A) \[1.55\,{{m}^{2}}\]
B) \[\sqrt{{}}(1.55){{m}^{2}}\]
C) \[1.55\,m\]
D) \[1.25\,{{m}^{2}}\]
Correct Answer: A
Solution :
Given that: \[{{m}_{1}}=2kg,\,{{\vec{r}}_{1}}=-2\hat{i};\,{{m}_{2}}=2kg,\,{{\vec{r}}_{2}}=-3\hat{j};\]\[{{m}_{3}}=2kg,\,{{\vec{r}}_{3}}=\vec{k}\] \[\therefore \]\[{{r}_{cm}}=\frac{{{m}_{1}}{{{\vec{r}}}_{1}}+{{m}_{2}}{{{\vec{r}}}_{2}}+{{m}_{3}}{{{\vec{r}}}_{3}}+}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}=\frac{2(-2\hat{i})+2(-3\hat{j})+2(\hat{k})}{6}\] \[=\frac{-2\hat{i}-3\hat{j}+k}{3}\] \[\therefore \]\[\left| {{r}_{cm}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=\sqrt{\frac{4}{9}+\frac{9}{9}+\frac{1}{9}}=\sqrt{\frac{14}{9}}\] \[\Rightarrow \]\[{{\left| {{{\vec{r}}}_{cm}} \right|}^{2}}=\frac{14}{9}=1.55\,{{m}^{2}}\] Hence, the correction option is [a].
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