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 EF (in the plane of cross) is
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a
ML26
b
ML24
c
ML212
d
ML23
answer is C.
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Detailed Solution
We have to find I1? I1 = I2----------(1) I0 = ML212+ML212 = ML26now I1 +I2 = I0--------(2)From (1) and (2), I1 = I02 = ML212
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 EF (in the plane of cross) is