Two masses of 8 kg & 4 kg are connected by a light rod of length 30 cm. Moment of inertia of system about an axis perpendicular to rod is I. Minimum value of I is
A
24×10−2kgm2
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B
36×102kgm2
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C
48×10−2kgm2
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D
108×10−2kgm2
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Solution
The correct option is A24×10−2kgm2 I=ICM+md2 {Minimum when system rotates about axis through Center of Mass(CM) i.e. d=0} ICM=m1x21+m2x22
By def 8x1=4x2 8x1=4(0.3−x1) 2x1=0.3−x1 3x1=0.3 x1=0.1