question_answer

AB and CD are two identical rods each of length and mass m joined to form a cross as shown in the figure. The moment of inertia of these two rods about a bisector (XY)of angle between the rods is

A)  \[\frac{m{{\ell }^{2}}}{12}\]                     

B) \[\frac{m{{\ell }^{2}}}{3}\]

C) \[\frac{2m{{\ell }^{2}}}{3}\]                      

D) \[\frac{m{{\ell }^{2}}}{6}\] 

Correct Answer: A

Solution :

Moment of inertia of rods AB and CD about an axis passing through their point of intersection 0 and perpendicular to the plane of the rods is \[{{I}_{0}}=\frac{m{{\ell }^{2}}}{12}+\frac{m{{\ell }^{2}}}{6}\] In the figure. XY and X'Y? are two mutually perpendicular axes passing through 0 lying in the plane of the two rods. By symmetry \[{{I}_{XY}}={{I}_{X'Y'}}=I,\]say From the theorem of perpendicular axes \[{{I}_{XY}}+{{I}_{X'Y'}}={{I}_{0}}\] \[\therefore \]    \[2I=\frac{m{{\ell }^{2}}}{6}\]      \[I=\frac{m{{\ell }^{2}}}{12}\] Hence, the correction option is [a].