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`        3particals of masses 100gmeach kept at corners of equilatral triangle of side 2m the M.I.of systemabout atrancevers axis passing through it's centroid is....? A.0.4kg-m2      B.0.6 kg-m2 C.0.8kg-m2  D.1.2kg-m2`
7 years ago

510 Points
```										distance of any corner from center is d =a/sqrt3...            (a is side of triangle)
moment of inertia of system is I=(I1+I2+I3)
mass and distance from axis of rotation is same for all hree particles  so I1=I2=I3
total moment of inertia    I = 3I1 = md2 =3(0.1. a2 /3)=0.4kg-m2
hence option A is correct
```
7 years ago
Devasish Bindani
45 Points
```										all the masses are kept at a distance of 2/31/2 m from the axis at the centroid.
there fore m.i. of each mass at the axis willl be same i.e.=0.4/3kgm2 on adding all the three moments of inertia it =0.4kgm2
thus according to me the answer is <a> if its incorrect please send me the correct explanation if possible
```
7 years ago
Priya Gorakshnath Anandkar
18 Points
```										plz solve my problem &give me list of reagents in org chem with there spatial feachers.  !!!!!!!!!!!!!!!!!!!
plz help me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
```
7 years ago
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