A thin rod of length 41 and mass 4 ~m is bent at the points as...

Total moment of inertia =$\mathrm{I}_{1}+$\mathrm{I}$_{2}+$\mathrm{I}$_{3}+$\mathrm{I}$_{4}=$2 $\mathrm{I}_{1}+$2 $\mathrm{I}_{2}$$\begin{array}{l} =2\left(l_{1}+l_{2}\right)\left\lceil I_{3}=I_{1}, I_{1}=I_{4}\right] \\ \text{ Now, } I_{2}=I_{3}=\frac{M I^{2}}{3} \end{array}$$$Using parallel axes theorem, we have$$\begin{array}{l} \mathrm{I}=\mathrm{I}_{\mathrm{CM}}+\mathrm{Mx}^{2} \text{\text{ and } } \mathrm{x}=\sqrt{l^{2}+\frac{l^{2}}{4}} \\ \mathrm{I}_{1}=\mathrm{I}_{4}=\frac{\mathrm{M} l^{2}}{12}+\mathrm{M}\left[\sqrt{l^{2}+\left(\frac{l}{2}\right)^{2}}\right]^{2} \end{array}$$Putting all values we get Moment of inertia,$\mathrm{I}=10\left(\frac{\mathrm{Ml}^{2}}{3}\right)$