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].You need to login to perform this action.
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