A) \[(-1,-1,-1)\]
B) \[(-2,-2,-2)\]
C) \[(2,\,2,\,2)\]
D) \[(1,\,1,\,1)\]
Correct Answer: B
Solution :
Centre of mass of a solid body is given by \[{{x}_{CM}}=\frac{\sum\limits_{i=1}^{n}{\Delta {{m}_{i}}{{x}_{i}}}}{\sum\limits_{j=1}^{n}{\Delta {{M}_{j}}}},{{y}_{CM}}=\frac{\sum\limits_{i=1}^{n}{\Delta {{m}_{i}}{{y}_{i}}}}{\sum\limits_{j=1}^{n}{\Delta {{M}_{j}}}}\] \[{{z}_{CM}}=\frac{\sum\limits_{i=1}^{n}{\Delta {{m}_{i}}{{z}_{i}}}}{\sum\limits_{j=1}^{n}{\Delta {{\Mu }_{j}}}}\] \[1\times {{x}_{1}}+2\times {{x}_{2}}+3\times {{x}_{3}}=(1+2+3)3\] ?(i) and \[{{x}_{1}}={{x}_{2}}={{x}_{3}}=3\] \[{{x}_{CM}}={{y}_{CM}}={{z}_{CM}}=1\,(given)\] \[1(1+2+3+4)=1{{x}_{1}}+1{{x}_{2}}+3{{x}_{3}}+4{{x}_{4}}\] ?(ii) Solving Eqs, (i) and (ii), we get \[4{{x}_{4}}=10-18\Rightarrow {{x}_{4}}=-2\] Similarly, \[{{y}_{4}}=-2,{{z}_{4}}=-2\] The fourth particle must be placed at the point \[(-2,-2,-2).\]You need to login to perform this action.
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