Question

Two uniform thin rods each of mass ${}^{\prime }{m}^{\prime }$ and length ${}^{\prime }{L}^{\prime }$ are arranged to form a cross. The moment of inertia of the system about an angular bisector is

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

Answer 1

The correct option is A $\frac{m{L}^2}{12}$

Moment of inertia of the rod about its centre is I=$\frac{M{L}^2}{12}$

Moment of inertia of the system of rods which form a cross about an axis passing through the centre and perpendicular to the rods is I=$I_x+I_y=2I$=$\frac{2M{L}^2}{12}$

Moment of inertia of the system through angular bisectors of rods = $\frac{I}{2}=\frac{M{L}^2}{12}$