3 particles of masses $2 kg$ each, are placed such that $1^{s t}$ one lies on negative $x -$ axis, $2^{n d}$ one lies on negative $y -$ axis and the third one lies on positive $z -$ axis at distances of $2 m, 3 m$ and $1 m$ respectively from the origin. Then the square of the distance of centre of mass of the system from the origin is

1.55 m^{2}

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\sqrt{1.55} m^{2}

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1.55 m

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1.25 m

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Solution

The correct option is A $1.55 m^{2}$
Given that;
$m_{1} = 2 kg;_{1} = - 2^i, m_{2} = 2 kg;_{2} = - 3^j$
$m_{3} = 2 kg;_{3} =^k$
$∴ {\to r}_{C M} = \frac{m_{1}_{1} + m_{2}_{2} + m_{3}_{3}}{m_{1} + m_{2} + m_{3}}$
$= \frac{2 (- 2^i) + 2 (- 3^j) + 2 (k)}{6} = \frac{- 2^i - 3^j +^k}{3}$
$∴ | {\to r}_{C M} | = \sqrt{x^{2} + y^{2} + z^{2}} = \sqrt{\frac{4}{9} + \frac{9}{9} + \frac{1}{9}} = \sqrt{\frac{14}{9}}$
$∴ | {\to r}_{C M} |^{2} = \frac{14}{9} = 1.55 m^{2}$
Hence, the correct answer is option (a).

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The correct option is A 1.55 m2Given that; m1=2 kg;→r1=−2^i,m2=2 kg;→r2=−3^j m3=2 kg;→r3=^k ∴→rCM=m1→r1+m2→r2+m3→r3m1+m2+m3 =2(−2^i)+2(−3^j)+2(k)6=−2^i−3^j+^k3 ∴|→rCM|=√x2+y2+z2=√49+99+19=√149 ∴|→rCM|2=149=1.55 m2 Hence, the correct answer is option (a).