac{M{l}^2}{12}$

Moment of inertia of system about an axes which is perpendicular to plane of rods and passing through the common centre of rods $I_z=\frac{M{l}^2}{12}+\frac{M{l}^2}{12}=\frac{M{l}^2}{6}$
Again from perpendicular axes theorem $I_2=I_{B_1}+I_{B_2}=2I_{B_1}=2I_{B_2}=\frac{M{l}^2}{6}$ [As $I_{B_1}=I_{B_2}]$
$\therefore I_{B_1}=I_{B_2}=\frac{M{l}^2}{12}.$