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Q.

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
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