Three identical spheres, each of mass $1$ kg are placed touching each other with their centres on a straight line. Their centers are marked P, Q and R respectively. The distance of center of mass of the system from P is:

\frac{P Q + P R + Q R}{3}

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\frac{P Q + P R}{3}

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\frac{P Q + Q R}{3}

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\frac{P R + Q R}{3}

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Solution

The correct option is B $\frac{P Q + P R}{3}$
The centre of mass of from P
$\begin{matrix} x = \frac{M_{p} (x_{p}) + M_{q} (x_{q}) + M_{R} (x_{R})}{M_{p} + M_{q} + M_{R}} = \frac{M (0) + M (P Q) + M (P R)}{M + M + M} = \frac{M (P Q + P R)}{3 M} = \frac{P Q + P R}{3} \end{matrix}$
Option B.

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| A |

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