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’)

# 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

$\stackrel{^}{i}+\stackrel{^}{j}$

2. B

$\frac{1}{2}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\right)$

3. C

$\left(\stackrel{^}{i}-\stackrel{^}{j}\right)$

4. D

$\frac{1}{2}\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\right)$

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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-

$COM=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}+{m}_{3}{x}_{3}+\dots \dots \dots ..+{m}_{n}{x}_{n}}{{m}_{1}+{m}_{2}+{m}_{3}+\dots \dots \dots +{m}_{n}}$

Substituting the values, we get,

$\mathrm{Xcom}=\frac{\sum {\mathrm{x}}_{\mathrm{i}}}{4}=\frac{1+1+0+0}{4}=\frac{1}{2}$

${\mathrm{Y}}_{\mathrm{com}}=\frac{\sum {\mathrm{Y}}_{\mathrm{i}}}{4}=\frac{0+0+1+1}{4}=\frac{1}{2}$

Position value of

Hence, the correct option is  (B).

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