# AB and CD are two identical rods each of length l and mass m joined to form a cross. The moment of inertia of three two rods about a bisector XY of angles between the rods is

$\begin{array}{1 1} (a)\;\frac{ml^2}{12} & \quad (b)\;\frac{ml^3}{3} \\ (c)\;\frac{2ml}{3} & \quad (d)\;\frac{ml^2}{6} \end{array}$

## 1 Answer

$2 \times \large\frac{1}{12}$$ml^2 \sin ^2 45=\large\frac{ml^2}{12}$
answered Nov 27, 2013 by

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