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Question

# Two identical solid sphere each of mass m=5 kg and radius r=1 m are joined at the ends of a light rod of length l=2 m to form a system as shown in figure. Radius of gyration of the system about an axis perpendicular to the length of rod and passing through center of mass of system is

A
44 m
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B
2.4 m
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C
3.2 m
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D
4.4 m
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Solution

## The correct option is D √4.4 m Given: Mass of each sphare m=5 kg Radius of sphere r=1 m We know, moment of inertia of the sphere about its center of mass Icom(sphere)=25mr2 Using parallel axes theorem, moment of inertia of individual sphere about the rotational axis Icom(system)=Icom(sphere)+md2 Here d=(r+l2)=1+22=2 ⇒Icom(system)=25mr2+m×22 =25×5×(1)2+5×(2)2=2+20=22 kg-m2 As there are two identical sphere having same axis of rotation, total moment of inertia about Icom(system) is =2×22=44 kg-m2 We know , Radius of gyration K=√Icom(system)Mass of systemK=√442×5 m=√4.4 m

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