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-m2B.0.6 kg-m2 C.0.8kg-m2D.1.2kg-m2

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

Approved

all the masses are kept at a distance of 2/3^_1/2 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

Approved

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