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Question

A system consists of particles of mass 4 kg placed at (4,0),6 kg placed at (0,4) and 3 kg placed at (3,4). How far from the origin must a particle of mass 10 kg be placed so that the moment of inertia of the system about the origin becomes 300 kgm2?

A
25 m
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B
2.5 m
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C
10 m
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D
1 m
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Solution

The correct option is B 2.5 m
Let the particles be A,B,C and D
Mass of A=mA=4 kg
Position of A=(4,0)
Distance of A from origin =rA=(40)2+(00)2
rA=4 m
Mass of B=mB=6 kg
Position of B=(0,4)
Distance of B from origin =rB=(00)2+(40)2
rB=4 m
Mass of C=3 kg
Position of C=(3,4)
Distance of C from origin =(30)2+(40)2
rC=9+16=25=5 m
Mass of D=10 kg
Moment of inertia of the system,
I=mAr2A+mBr2B+mCr2C+mDr2D
300=4×42+6×42+3×52+10×r2D
300=64+96+75+10r2D
10r2D=65r2D=6.5
rD=6.5 m=2.5 m

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