First, recall the formula of the center of mass of n particle system,
rcm=m1+m2+m3+..+mnm1r1+m2r2+m3r3+...+mnrn

⇒xcom=3MM×0+M×2+M×0=32
⇒ycom =3MM×0+M+2+M×0=32
So the position vector of center of mass is, (32i^+32j^).
NEET UG 2020 — Physics Mechanics
Three identical spheres, each of mass M, are placed at the corners of a right angle triangle with mutually perpendicular sides equal to 2m (see figure). Taking the point of intersection of the two mutually perpendicular sides as the origin, find the position vector of centre of mass.

Held on 30 Apr 2020 · Verified 9 Jul 2026.
2(i^+j^)
(i^+j^)
32(i^+j^)
34(i^+j^)
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