
Position vector rCOM=m1+m2+m3m1r1+m2r2+m3r3
rCOM =6M2M×0+2M×4i^+2M×4j^
⇒r=34i^+34j^
∣r∣=(34)2+(34)2=342
Therefore, x=3.
The identical spheres each of mass 2M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is x42, where the value of x is ________.
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Work through every JEE Main Mechanics PYQ, year by year.