Question

Two uniform, thin identical rods each of mass $M$ and length $L$ are joined at middle so as to form a cross as shown. The moment of inertia of the cross about a bisector line $E F$ (in the plane of cross) is

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

Answer 1

Answer: (C)

We have to find $I_{1} ?$
$ I_{1}=I_{2} \ldots$ (1)
$I_{0}=\frac{M L^{2}}{12}+\frac{M L^{2}}{12}=\frac{M L^{2}}{6}$
now $I_{1}+I_{2}=I_{0}\ldots$ (2)
From (1) and (2), $I_{1}=\frac{I_{0}}{2}=\frac{M L^{2}}{12}$