Three particle each of 1kg mass are placed at corners of a right angled triangle AOB, ‘O’ being the origin of coordinate system (OA and OB along the x-direction and +ve y-direction). If OA = OB =1m, the position vector of centre of mass (in metres) is

Three particle each of 1kg mass are placed at corners of a right angled triangle AOB, ‘O’ being the origin of coordinate system (OA and OB along the x-direction and +ve y-direction). If OA = OB =1m, the position vector of centre of mass (in metres) is

1. A

$\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}}{3}$

2. B

$\frac{\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}}{3}$

3. C

$\frac{2\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\right)}{3}$

4. D

$\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}$

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Solution:

From the question,

$\mathrm{OA}=\mathrm{OB}=1\mathrm{m}$

${\mathrm{m}}_{1}={\mathrm{m}}_{2}={\mathrm{m}}_{3}=1\mathrm{kg}$

coordinate of O = (0,0)

coordinate of A = (1,0)

coordinate of B = (0,1)

${\mathrm{X}}_{\mathrm{COM}}=\frac{\left({\mathrm{m}}_{1}{\mathrm{x}}_{1}+{\mathrm{m}}_{2}{\mathrm{x}}_{2}+{\mathrm{m}}_{3}{\mathrm{x}}_{3}\right)}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}+{\mathrm{m}}_{3}}=\frac{0+1+0}{3}=\frac{1}{3}$

${\mathrm{Y}}_{\mathrm{COM}}=\frac{\left({\mathrm{m}}_{1}{\mathrm{y}}_{1}+{\mathrm{m}}_{2}{\mathrm{y}}_{2}+{\mathrm{m}}_{3}{\mathrm{y}}_{3}\right)}{{\mathrm{m}}_{1}+{\mathrm{m}}_{2}+{\mathrm{m}}_{3}}=\frac{0+0+1}{3}=\frac{1}{3}$

Hence the correct answer is $\frac{\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}}{3}$.

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