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Three identicle particle each of mass `1 kg` are placed with their centres on a straight line. Their centres are marked `A, B and C` respectively. The distance of centre of mass of the system from `A` is.A. `(PQ + PR + QR)/(3)`B. `(PQ + PR)/(3)`C. `(PQ + QR)/(3)`D. `(PQ + QR + PR)/(3)` |
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Answer» Correct Answer - B Clearly `CM` of the system is at `Q`, its distance from `P`, i.e. `PQ = (mxx0+mxxPQ+mxxPR)/(3m) = (PQ + PR)/(3)` |
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