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Question


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
24×102 kgm2
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B
36×102 kgm2
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C
48×102 kgm2
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D
108×102 kgm2
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Solution

The correct option is A 24×102 kgm2
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(0.3x1)
2x1=0.3x1
3x1=0.3
x1=0.1

Icm=8×(0.1)2+4×(0.2)2
=8×102+16×102
=24×102 kgm2

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