Two uniform thin identical rods. AB and CD each of mass M and length I are joined so as to form a cross as shown in fig. The moment of inertia of the crossabout the bisector line EF is
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a
ML26
b
ML24
c
ML212
d
ML23
answer is C.
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Detailed Solution
The line EF makes an angle of 450 with each rod .Therefore I = I1 +I2 =( 112ML2sin2 450 )×2=112ML2