PhysicsFour particles, each of mass 1kg, are placed at the corners of a square OABC of side 1m. ‘O’ is at the origin of the coordinate system. OA and OC are aligned along positive x-axis and positive y-axis respectively. The position vector of the centre of mass is (in ‘m’)

Four particles, each of mass 1kg, are placed at the corners of a square OABC of side 1m. ‘O’ is at the origin of the coordinate system. OA and OC are aligned along positive x-axis and positive y-axis respectively. The position vector of the centre of mass is (in ‘m’)

  1. A

    i^+j^

  2. B

    12(i^+j^)

  3. C

    (i^-j^)

  4. D

    12(i^-j^)

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    Solution:

    Given data:   4 particles of mass 1kg, are placed at the corners of a square OABC of side 1m. ‘O’ represents the origin. OA and OC are aligned along positive x-axis and positive y-axis respectively.

    Concept Used:  Systems of Particles

    Detailed Solution:

    We know, formula to calculate the center of mass of discrete objects is-

    COM=m1x1+m2x2+m3x3+..+mnxnm1+m2+m3++mn

    Substituting the values, we get,

    Xcom=xi4=1+1+0+04=12

    Ycom=Yi4=0+0+1+14=12

    Position value of COM= A^2+1A^2=12(i^+j^)

    Hence, the correct option is  (B).

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